EquilibriumClass 11 Chemistry NCERT Solutions
73 Solutions
Generated by KedovoAI
Solution 1 of 73
Q1EXERCISES
A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased. a) What is the initial effect of the change on vapour pressure? b) How do rates of evaporation and condensation change initially? c) What happens when equilibrium is restored finally and what will be the final vapour pressure?
Solution
The equilibrium is: Liquid Vapour
a) When the volume of the container is suddenly increased, the number of vapour molecules per unit volume decreases. As a result, the vapour pressure initially decreases.
b) The rate of evaporation depends on the temperature and surface area of the liquid, so it remains constant. The rate of condensation depends on the concentration (or partial pressure) of the vapour molecules. Since the vapour pressure has decreased, the rate of condensation decreases initially.
c) Since the rate of evaporation is greater than the rate of condensation, more liquid will evaporate to form vapour. This continues until the vapour pressure increases to its original value, at which point the rate of condensation becomes equal to the rate of evaporation again, and a new equilibrium is restored. The final vapour pressure will be the same as the initial vapour pressure because the equilibrium vapour pressure of a liquid depends only on the temperature, not on the volume of the container.
Q2EXERCISES
What is for the following equilibrium when the equilibrium concentration of each substance is: and ?
Solution
Given:
Equilibrium concentrations:
Reaction:
Formula:
The expression for the equilibrium constant, , is:
Calculation:
Substituting the given equilibrium concentrations into the expression:
Final Answer: The value of for the equilibrium is approximately .
Q3EXERCISES
At a certain temperature and total pressure of , iodine vapour contains by volume of I atoms Calculate for the equilibrium.
Solution
Given:
Total pressure,
Volume percentage of I atoms =
Reaction:
Calculation:
According to Dalton's law of partial pressures, the partial pressure of a gas is equal to its mole fraction multiplied by the total pressure. For gases, the volume percentage is equal to the mole percentage.
Volume % of I atoms = 40%
Therefore, mole fraction of I atoms,
Partial pressure of I atoms,
Volume % of molecules = 100% - 40% = 60%
Therefore, mole fraction of molecules,
Partial pressure of molecules,
Formula:
The expression for the equilibrium constant, , is:
Calculation of Kp:
Substituting the values of partial pressures:
Final Answer: The value of for the equilibrium is .
Q4EXERCISES
Write the expression for the equilibrium constant, for each of the following reactions:
(i)
(ii)
(iii)
(iv)
(v)
Solution
The expression for the equilibrium constant, , is written as the ratio of the product of the molar concentrations of the products to the product of the molar concentrations of the reactants, with each concentration term raised to the power of its stoichiometric coefficient. The concentrations of pure solids and pure liquids are taken as unity and are not included in the expression.
(i)
(ii)
Since and are pure solids, their concentrations are constant and taken as 1.
(iii)
In this reaction, water is a reactant and not just the solvent. However, if the reaction occurs in a dilute aqueous solution, the concentration of water is considered constant and is omitted. If it is not a dilute solution (as in esterification), its concentration is included. Given the states, we assume it is included.
(Note: In many contexts for dilute aqueous solutions, is omitted. But for reactions like ester hydrolysis where water is a reactant, it is included unless specified to be in large excess.)
(iv)
Since is a pure solid, its concentration is constant and taken as 1.
(v)
(assuming and are gases)
Since is a pure solid, its concentration is constant and taken as 1.
Q5EXERCISES
Find out the value of for each of the following equilibria from the value of :
(i)
at
(ii)
at
Solution
The relationship between and is given by the equation:
where:
= Ideal gas constant ()
= Temperature in Kelvin
= (moles of gaseous products) - (moles of gaseous reactants)
Therefore, or
(i)
Given:
(assuming units of bar)
Calculation of :
Calculation of :
(ii)
Given:
(assuming units of bar)
Calculation of :
Only is a gas.
Calculation of :
Final Answers:
(i)
(ii)
Q6EXERCISES
For the following equilibrium, at Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is , for the reverse reaction?
Solution
Given:
Forward reaction:
Equilibrium constant for the forward reaction, at .
To Find:
The equilibrium constant, , for the reverse reaction.
Reverse Reaction:
Relationship between Equilibrium Constants:
The equilibrium constant for a reverse reaction is the reciprocal of the equilibrium constant for the corresponding forward reaction.
Calculation:
Final Answer: The equilibrium constant, , for the reverse reaction is .
Q7EXERCISES
Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression?
Solution
Pure liquids and solids can be ignored while writing the equilibrium constant expression because their concentrations (or more accurately, their activities) remain essentially constant during the course of a reaction.
The concentration of a substance is defined as its amount (moles) per unit volume. For a pure solid or a pure liquid, its density is constant at a given temperature. The molar concentration is calculated as:
Concentration =
Since both density and molar mass are constant for a pure substance, its molar concentration is also a constant value, regardless of how much of the substance is present.
For example, the concentration of water is about and this value does not change significantly in most aqueous reactions.
In the equilibrium constant expression, these constant concentration terms are incorporated into the equilibrium constant itself. For a general reaction involving a solid 'S':
The equilibrium expression would be:
Since is a constant, we can define a new equilibrium constant, :
Thus, the constant terms for pure solids and liquids are conventionally included within the value of or , and they do not appear explicitly in the final expression.
Q8EXERCISES
Reaction between and takes place as follows: If a mixture of and of is placed in a reaction vessel and allowed to form at a temperature for which , determine the composition of equilibrium mixture.
Solution
Given:
Initial moles of
Initial moles of
Volume of vessel,
Equilibrium constant,
Reaction:
Initial Concentrations:
ICE Table (Initial, Change, Equilibrium):
Let be the change in concentration of .
| Species | Initial (M) | Change (M) | Equilibrium (M) |
|---|---|---|---|
| 0.0482 | |||
| 0.0933 | |||
| 0 |
Equilibrium Constant Expression:
Calculation:
Since the value of is extremely small (), the reaction proceeds to a very negligible extent. This means that the value of will be very small. Therefore, we can make the approximation:
Now, the expression simplifies to:
The value of is indeed very small, so our approximation is valid.
Equilibrium Concentrations:
Final Answer:
The composition of the equilibrium mixture is:
Concentration of
Concentration of
Concentration of
Q9EXERCISES
Nitric oxide reacts with and gives nitrosyl bromide as per reaction given below: When of and of are mixed in a closed container at constant temperature, of is obtained at equilibrium. Calculate equilibrium amount of and .
Solution
Given:
Initial moles of NO =
Initial moles of
Equilibrium moles of NOBr =
Reaction:
Calculation:
From the stoichiometry of the reaction, 2 moles of NO react with 1 mole of to form 2 moles of NOBr.
This means that to form of NOBr, the amount of reactants consumed is:
Moles of NO reacted =
Moles of reacted =
Now, we can calculate the amount of each reactant remaining at equilibrium.
Equilibrium moles of NO = Initial moles of NO - Moles of NO reacted
Equilibrium moles of = Initial moles of - Moles of reacted
Final Answer:
The equilibrium amounts are:
Amount of NO =
Amount of
Q10EXERCISES
At for the given reaction at equilibrium. What is at this temperature ?
Solution
Given:
Reaction:
Formula:
The relationship between and is:
So,
Calculation of :
= (moles of gaseous products) - (moles of gaseous reactants)
Calculation of :
Final Answer: The value of at this temperature is .
Q11EXERCISES
A sample of is placed in flask at a pressure of . At equilibrium the partial pressure of is . What is for the given equilibrium ?
Solution
Given:
Initial pressure of HI,
Equilibrium partial pressure of HI,
Reaction:
ICE Table (in terms of pressure):
Let be the decrease in pressure of HI.
| Species | Initial (atm) | Change (atm) | Equilibrium (atm) |
|---|---|---|---|
| HI | 0.2 | ||
| H | 0 | ||
| I | 0 |
Calculation of x:
We are given that the equilibrium pressure of HI is .
Equilibrium Partial Pressures:
Formula for :
Calculation of :
Final Answer: The value of for the given equilibrium is 4.
Q12EXERCISES
A mixture of of , of and of is introduced into a reaction vessel at . At this temperature, the equilibrium constant, for the reaction is . Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?
Solution
Given:
Moles of
Moles of
Moles of
Volume,
Equilibrium constant,
Temperature,
Reaction:
Step 1: Calculate the initial concentrations.
Step 2: Calculate the reaction quotient, .
The expression for the reaction quotient is the same as for the equilibrium constant, but using the current concentrations instead of equilibrium concentrations.
Step 3: Compare with .
Value of
Value of
Here, .
Conclusion:
Since the reaction quotient is greater than the equilibrium constant , the reaction mixture is not at equilibrium.
To reach equilibrium, the value of must decrease until it becomes equal to . This requires the numerator (concentration of products) to decrease and the denominator (concentration of reactants) to increase. Therefore, the net reaction will proceed in the reverse direction (from right to left).
Final Answer:
No, the reaction mixture is not at equilibrium. The net reaction will proceed in the reverse direction, i.e., towards the formation of and .
Q13EXERCISES
The equilibrium constant expression for a gas reaction is, Write the balanced chemical equation corresponding to this expression.
Solution
The equilibrium constant expression is given by:
In the expression for , the substances in the numerator are the products, and the substances in the denominator are the reactants. The exponents of the concentration terms correspond to the stoichiometric coefficients in the balanced chemical equation.
From the expression:
Reactants:
- NO, with a stoichiometric coefficient of 4.
- , with a stoichiometric coefficient of 6.
Products:
- , with a stoichiometric coefficient of 4.
- , with a stoichiometric coefficient of 5.
Using this information, we can write the balanced chemical equation:
Final Answer:
The balanced chemical equation corresponding to the given equilibrium constant expression is:
Q14EXERCISES
One mole of and one mole of are taken in vessel and heated to . At equilibrium of water (by mass) reacts with according to the equation, Calculate the equilibrium constant for the reaction.
Solution
Given:
Initial moles of
Initial moles of CO =
Volume of vessel,
Temperature,
Percentage of water reacted = 40%
Reaction:
Step 1: Calculate moles reacted and moles at equilibrium.
Since percentage of water reacted is 40% by mass, it is also 40% by moles.
Moles of reacted = of .
From the stoichiometry of the reaction (1:1:1:1 ratio):
Moles of CO reacted =
Moles of formed =
Moles of formed =
Now, calculate the moles of each substance at equilibrium:
Moles of at equilibrium = Initial moles - Moles reacted =
Moles of CO at equilibrium = Initial moles - Moles reacted =
Moles of at equilibrium =
Moles of at equilibrium =
Step 2: Calculate equilibrium concentrations.
Volume,
Step 3: Calculate the equilibrium constant, .
Final Answer:
The equilibrium constant for the reaction is approximately .
Q15EXERCISES
At , equilibrium constant for the reaction: is . If of is present at equilibrium at , what are the concentration of and assuming that we initially started with and allowed it to reach equilibrium at ?
Solution
Given:
Equilibrium constant,
Temperature,
Equilibrium concentration of HI,
Reaction:
Since the reaction starts with HI and reaches equilibrium, we should consider the reverse reaction for setting up the changes:
Let the initial concentration of HI be 'c'.
Let be the change in concentration of HI.
| Species | Initial (M) | Change (M) | Equilibrium (M) |
|---|---|---|---|
| HI | c | ||
| H | 0 | ||
| I | 0 |
From the stoichiometry of the decomposition of HI, the concentrations of and formed at equilibrium will be equal.
Let .
Equilibrium Constant Expression:
For the forward reaction:
Calculation:
Substitute the known equilibrium values into the expression:
So, the equilibrium concentrations are:
Final Answer:
The concentrations of and at equilibrium are both .
Q16EXERCISES
What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was ?
Solution
Given:
Initial concentration of ICl,
Equilibrium constant,
Reaction:
ICE Table (Initial, Change, Equilibrium):
Let be the change in concentration of ICl.
| Species | Initial (M) | Change (M) | Equilibrium (M) |
|---|---|---|---|
| ICl | 0.78 | ||
| I | 0 | ||
| Cl | 0 |
Equilibrium Constant Expression:
Calculation:
Substitute the equilibrium concentrations from the ICE table into the expression:
Take the square root of both sides:
Now, solve for :
Equilibrium Concentrations:
Final Answer:
The equilibrium concentrations are:
Q17EXERCISES
at for the equilibrium shown below. What is the equilibrium concentration of when it is placed in a flask at pressure and allowed to come to equilibrium?
Solution
Given:
Initial pressure of ,
Temperature,
Reaction:
ICE Table (in terms of pressure):
Let be the decrease in pressure of .
| Species | Initial (atm) | Change (atm) | Equilibrium (atm) |
|---|---|---|---|
| 4.0 | |||
| 0 | |||
| 0 |
Equilibrium Constant Expression:
Calculation:
Substitute the equilibrium pressures from the ICE table into the expression:
This is a quadratic equation of the form . We can solve for using the quadratic formula:
Here, , , .
Since pressure () cannot be negative, we take the positive root:
Equilibrium Concentration of :
The question asks for the equilibrium concentration, but provides pressure data. We will first find the equilibrium partial pressure of .
To find the molar concentration, we use the ideal gas law, , which gives .
Equilibrium concentration of ,
Final Answer:
The equilibrium partial pressure of is .
The equilibrium concentration of is .
Q18EXERCISES
Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as:
(i)
Write the concentration ratio (reaction quotient), , for this reaction (note: water is not in excess and is not a solvent in this reaction)
(ii)
At , if one starts with of acetic acid and of ethanol, there is of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.
(iii)
Starting with of ethanol and of acetic acid and maintaining it at of ethyl acetate is found after sometime. Has equilibrium been reached?
Solution
(i) Concentration Ratio ()
For the reaction:
Since all species are liquids and water is not in excess (it is a product), the concentration of each component is relevant. Assuming the volume V is constant for the mixture, the concentration ratio is:
Since the volume V is the same for all species, we can use moles directly in the ratio.
(ii) Calculation of Equilibrium Constant ()
Initial moles:
(assuming we start with anhydrous reactants)
Equilibrium moles:
(given)
From stoichiometry, moles of formed = moles of ethyl acetate formed = .
Moles of reacted = .
Moles of reacted = .
Equilibrium moles of .
Equilibrium moles of .
Calculation of :
(iii) Has equilibrium been reached?
Initial moles:
Moles at 'sometime':
Moles of formed = .
Moles of reacted = , remaining = .
Moles of reacted = , remaining = .
Calculate at this time:
Compare with :
from part (ii) is .
Current is .
Since , the reaction has not yet reached equilibrium. The reaction will continue to proceed in the forward direction to form more products.
Final Answers:
(i)
(ii)
The equilibrium constant is .
(iii)
No, equilibrium has not been reached because .
Q19EXERCISES
A sample of pure was introduced into an evacuated vessel at . After equilibrium was attained, concentration of was found to be . If value of is , what are the concentrations of and at equilibrium?
Solution
Given:
Equilibrium concentration of ,
Equilibrium constant,
Temperature,
Reaction:
Let's set up the concentrations at equilibrium:
From the stoichiometry of the reaction, when one mole of decomposes, one mole of and one mole of are formed. Therefore, at equilibrium, the concentrations of and will be equal.
Let .
Equilibrium Constant Expression:
Calculation:
Substitute the known values into the expression:
So, the equilibrium concentrations are:
Final Answer:
The concentrations at equilibrium are:
Q20EXERCISES
One of the reaction that takes place in producing steel from iron ore is the reduction of iron(II) oxide by carbon monoxide to give iron metal and . What are the equilibrium partial pressures of and at if the initial partial pressures are: and ?
Solution
Given:
Initial partial pressure of CO,
Initial partial pressure of ,
Equilibrium constant,
Temperature,
Reaction:
Step 1: Calculate the reaction quotient, .
Since FeO and Fe are solids, they are not included in the expression.
Step 2: Compare with .
Since , the reaction will proceed in the reverse direction (from right to left) to reach equilibrium.
Step 3: Set up the ICE table.
Let be the change in partial pressure.
| Species | Initial (atm) | Change (atm) | Equilibrium (atm) |
|---|---|---|---|
| CO | 1.4 | ||
| CO | 0.80 |
Step 4: Use the equilibrium constant expression to solve for x.
Step 5: Calculate the equilibrium partial pressures.
Final Answer:
The equilibrium partial pressures are:
Q21EXERCISES
Equilibrium constant, for the reaction At a particular time, the analysis shows that composition of the reaction mixture is , and . Is the reaction at equilibrium? If not in which direction does the reaction tend to proceed to reach equilibrium?
Solution
Given:
Equilibrium constant,
At a particular time:
Reaction:
Step 1: Calculate the reaction quotient, .
Substitute the given concentrations:
Step 2: Compare with .
Here, .
Conclusion:
Since the reaction quotient is less than the equilibrium constant , the reaction is not at equilibrium.
To reach equilibrium, the value of must increase until it becomes equal to . This requires the numerator (concentration of products) to increase and the denominator (concentration of reactants) to decrease. Therefore, the reaction will proceed in the forward direction (from left to right) to produce more .
Final Answer:
No, the reaction is not at equilibrium. The reaction will proceed in the forward direction to reach equilibrium.
Q22EXERCISES
Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium: for which at . If initially pure BrCl is present at a concentration of , what is its molar concentration in the mixture at equilibrium?
Solution
Given:
Initial concentration of BrCl,
Equilibrium constant,
Temperature,
Reaction:
ICE Table (Initial, Change, Equilibrium):
Let be the change in concentration of BrCl.
| Species | Initial (M) | Change (M) | Equilibrium (M) |
|---|---|---|---|
| BrCl | |||
| Br | 0 | ||
| Cl | 0 |
Equilibrium Constant Expression:
Calculation:
Substitute the equilibrium concentrations from the ICE table into the expression:
Take the square root of both sides:
Now, solve for :
Equilibrium Concentration of BrCl:
Final Answer:
The molar concentration of BrCl in the mixture at equilibrium is (or approximately ).
Q23EXERCISES
At and pressure, a gaseous mixture of and in equilibrium with soild carbon has by mass Calculate for this reaction at the above temperature.
Solution
Given:
Temperature,
Total pressure,
Mass % of CO = 90.55%
Reaction:
Step 1: Calculate mole fractions from mass percentages.
Let the total mass of the gaseous mixture be 100 g.
Mass of CO = 90.55 g
Mass of = 100 - 90.55 = 9.45 g
Molar mass of CO = 12 + 16 = 28 g/mol
Molar mass of = 12 + 2(16) = 44 g/mol
Moles of CO,
Moles of ,
Total moles of gas,
Mole fraction of CO,
Mole fraction of ,
Step 2: Calculate partial pressures.
Step 3: Calculate .
Step 4: Calculate from .
For this reaction, .
Final Answer:
for the reaction at is .
Q24EXERCISES
Calculate a) and b) the equilibrium constant for the formation of from and at where & \Delta_{\mathrm{f}} G^{\ominus}\left(\mathrm{NO}_{2}\right)=52.0 \mathrm{~kJ} / \mathrm{mol} \n& \Delta_{\mathrm{f}} G^{\ominus}(\mathrm{NO})=87.0 \mathrm{~kJ} / \mathrm{mol} \n& \Delta_{\mathrm{f}} G^{\ominus}\left(\mathrm{O}_{2}\right)=0 \mathrm{~kJ} / \mathrm{mol} \end{aligned}$$
Solution
Given:
Temperature,
Gas constant,
Reaction:
a) Calculation of
The standard Gibbs free energy change for the reaction is calculated as:
b) Calculation of the equilibrium constant (K)
The relationship between the standard Gibbs free energy change and the equilibrium constant ( for gas-phase reactions) is:
Rearranging to solve for :
Substitute the values (ensure units are consistent, using J/mol for energy):
Now, solve for :
Final Answer:
a)
b) The equilibrium constant, , is .
Q25EXERCISES
Does the number of moles of reaction products increase, decrease or remain same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume?
(a)
(b)
(c)
Solution
According to Le Chatelier's principle, when the pressure on a system at equilibrium is decreased (by increasing the volume), the equilibrium will shift in the direction that results in an increase in the number of moles of gas. This counteracts the decrease in pressure.
To determine the effect, we need to calculate , the change in the number of moles of gas for each reaction.
= (moles of gaseous products) - (moles of gaseous reactants)
(a)
Moles of gaseous reactants = 1
Moles of gaseous products = 1 + 1 = 2
Since is positive, a decrease in pressure will shift the equilibrium to the right (forward direction), which has more moles of gas. Therefore, the number of moles of products ( and ) will increase.
(b)
Moles of gaseous reactants = 1
Moles of gaseous products = 0
Since is negative, a decrease in pressure will shift the equilibrium to the left (reverse direction), which has more moles of gas. Therefore, the number of moles of product () will decrease.
(c)
Moles of gaseous reactants = 4
Moles of gaseous products = 4
Since is zero, a change in pressure (or volume) will have no effect on the position of the equilibrium. Therefore, the number of moles of products will remain the same.
Q26EXERCISES
Which of the following reactions will get affected by increasing the pressure? Also, mention whether change will cause the reaction to go into forward or backward direction.
(i)
(ii)
(iii)
(iv)
(v)
(vi)
Solution
According to Le Chatelier's principle, increasing the pressure on a system at equilibrium will cause the equilibrium to shift in the direction that has fewer moles of gas. A reaction is affected by pressure change only if the number of moles of gaseous reactants is different from the number of moles of gaseous products (i.e., ).
(i)
. Since , the reaction is affected. Increasing pressure will shift it to the side with fewer gas moles, i.e., the backward direction.
(ii)
. Since , the reaction is not affected by a change in pressure.
(iii)
. Since , the reaction is affected. Increasing pressure will shift it to the side with fewer gas moles, i.e., the backward direction.
(iv)
. Since , the reaction is affected. Increasing pressure will shift it to the side with fewer gas moles, i.e., the forward direction.
(v)
. Since , the reaction is affected. Increasing pressure will shift it to the side with fewer gas moles, i.e., the backward direction.
(vi)
. Since , the reaction is affected. Increasing pressure will shift it to the side with fewer gas moles, i.e., the backward direction.
Q27EXERCISES
The equilibrium constant for the following reaction is at Find the equilibrium pressure of all gases if of is introduced into a sealed container at .
Solution
Given:
Equilibrium constant,
Initial pressure of HBr,
Temperature,
Reaction:
Since we start with only the product, the reaction will proceed in the reverse direction to reach equilibrium.
Let's consider the reverse reaction for the ICE table:
The equilibrium constant for this reverse reaction, , is:
ICE Table (in terms of pressure):
Let be the decrease in pressure of HBr.
| Species | Initial (bar) | Change (bar) | Equilibrium (bar) |
|---|---|---|---|
| HBr | 10.0 | ||
| H | 0 | ||
| Br | 0 |
Equilibrium Constant Expression for Reverse Reaction:
Calculation:
Since is very small, the reaction proceeds to a very small extent. Thus, will be very small compared to 10.0. We can approximate:
The approximation is valid as , which is much smaller than 10.0.
Equilibrium Pressures:
Final Answer:
The equilibrium pressures are:
Q28EXERCISES
Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction: (a) Write as expression for Kp for the above reaction. (b) How will the values of Kp and composition of equilibrium mixture be affected by
(i)
increasing the pressure
(ii)
increasing the temperature
(iii)
using a catalyst?
Solution
Reaction:
(a) Expression for
The equilibrium constant in terms of partial pressures, , is given by:
(b) Effects of changing conditions
(i) Increasing the pressure
- Effect on : The value of the equilibrium constant depends only on temperature. Therefore, increasing the pressure will not change the value of .
- Effect on composition: According to Le Chatelier's principle, if pressure is increased, the equilibrium will shift to the side with fewer moles of gas to counteract the change. Moles of gaseous reactants = Moles of gaseous products = The forward reaction increases the number of gas moles. Therefore, increasing the pressure will shift the equilibrium to the left (reverse direction). The composition will change to have more reactants ( and ) and fewer products (CO and ).
(ii) Increasing the temperature
- Effect on : The reaction is given as endothermic (). For an endothermic reaction, increasing the temperature shifts the equilibrium to the right (in the direction of product formation). This means the value of will increase.
- Effect on composition: Since the equilibrium shifts to the right (forward direction), the composition of the equilibrium mixture will change to have fewer reactants ( and ) and more products (CO and ).
(iii) Using a catalyst
- Effect on : A catalyst does not affect the position of equilibrium or the equilibrium constant. Therefore, remains unchanged.
- Effect on composition: A catalyst increases the rate of both the forward and reverse reactions equally. It helps the system reach equilibrium faster, but it does not change the composition of the equilibrium mixture.
Q29EXERCISES
Describe the effect of: a) addition of b) addition of c) removal of d) removal of on the equilibrium of the reaction:
Solution
The given equilibrium is:
We will use Le Chatelier's principle, which states that if a change of condition is applied to a system in equilibrium, the system will shift in a direction that counteracts the change.
a) Addition of
Hydrogen () is a reactant. If more is added, the system will try to consume the added . To do this, the equilibrium will shift to the right (forward direction), favoring the formation of the product, methanol (). The concentrations of CO will decrease and will increase.
b) Addition of
Methanol () is the product. If more is added, the system will try to consume the added product. The equilibrium will shift to the left (reverse direction), favoring the formation of reactants, hydrogen () and carbon monoxide (CO).
c) Removal of CO
Carbon monoxide (CO) is a reactant. If CO is removed, the system will try to replenish it. The equilibrium will shift to the left (reverse direction) to produce more CO. This will also produce more and consume .
d) Removal of
Methanol () is the product. If is removed, the system will try to replenish it. The equilibrium will shift to the right (forward direction) to produce more . This will consume the reactants, and CO.
Q30EXERCISES
At , equilibrium constant for decomposition of phosphorus pentachloride, is . If decomposition is depicted as, a) write an expression for Kc for the reaction. b) what is the value of Kc for the reverse reaction at the same temperature? c) what would be the effect on if (i) more is added (ii) pressure is increased (iii) the temperature is increased ?
Solution
Reaction:
a) Expression for
The equilibrium constant expression for the given reaction is:
b) Value of for the reverse reaction
The reverse reaction is:
The equilibrium constant for the reverse reaction, , is the reciprocal of the equilibrium constant for the forward reaction, .
So, for the reverse reaction is approximately .
c) Effect on
The value of the equilibrium constant for a given reaction depends only on temperature.
(i) More is added:
Adding more will disturb the equilibrium, causing the reaction to shift forward to consume the added reactant. However, the value of the equilibrium constant will remain unchanged because the temperature is constant.
(ii) Pressure is increased:
Increasing the pressure will shift the equilibrium to the side with fewer moles of gas (the reverse direction). However, as long as the temperature is constant, the value of will remain unchanged.
(iii) The temperature is increased:
The forward reaction is endothermic, as indicated by the positive value of . According to Le Chatelier's principle, if the temperature of an endothermic reaction is increased, the equilibrium will shift in the forward direction to absorb the added heat. A shift in the forward direction means the ratio of products to reactants increases. Therefore, the value of will increase.
Q31EXERCISES
Dihydrogen gas used in Haber's process is produced by reacting methane from natural gas with high temperature steam. The first stage of two stage reaction involves the formation of and . In second stage, formed in first stage is reacted with more steam in water gas shift reaction, If a reaction vessel at is charged with an equimolar mixture of and steam such that bar, what will be the partial pressure of at equilibrium? at
Solution
Given:
Initial pressure of CO,
Initial pressure of ,
Equilibrium constant,
Temperature,
Reaction:
ICE Table (in terms of pressure):
Let be the change in partial pressure.
| Species | Initial (bar) | Change (bar) | Equilibrium (bar) |
|---|---|---|---|
| CO | 4.0 | ||
| HO | 4.0 | ||
| CO | 0 | ||
| H | 0 |
Equilibrium Constant Expression:
Calculation:
Substitute the equilibrium pressures from the ICE table into the expression:
Take the square root of both sides:
Now, solve for :
Partial Pressure of H at Equilibrium:
The partial pressure of at equilibrium is equal to .
Final Answer:
The partial pressure of at equilibrium will be .
Q32EXERCISES
Predict which of the following reaction will have appreciable concentration of reactants and products: a) b) c)
Solution
The magnitude of the equilibrium constant, , indicates the extent to which a reaction proceeds at equilibrium.
- If is very large (), the reaction proceeds nearly to completion, and the equilibrium mixture consists mainly of products.
- If is very small (), the reaction hardly proceeds, and the equilibrium mixture consists mainly of reactants.
- If is in the range of to , the equilibrium mixture contains appreciable concentrations of both reactants and products.
Let's analyze each reaction:
a)
The value of is extremely small. This indicates that the equilibrium lies far to the left. The reaction proceeds to a negligible extent, and the equilibrium mixture will contain almost entirely reactants (). There will not be appreciable concentrations of both reactants and products.
b)
The value of is very large. This indicates that the equilibrium lies far to the right. The reaction proceeds almost to completion, and the equilibrium mixture will contain almost entirely products (NOCl). There will not be appreciable concentrations of both reactants and products.
c)
The value of is close to 1 and lies within the range of to . This indicates that the equilibrium lies in the middle. The reaction proceeds to a significant extent, and the equilibrium mixture will contain comparable and appreciable concentrations of both reactants (, ) and products ().
Final Answer:
Reaction (c) will have appreciable concentrations of reactants and products.
Q33EXERCISES
The value of for the reaction is at . If the equilibrium concentration of in air at is , what is the concentration of ?
Solution
Given:
Equilibrium constant,
Equilibrium concentration of ,
Temperature,
Reaction:
Equilibrium Constant Expression:
Calculation:
We need to find the concentration of ozone, . We can rearrange the equilibrium expression:
Substitute the given values:
Now, take the square root to find :
Final Answer:
The concentration of is .
Q34EXERCISES
The reaction, is at equilibrium at in a flask. It also contain of , of and of and an unknown amount of in the flask. Determine the concentration of in the mixture. The equilibrium constant, for the reaction at the given temperature is .
Solution
Given:
Volume of flask,
Equilibrium moles of CO =
Equilibrium moles of
Equilibrium moles of
Equilibrium constant,
Temperature,
Reaction:
Step 1: Calculate equilibrium concentrations.
Since the volume is 1 L, the concentration in mol/L is numerically equal to the number of moles.
Let the concentration of methane be .
Step 2: Use the equilibrium constant expression.
Step 3: Solve for the unknown concentration.
Substitute the known values into the expression:
Now, solve for :
Final Answer:
The concentration of in the mixture is .
Q35EXERCISES
What is meant by the conjugate acid-base pair? Find the conjugate acid/base for the following species: , and
Solution
Conjugate Acid-Base Pair:
A conjugate acid-base pair consists of two species that differ from each other by a single proton (H). According to the Brönsted-Lowry theory, when an acid donates a proton, the remaining species is its conjugate base. When a base accepts a proton, the new species formed is its conjugate acid.
- Conjugate Base: Formed by removing one H from an acid.
- Conjugate Acid: Formed by adding one H to a base.
Finding the Conjugate Acids/Bases:
-
(Nitrous acid): This is a Brönsted acid. To find its conjugate base, we remove one H. Conjugate Base: (Nitrite ion)
-
(Cyanide ion): This is a Brönsted base. To find its conjugate acid, we add one H. Conjugate Acid: HCN (Hydrogen cyanide)
-
(Perchloric acid): This is a Brönsted acid. To find its conjugate base, we remove one H. Conjugate Base: (Perchlorate ion)
-
(Fluoride ion): This is a Brönsted base. To find its conjugate acid, we add one H. Conjugate Acid: HF (Hydrogen fluoride)
-
(Hydroxide ion): This is a Brönsted base. To find its conjugate acid, we add one H. Conjugate Acid: (Water)
-
(Carbonate ion): This is a Brönsted base. To find its conjugate acid, we add one H. Conjugate Acid: (Bicarbonate ion)
-
(Sulfide ion): This is a Brönsted base. To find its conjugate acid, we add one H. Conjugate Acid: (Hydrosulfide ion)
Q36EXERCISES
Which of the followings are Lewis acids? , and
Solution
A Lewis acid is defined as a species (atom, ion, or molecule) that can accept a pair of electrons.
Let's analyze each species:
-
(Water): The oxygen atom in water has two lone pairs of electrons which it can donate. Therefore, acts as a Lewis base, not a Lewis acid.
-
(Boron trifluoride): The boron atom in has only six electrons in its valence shell, making it electron-deficient. It has an empty p-orbital and can accept a lone pair of electrons (e.g., from ammonia, ) to complete its octet. Therefore, is a Lewis acid.
-
(Proton): A proton has no electrons and an empty 1s orbital. It can readily accept a pair of electrons from a Lewis base (e.g., from to form ). Therefore, is a Lewis acid.
-
(Ammonium ion): The nitrogen atom in the ammonium ion has a complete octet and no lone pairs to donate. All its valence electrons are involved in covalent bonds with hydrogen atoms. It cannot accept an electron pair. While it acts as a Brönsted-Lowry acid by donating a proton, it does not fit the definition of a Lewis acid. Some might argue that the H atoms are electron-deficient, but the species as a whole does not accept an electron pair into a vacant orbital on the central atom. The primary Lewis acids in this list are and .
Final Answer:
The Lewis acids from the list are and .
Q37EXERCISES
What will be the conjugate bases for the Brönsted acids: and ?
Solution
A conjugate base is formed when a Brönsted acid donates a proton (H). To find the conjugate base, we remove one H from the acid and decrease the charge by one.
-
Brönsted Acid: HF (Hydrogen fluoride) When HF donates a proton, it forms the fluoride ion. Conjugate Base:
-
Brönsted Acid: (Sulfuric acid) When donates its first proton, it forms the hydrogen sulfate (or bisulfate) ion. Conjugate Base:
-
Brönsted Acid: (Bicarbonate ion) When the bicarbonate ion acts as an acid and donates its proton, it forms the carbonate ion. Conjugate Base:
Q38EXERCISES
Write the conjugate acids for the following Brönsted bases: and .
Solution
A conjugate acid is formed when a Brönsted base accepts a proton (H). To find the conjugate acid, we add one H to the base and increase the charge by one.
-
Brönsted Base: (Amide ion) When accepts a proton, it forms ammonia. Conjugate Acid:
-
Brönsted Base: (Ammonia) When accepts a proton, it forms the ammonium ion. Conjugate Acid:
-
Brönsted Base: (Formate ion) When accepts a proton, it forms formic acid. Conjugate Acid: HCOOH
Q39EXERCISES
The species: and can act both as Brönsted acids and bases. For each case give the corresponding conjugate acid and base.
Solution
Species that can act as both Brönsted acids (proton donors) and Brönsted bases (proton acceptors) are called amphiprotic.
- Conjugate Acid: Formed by adding H.
- Conjugate Base: Formed by removing H.
1. Species: (Water)
- As a Brönsted acid (donates H): Conjugate Base:
- As a Brönsted base (accepts H): Conjugate Acid:
2. Species: (Bicarbonate ion)
- As a Brönsted acid (donates H): Conjugate Base:
- As a Brönsted base (accepts H): Conjugate Acid:
3. Species: (Hydrogen sulfate ion)
- As a Brönsted acid (donates H): Conjugate Base:
- As a Brönsted base (accepts H): Conjugate Acid:
4. Species: (Ammonia)
- As a Brönsted acid (donates H): Conjugate Base:
- As a Brönsted base (accepts H): Conjugate Acid:
Q40EXERCISES
Classify the following species into Lewis acids and Lewis bases and show how these act as Lewis acid/base: (a) (b) (c) (d) .
Solution
A Lewis acid is an electron-pair acceptor. A Lewis base is an electron-pair donor.
(a) (Hydroxide ion)
- Classification: Lewis Base.
- Reason: The oxygen atom in the hydroxide ion has three lone pairs of electrons. It can donate one of these electron pairs to a Lewis acid.
- Example: It can donate an electron pair to a proton (H), a Lewis acid, to form water.
(b) (Fluoride ion)
- Classification: Lewis Base.
- Reason: The fluoride ion has four lone pairs of electrons in its valence shell. It can donate one of these electron pairs.
- Example: It can donate an electron pair to , a Lewis acid, to form the tetrafluoroborate ion.
(c) (Proton)
- Classification: Lewis Acid.
- Reason: A proton has an empty 1s orbital and no electrons, making it capable of accepting a pair of electrons.
- Example: It accepts an electron pair from a Lewis base like ammonia () to form the ammonium ion ().
(d) (Boron trichloride)
- Classification: Lewis Acid.
- Reason: The central boron atom in has only six electrons in its valence shell, so it is electron-deficient and has a vacant p-orbital. It can accept a lone pair of electrons to complete its octet.
- Example: It accepts an electron pair from a Lewis base like ammonia () to form an adduct.
Q41EXERCISES
The concentration of hydrogen ion in a sample of soft drink is . What is its ?
Solution
Given:
Concentration of hydrogen ion,
To Find:
The pH of the soft drink.
Formula:
The pH of a solution is defined as the negative logarithm to the base 10 of the hydrogen ion concentration.
Calculation:
Using the logarithm property :
We know that and . will be between these values, approximately 0.58.
Final Answer:
The pH of the soft drink is 2.42.
Q42EXERCISES
The of a sample of vinegar is . Calculate the concentration of hydrogen ion in it.
Solution
Given:
pH of vinegar = 3.76
To Find:
The concentration of hydrogen ion, .
Formula:
The pH is related to the hydrogen ion concentration by the equation:
To find , we can rearrange this equation:
Calculation:
To solve this, we can write:
We need to find the antilog of 0.24.
We know that and . So, is a value between 1 and 2, closer to 2. A calculator gives antilog(0.24) .
Final Answer:
The concentration of hydrogen ion in the vinegar sample is .
Q43EXERCISES
The ionization constant of and at are , and respectively. Calculate the ionization constants of the corresponding conjugate base.
Solution
Given:
Ionization constant of HF,
Ionization constant of HCOOH,
Ionization constant of HCN,
Ionic product of water, at .
Formula:
For any conjugate acid-base pair, the product of the acid ionization constant () and the base ionization constant () is equal to the ionic product of water ().
Therefore, .
1. Conjugate base of HF is
2. Conjugate base of HCOOH is
3. Conjugate base of HCN is
Final Answer:
The ionization constants of the conjugate bases are:
- For ,
- For ,
- For ,
Q44EXERCISES
The ionization constant of phenol is . What is the concentration of phenolate ion in solution of phenol? What will be its degree of ionization if the solution is also in sodium phenolate?
Solution
Part 1: Ionization of 0.05 M Phenol
Given:
Ionization constant of phenol,
Initial concentration of phenol,
Reaction:
Let phenol be represented as HPh.
where is the phenolate ion.
ICE Table:
| Species | Initial (M) | Change (M) | Equilibrium (M) |
|---|---|---|---|
| HPh | 0.05 | ||
| H | 0 | ||
| Ph | 0 |
Calculation:
Since is very small, we can assume , so .
The concentration of phenolate ion, .
Part 2: Ionization in presence of Sodium Phenolate (Common Ion Effect)
Given:
Initial concentration of phenol [HPh] =
Concentration of sodium phenolate = . Since sodium phenolate is a salt of a strong base, it dissociates completely, so .
ICE Table:
Let be the degree of ionization of phenol.
Equilibrium concentrations:
(since will be very small)
Calculation:
Final Answer:
- The concentration of phenolate ion in phenol solution is .
- The degree of ionization of phenol in the solution containing sodium phenolate is .
Q45EXERCISES
The first ionization constant of is . Calculate the concentration of ion in its solution. How will this concentration be affected if the solution is in also? If the second dissociation constant of is , calculate the concentration of under both conditions.
Solution
Given:
First ionization constant,
Second ionization constant,
Initial concentration of ,
Condition 1: 0.1 M solution
First Ionization:
| Species | Initial (M) | Change (M) | Equilibrium (M) |
|---|---|---|---|
| HS | 0.1 | ||
| H | 0 | ||
| HS | 0 |
Since is small, , so .
So, and .
Second Ionization:
Since , we can assume . Therefore, .
.
Condition 2: 0.1 M and 0.1 M HCl solution (Common Ion Effect)
HCl is a strong acid, so it completely ionizes: .
This high concentration of H from HCl will suppress the dissociation of the weak acid .
First Ionization:
(from HCl)
Let .
.
So, . The concentration of is significantly decreased due to the common ion effect.
Second Ionization:
.
Final Answer:
- In 0.1 M solution:
- In 0.1 M + 0.1 M HCl solution:
- (concentration is decreased)
Q46EXERCISES
The ionization constant of acetic acid is . Calculate the degree of dissociation of acetic acid in its solution. Calculate the concentration of acetate ion in the solution and its .
Solution
Given:
Ionization constant of acetic acid,
Initial concentration of acetic acid,
Reaction:
Let acetic acid be HAc.
Let be the degree of dissociation.
Equilibrium concentrations:
Step 1: Calculate the degree of dissociation ()
Since is small, we can assume is small, so .
Degree of dissociation, .
Step 2: Calculate the concentration of acetate ion
Concentration of acetate ion,
Step 3: Calculate the pH of the solution
First, find the hydrogen ion concentration, .
Now, calculate pH:
Final Answer:
- Degree of dissociation of acetic acid: (or )
- Concentration of acetate ion:
- pH of the solution:
Q47EXERCISES
It has been found that the of a solution of an organic acid is . Calculate the concentration of the anion, the ionization constant of the acid and its .
Solution
Given:
Initial concentration of the organic acid (HA),
pH of the solution = 4.15
Step 1: Calculate the hydrogen ion concentration,
Step 2: Calculate the concentration of the anion,
The ionization of the weak acid is:
From the stoichiometry, at equilibrium, .
Therefore, the concentration of the anion, .
Step 3: Calculate the ionization constant of the acid,
First, find the equilibrium concentration of the undissociated acid, [HA].
Now, use the expression for :
Step 4: Calculate the
Final Answer:
- Concentration of the anion:
- Ionization constant of the acid, :
- of the acid:
Q48EXERCISES
Assuming complete dissociation, calculate the of the following solutions:
(a)
(b)
(c)
(d)
Solution
For strong acids and strong bases, we assume 100% (complete) dissociation in water.
(a)
HCl is a strong acid.
So, .
pH = 2.52
(b)
NaOH is a strong base.
So, .
First, calculate pOH:
Now, calculate pH using the relation (at 298 K).
pH = 11.70
(c)
HBr is a strong acid.
So, .
pH = 2.70
(d)
KOH is a strong base.
So, .
First, calculate pOH:
Now, calculate pH:
pH = 11.30
Q49EXERCISES
Calculate the of the following solutions: a) of dissolved in water to give of solution. b) of dissolved in water to give of solution. c) of dissolved in water to give of solution. d) of is diluted with water to give of solution.
Solution
a) 2 g of TlOH in 2 L of solution
Molar mass of TlOH = 204.4 (Tl) + 16.0 (O) + 1.0 (H) = 221.4 g/mol
Moles of TlOH =
Molarity of TlOH =
TlOH is a strong base: . So, .
pOH =
pH =
pH = 11.66
b) 0.3 g of in 500 mL of solution
Molar mass of = 40.1 + 2(16.0 + 1.0) = 74.1 g/mol
Moles of =
Molarity of =
is a strong base: . So, .
pOH =
pH =
pH = 12.21
c) 0.3 g of NaOH in 200 mL of solution
Molar mass of NaOH = 23.0 + 16.0 + 1.0 = 40.0 g/mol
Moles of NaOH =
Molarity of NaOH =
NaOH is a strong base: . So, .
pOH =
pH =
pH = 12.57
d) 1 mL of 13.6 M HCl diluted to 1 L
We use the dilution formula:
,
HCl is a strong acid: . So, .
pH =
pH = 1.87
Q50EXERCISES
The degree of ionization of a bromoacetic acid solution is . Calculate the of the solution and the of bromoacetic acid.
Solution
Given:
Initial concentration of bromoacetic acid (HBrAc),
Degree of ionization,
Reaction:
Step 1: Calculate the hydrogen ion concentration,
For a weak acid,
Step 2: Calculate the pH of the solution
Step 3: Calculate the ionization constant,
Step 4: Calculate the
Final Answer:
- The pH of the solution is .
- The of bromoacetic acid is .
Q51EXERCISES
The of codeine solution is . Calculate its ionization constant and .
Solution
Given:
Initial concentration of codeine (a weak base),
pH of the solution = 9.95
Let codeine be represented as 'B'.
Reaction:
Step 1: Calculate pOH and
Using the relation :
Now, calculate the hydroxide ion concentration:
Step 2: Determine equilibrium concentrations
From the stoichiometry, .
The equilibrium concentration of the base is:
Step 3: Calculate the ionization constant,
Step 4: Calculate
Final Answer:
- The ionization constant, , is .
- The is .
Q52EXERCISES
What is the of aniline solution? The ionization constant of aniline can be taken from Table 6.7. Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.
Solution
Given:
Initial concentration of aniline (),
From Table 6.7, ionization constant of aniline, .
Reaction:
Step 1: Calculate and degree of ionization ()
Let . Then and .
Since is very small, we can assume , so .
So, .
Degree of ionization, .
Step 2: Calculate pH
pOH =
pH = .
Step 3: Calculate the ionization constant of the conjugate acid
The conjugate acid of aniline () is the anilinium ion ().
We use the relation , where .
Final Answer:
- The pH of the solution is .
- The degree of ionization of aniline is .
- The ionization constant of the conjugate acid (anilinium ion) is .
Q53EXERCISES
Calculate the degree of ionization of acetic acid if its value is . How is the degree of dissociation affected when its solution also contains (a) (b) in ?
Solution
Given:
Concentration of acetic acid (HAc),
First, calculate :
Part 1: Degree of ionization in 0.05 M HAc
Let be the degree of ionization.
Part 2: Effect of HCl (Common Ion Effect)
HCl is a strong acid, providing a common ion H. This will suppress the ionization of the weak acetic acid.
(a) In the presence of 0.01 M HCl
Let be the new degree of ionization.
(b) In the presence of 0.1 M HCl
Let be the new degree of ionization.
Conclusion:
The degree of dissociation decreases significantly in the presence of a strong acid due to the common ion effect. The decrease is more pronounced as the concentration of the common ion (H from HCl) increases.
Final Answer:
- Degree of ionization in 0.05 M acetic acid: .
- Degree of ionization in the presence of 0.01 M HCl: .
- Degree of ionization in the presence of 0.1 M HCl: .
Q54EXERCISES
The ionization constant of dimethylamine is . Calculate its degree of ionization in its solution. What percentage of dimethylamine is ionized if the solution is also in ?
Solution
Given:
Ionization constant of dimethylamine,
Concentration of dimethylamine,
Let dimethylamine be B.
Reaction:
Part 1: Degree of ionization in 0.02 M solution
Let be the degree of ionization.
Since is not extremely small compared to c, we should solve the quadratic equation.
Divide by 0.02:
Using quadratic formula:
Since must be positive,
Part 2: Ionization in presence of 0.1 M NaOH (Common Ion Effect)
NaOH is a strong base, so . This will suppress the ionization of the weak base dimethylamine.
Let be the new degree of ionization.
Percentage ionization =
Final Answer:
- The degree of ionization in 0.02 M solution is .
- The percentage of dimethylamine ionized in the presence of 0.1 M NaOH is .
Q55EXERCISES
Calculate the hydrogen ion concentration in the following biological fluids whose are given below:
(a)
Human muscle-fluid,
(b)
Human stomach fluid,
(c)
Human blood,
(d)
Human saliva, .
Solution
The relationship between pH and hydrogen ion concentration is given by the formula:
(a) Human muscle-fluid, pH = 6.83
(b) Human stomach fluid, pH = 1.2
(c) Human blood, pH = 7.38
(d) Human saliva, pH = 6.4
Final Answer:
(a) Human muscle-fluid:
(b) Human stomach fluid:
(c) Human blood:
(d) Human saliva:
Q56EXERCISES
The of milk, black coffee, tomato juice, lemon juice and egg white are and respectively. Calculate corresponding hydrogen ion concentration in each.
Solution
The relationship between pH and hydrogen ion concentration is given by the formula:
1. Milk, pH = 6.8
2. Black coffee, pH = 5.0
3. Tomato juice, pH = 4.2
4. Lemon juice, pH = 2.2
5. Egg white, pH = 7.8
Final Answer:
- Milk:
- Black coffee:
- Tomato juice:
- Lemon juice:
- Egg white:
Q57EXERCISES
If of is dissolved in water to give of solution at . Calculate the concentrations of potassium, hydrogen and hydroxyl ions. What is its ?
Solution
Given:
Mass of KOH = 0.561 g
Volume of solution = 200 mL = 0.2 L
Step 1: Calculate the molarity of KOH solution.
Molar mass of KOH = 39.1 (K) + 16.0 (O) + 1.0 (H) = 56.1 g/mol
Moles of KOH =
Molarity of KOH =
Step 2: Calculate ion concentrations.
KOH is a strong base and dissociates completely in water:
From the stoichiometry, one mole of KOH produces one mole of K and one mole of OH.
Concentration of potassium ions, .
Concentration of hydroxyl ions, .
Step 3: Calculate hydrogen ion concentration.
Using the ionic product of water, at 298 K.
Step 4: Calculate the pH.
Alternatively, pOH = .
pH = .
Final Answer:
- Concentration of potassium ions, .
- Concentration of hydroxyl ions, .
- Concentration of hydrogen ions, .
- The pH of the solution is .
Q58EXERCISES
The solubility of at is of solution. Calculate the concentrations of strontium and hydroxyl ions and the of the solution.
Solution
Given:
Solubility of
Step 1: Calculate the molar solubility.
Molar mass of = 87.6 (Sr) + 2(16.0 + 1.0) = 121.6 g/mol
Molar solubility (S) =
Step 2: Calculate ion concentrations.
is a strong base and dissociates completely in water according to its solubility:
From the stoichiometry:
Concentration of strontium ions, .
Concentration of hydroxyl ions, .
Step 3: Calculate the pH.
First, calculate pOH:
Now, calculate pH using the relation (at 298 K).
Final Answer:
- Concentration of strontium ions, .
- Concentration of hydroxyl ions, .
- The pH of the solution is .
Q59EXERCISES
The ionization constant of propanoic acid is . Calculate the degree of ionization of the acid in its solution and also its . What will be its degree of ionization if the solution is in also?
Solution
Given:
Ionization constant of propanoic acid,
Concentration of propanoic acid,
Part 1: In 0.05 M propanoic acid solution
Let be the degree of ionization.
Degree of ionization, .
Now, calculate pH:
Part 2: In solution with 0.01 M HCl (Common Ion Effect)
HCl is a strong acid, so .
Let be the new degree of ionization.
Degree of ionization, .
Final Answer:
- In 0.05 M solution: Degree of ionization is and pH is .
- In the presence of 0.01 M HCl: The degree of ionization is .
Q60EXERCISES
The of solution of cyanic acid (HCNO) is . Calculate the ionization constant of the acid and its degree of ionization in the solution.
Solution
Given:
Initial concentration of cyanic acid (HCNO),
pH of the solution = 2.34
Step 1: Calculate the hydrogen ion concentration,
Step 2: Calculate the degree of ionization ()
For a weak acid, .
Step 3: Calculate the ionization constant,
The ionization is:
At equilibrium:
Alternatively, using the degree of ionization:
Final Answer:
- The ionization constant of cyanic acid, , is .
- Its degree of ionization in the solution is .
Q61EXERCISES
The ionization constant of nitrous acid is . Calculate the of sodium nitrite solution and also its degree of hydrolysis.
Solution
Given:
Ionization constant of nitrous acid (),
Concentration of sodium nitrite (),
Sodium nitrite is a salt of a weak acid () and a strong base (NaOH). The nitrite ion () will hydrolyze in water.
Hydrolysis Reaction:
Step 1: Calculate the hydrolysis constant, (which is for )
Step 2: Calculate the degree of hydrolysis (h)
Let h be the degree of hydrolysis.
At equilibrium:
Step 3: Calculate the pH
pOH =
pH =
Final Answer:
- The pH of the sodium nitrite solution is .
- The degree of hydrolysis is .
Q62EXERCISES
A solution of pyridinium hydrochloride has . Calculate the ionization constant of pyridine.
Solution
Given:
Concentration of pyridinium hydrochloride,
pH of the solution = 3.44
Pyridinium hydrochloride () is a salt of a weak base, pyridine (), and a strong acid, HCl. The pyridinium ion () will hydrolyze (act as an acid) in water.
Hydrolysis (Acid Ionization) Reaction:
This is the ionization of the conjugate acid of pyridine. We need to find for this reaction.
Step 1: Calculate from pH
Step 2: Calculate for the pyridinium ion
At equilibrium:
Step 3: Calculate the ionization constant of pyridine ()
We use the relation , where .
is for the conjugate acid, and is for the base (pyridine).
Final Answer:
The ionization constant of pyridine () is .
Q63EXERCISES
Predict if the solutions of the following salts are neutral, acidic or basic: and
Solution
The acidity or basicity of a salt solution depends on the nature of the acid and base from which the salt is formed.
-
NaCl (Sodium chloride): Formed from a strong base (NaOH) and a strong acid (HCl). Neither Na nor Cl hydrolyzes. The solution will be neutral (pH = 7).
-
KBr (Potassium bromide): Formed from a strong base (KOH) and a strong acid (HBr). Neither K nor Br hydrolyzes. The solution will be neutral (pH = 7).
-
NaCN (Sodium cyanide): Formed from a strong base (NaOH) and a weak acid (HCN). The cyanide ion (CN) is the conjugate base of a weak acid and will hydrolyze to produce OH. The solution will be basic (pH > 7).
-
(Ammonium nitrate): Formed from a weak base () and a strong acid (). The ammonium ion () is the conjugate acid of a weak base and will hydrolyze to produce H. The solution will be acidic (pH < 7).
-
(Sodium nitrite): Formed from a strong base (NaOH) and a weak acid (, nitrous acid). The nitrite ion () will hydrolyze to produce OH. The solution will be basic (pH > 7).
-
KF (Potassium fluoride): Formed from a strong base (KOH) and a weak acid (HF). The fluoride ion (F) will hydrolyze to produce OH. The solution will be basic (pH > 7).
Q64EXERCISES
The ionization constant of chloroacetic acid is . What will be the of acid and its sodium salt solution?
Solution
Part 1: pH of 0.1 M chloroacetic acid solution
Given: ,
Let the acid be HA.
Since is not extremely small, we solve the quadratic equation:
Using the quadratic formula, :
Taking the positive root, .
Part 2: pH of 0.1 M sodium chloroacetate solution
This is a salt of a weak acid and a strong base. The chloroacetate ion (A) will hydrolyze.
First, calculate the hydrolysis constant, .
Now, calculate in the 0.1 M salt solution.
pOH =
pH =
Final Answer:
- The pH of the 0.1 M chloroacetic acid solution is .
- The pH of the 0.1 M sodium chloroacetate solution is .
Q65EXERCISES
Ionic product of water at is . What is the of neutral water at this temperature?
Solution
Given:
Ionic product of water, at .
Concept of Neutral Water:
In neutral water, the concentration of hydrogen ions is equal to the concentration of hydroxide ions.
Calculation:
The ionic product of water is given by:
For neutral water, we can substitute with :
Now, calculate the pH:
Final Answer:
The pH of neutral water at is .
Q66EXERCISES
Calculate the of the resultant mixtures: a) of of b) of of c) of of
Solution
a) of of
Millimoles of OH from
Millimoles of H from HCl =
Net mmol of OH remaining =
Total volume =
Final
pOH =
pH =
b) of of
Millimoles of H from
Millimoles of OH from
Since the millimoles of H and OH are equal, the solution is completely neutralized.
The resulting solution contains water and the salt . Since is a salt of a strong acid and a strong base, it does not hydrolyze. The solution will be neutral.
pH = 7.00
c) of of
Millimoles of H from
Millimoles of OH from KOH =
Net mmol of H remaining =
Total volume =
Final
pH =
Final Answers:
a) pH = 12.63
b) pH = 7.00
c) pH = 1.30
Q67EXERCISES
Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at from their solubility product constants given in Table 6.9. Determine also the molarities of individual ions.
Solution
Let S be the molar solubility in mol/L.
1. Silver Chromate ()
,
,
2. Barium Chromate ()
,
3. Ferric Hydroxide ()
,
,
4. Lead Chloride ()
,
,
5. Mercurous Iodide ()
,
,
Q68EXERCISES
The solubility product constant of and are and respectively. Calculate the ratio of the molarities of their saturated solutions.
Solution
Given:
Step 1: Calculate the molar solubility of
Let the molar solubility be .
At equilibrium, and .
Step 2: Calculate the molar solubility of AgBr
Let the molar solubility be .
At equilibrium, and .
Step 3: Calculate the ratio of the molarities
The ratio of the molarities of their saturated solutions is the ratio of their molar solubilities, .
Final Answer:
The ratio of the molarities of the saturated solutions of to is approximately .
Q69EXERCISES
Equal volumes of solutions of sodium iodate and cupric chlorate are mixed together. Will it lead to precipitation of copper iodate? (For cupric iodate ).
Solution
Given:
Initial concentration of sodium iodate () solution =
Initial concentration of cupric chlorate () solution =
for cupric iodate () =
Step 1: Determine the concentrations of ions in the mixture.
When equal volumes are mixed, the total volume doubles, and the concentration of each substance is halved.
From sodium iodate solution:
Concentration of in the mixture, .
From cupric chlorate solution:
Concentration of in the mixture, .
Step 2: Calculate the ionic product, , for copper iodate.
The possible precipitate is copper iodate, .
The equilibrium for its dissolution is:
The ionic product expression is:
Substitute the concentrations from the mixture:
Step 3: Compare with .
Here, .
Conclusion:
Since the ionic product () is less than the solubility product constant (), the solution is unsaturated. Therefore, no precipitation will occur.
Final Answer:
No, mixing the solutions will not lead to the precipitation of copper iodate.
Q70EXERCISES
The ionization constant of benzoic acid is and for silver benzoate is . How many times is silver benzoate more soluble in a buffer of compared to its solubility in pure water?
Solution
Given:
(benzoic acid) =
(silver benzoate) =
pH of buffer = 3.19
Step 1: Solubility in pure water (S1)
Let silver benzoate be AgBz.
Step 2: Solubility in buffer of pH 3.19 (S2)
In the acidic buffer, the benzoate ion (Bz) will react with H:
This reaction reduces the concentration of free Bz ions, thus increasing the solubility of AgBz.
First, find from pH:
Let be the new molar solubility. Then, .
The total concentration of benzoate from the dissolved salt is also . This benzoate exists as both Bz and HBz.
From the acid dissociation equilibrium, , we have .
Substitute this into the solubility equation:
From the expression, .
Substitute this back:
Step 3: Calculate the ratio of solubilities
Ratio =
Final Answer:
Silver benzoate is times more soluble in the buffer of pH 3.19 than in pure water.
Q71EXERCISES
What is the maximum concentration of equimolar solutions of ferrous sulphate and sodium sulphide so that when mixed in equal volumes, there is no precipitation of iron sulphide? (For iron sulphide, ).
Solution
Given:
for iron(II) sulfide (FeS) =
Solutions of ferrous sulfate () and sodium sulfide () are equimolar.
Equal volumes of the solutions are mixed.
Let the initial concentration of both and solutions be M.
When equal volumes are mixed, the volume doubles, and the concentration of each ion is halved.
Concentration of in the mixture,
Concentration of in the mixture,
The precipitation reaction is:
For no precipitation to occur, the ionic product () must be less than or equal to the solubility product constant (). The maximum concentration is achieved when .
Final Answer:
The maximum concentration of the equimolar solutions of ferrous sulphate and sodium sulphide is .
Q72EXERCISES
What is the minimum volume of water required to dissolve of calcium sulphate at ? (For calcium sulphate, is ).
Solution
Given:
Mass of calcium sulfate () = 1 g
for
Step 1: Calculate the molar mass of
Molar mass = 40.1 (Ca) + 32.1 (S) + 4(16.0) (O) = 136.2 g/mol
Step 2: Calculate the molar solubility (S) of
This is the maximum concentration of that can be dissolved in water to form a saturated solution.
Step 3: Calculate the moles of in 1 g
Moles =
Step 4: Calculate the minimum volume of water required
The molar solubility tells us that moles can be dissolved in 1 L of water.
We need to find the volume (V) required to dissolve moles.
Final Answer:
The minimum volume of water required to dissolve 1 g of calcium sulphate is .
Q73EXERCISES
The concentration of sulphide ion in solution saturated with hydrogen sulphide is . If of this is added to of solution of the following: and . in which of these solutions precipitation will take place?
Solution
Given:
Initial concentration of S in the stock solution,
Initial concentration of metal ion solutions,
Volume of S solution = 10 mL
Volume of M solution = 5 mL
Total volume of mixture = 10 mL + 5 mL = 15 mL
Step 1: Calculate the concentrations of ions in the final mixture.
Using the dilution formula :
Final concentration of S,
Final concentration of metal ions,
Step 2: Calculate the ionic product () for each metal sulfide (MS).
Step 3: Compare with the value for each sulfide.
Precipitation occurs if .
values from Table 6.9 in the textbook:
Comparison:
- FeS: . No precipitation.
- MnS: . No precipitation.
- ZnS: . Precipitation will occur.
- CdS: . Precipitation will occur.
Final Answer:
Precipitation will take place in the solutions containing and .