Q: Q5. Find out the value of K_c for each of the following reactions from the K_p values:\n(i) 2NO(g) + Cl_2(g) ⇌ 2NOCl(g); K_p = 1.8 × 10^(-2) at 500 K\n(ii) CaCO_3(s) ⇌ CaO(s) + CO_2(g)

(i) For 2NO(g) + Cl_2(g) ⇌ 2NOCl(g), K_c = 0.73 M^(-1).\n(ii) For CaCO_3(s) ⇌ CaO(s) + CO_2(g), K_c = 0.013 M.

Question 1

Q1. A liquid is in equilibrium with its vapour in a sealed container at a given temperature. The volume of the container is suddenly increased.
a) What is the initial effect of the ch

Accepted Answer

When a liquid is in equilibrium with its vapour in a sealed container, the rates of evaporation and condensation are equal, and the vapour exerts a constant pressure (vapour pressure) at a given temperature.

a) **Initial effect on vapour pressure:**
When the volume of the container is suddenly increased, the existing vapour molecules now occupy a larger volume. This leads to a decrease in the number of vapour molecules per unit volume. Consequently, the vapour pressure immediately drops below the equilibrium vapour pressure.

b) **Initial change in rates of evaporation and condensation:**
*   **Rate of evaporation:** The rate of evaporation depends on the surface area of the liquid and the temperature. Since neither of these factors changes immediately, the rate of evaporation remains constant initially.
*   **Rate of condensation:** The rate of condensation depends on the concentration of vapour molecules in the given volume. Since the volume has increased and the number of vapour molecules per unit volume has decreased, the rate of condensation decreases initially.

c) **Restoration of equilibrium and final vapour pressure:**
Because the rate of evaporation (constant) is now greater than the rate of condensation (decreased), more liquid molecules will evaporate into the vapour phase. This increases the concentration of vapour molecules in the container. This process continues until the rate of condensation again becomes equal to the rate of evaporation. At this point, a new equilibrium is established.

Since the temperature is constant, the final vapour pressure will be the same as the initial equilibrium vapour pressure. Vapour pressure is an intensive property and depends only on the nature of the liquid and temperature, not on the volume of the container (as long as both liquid and vapour phases are present at equilibrium). The system simply adjusts by evaporating more liquid to fill the increased volume until the equilibrium vapour pressure is re-established.

Question 2

Q2. What is K_c for the following equilibrium when equilibrium concentration of each substance is: [SO_2] = 0.60 M, [O_2] = 0.82 M and [SO_3] = 1.90 M ?
2SO_2(g) + O_2(g) ⇌ 2SO_3(g)

Accepted Answer

The equilibrium constant K_c for the given reaction is 12.23 M^(-1).

Question 3

Q3. At a certain temperature and total pressure of 10^5 Pa, iodine vapour contains 40% by volume of I atoms
I_2(g) ⇌ 2I(g)
Calculate K_p for the equilibrium.

Accepted Answer

The K_p for the equilibrium I_2(g) ⇌ 2I(g) is 2.67 × 10^4 Pa.

Question 4

Q4. Write the expression for the equilibrium constant, K_c for each of the following reactions:
(i) 2NOCl(g) ⇌ 2NO(g) + Cl_2(g)
(ii) 2Cu(NO_3)_2(s) ⇌ 2CuO(s) + 4NO_2(g) + O_2(g)
(ii

Accepted Answer

The expressions for the equilibrium constant, K_c, for the given reactions are as follows:

(i) For the reaction: 2NOCl(g) ⇌ 2NO(g) + Cl_2(g)
All reactants and products are in the gaseous phase, so all are included in the K_c expression.
K_c = ([NO]^2 [Cl_2]) / ([NOCl]^2)

(ii) For the reaction: 2Cu(NO_3)_2(s) ⇌ 2CuO(s) + 4NO_2(g) + O_2(g)
Pure solids (Cu(NO_3)_2 and CuO) are not included in the K_c expression because their concentrations are considered constant.
K_c = ([NO_2]^4 [O_2])

(iii) For the reaction: CH_3COOH(aq) + C_2H_5OH(aq) ⇌ CH_3COOC_2H_5(aq) + H_2O(l)
Reactants and products are in aqueous solution, except for water, which is a pure liquid. In reactions where water is a solvent and its concentration does not change significantly, it is generally omitted from the K_c expression. However, in esterification reactions, water is a product and its concentration changes significantly, so it should be included. But NCERT convention for K_c usually omits pure liquids/solids. Following the strict convention of omitting pure liquids, K_c would be:
K_c = ([CH_3COOC_2H_5]) / ([CH_3COOH] [C_2H_5OH])

*Self-correction/Clarification for (iii): While water is a product, if it's also the solvent, its concentration is very high and often considered constant, hence omitted. However, in esterification, water is produced. If the reaction occurs in a non-aqueous solvent or if water is explicitly treated as a product whose concentration changes, it should be included. Given the (l) state, NCERT typically omits pure liquids. Let's stick to the standard NCERT convention for K_c where pure liquids are omitted.*

(iv) For the reaction: Fe^(3+)(aq) + 3SCN^(-)(aq) ⇌ Fe(SCN)_3(aq)
All species are in the aqueous phase, so all are included in the K_c expression.
K_c = ([Fe(SCN)_3]) / ([Fe^(3+)] [SCN^(-)]^3)

Question 5

Q5. Find out the value of K_c for each of the following reactions from the K_p values:
(i) 2NO(g) + Cl_2(g) ⇌ 2NOCl(g); K_p = 1.8 × 10^(-2) at 500 K
(ii) CaCO_3(s) ⇌ CaO(s) + CO_2(g)

Accepted Answer

(i) For 2NO(g) + Cl_2(g) ⇌ 2NOCl(g), K_c = 0.73 M^(-1).
(ii) For CaCO_3(s) ⇌ CaO(s) + CO_2(g), K_c = 0.013 M.

Question 6

Q6. The value of K_c for the reaction 2A ⇌ B + C is 2 × 10⁻³ at a given time, the composition of reaction mixture is [A] = [B] = [C] = 3 × 10⁻⁴ M. In which direction the reaction will

Accepted Answer

To determine the direction in which the reaction will proceed, we need to calculate the reaction quotient (Q_c) and compare it with the equilibrium constant (K_c).

The given reaction is: 2A ⇌ B + C

The expression for the reaction quotient (Q_c) is:
Q_c = ([B][C]) / [A]²

Given concentrations at a particular time:
[A] = 3 × 10⁻⁴ M
[B] = 3 × 10⁻⁴ M
[C] = 3 × 10⁻⁴ M

Substitute these values into the Q_c expression:
Q_c = (3 × 10⁻⁴ M × 3 × 10⁻⁴ M) / (3 × 10⁻⁴ M)²
Q_c = (9 × 10⁻⁸ M²) / (9 × 10⁻⁸ M²)
Q_c = 1

Given equilibrium constant K_c = 2 × 10⁻³.

Now, compare Q_c with K_c:
Q_c = 1
K_c = 2 × 10⁻³ = 0.002

Since Q_c (1) > K_c (0.002), the reaction is not at equilibrium. For the reaction to reach equilibrium, the concentrations of products must decrease and the concentration of reactants must increase. This means the reaction will proceed in the **backward (reverse) direction**.

Question 7

Q7. Explain the effect of the following factors on the equilibrium of a reaction:
a) Temperature
b) Pressure
c) Catalyst

Accepted Answer

The effect of various factors on the equilibrium of a reaction can be explained using 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 relieves the stress.

### a) Effect of Temperature
Temperature is the only factor that changes the value of the equilibrium constant (K). The effect of temperature depends on whether the reaction is exothermic or endothermic.

1.  **For Exothermic Reactions (ΔH < 0):** Heat is considered a product. Increasing the temperature adds 'heat' to the system. According to Le Chatelier's Principle, the equilibrium will shift in the direction that consumes this added heat, i.e., the **backward (reverse) direction**. This leads to a decrease in the equilibrium constant (K).
    *Example: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) + Heat*
    *Increasing temperature shifts equilibrium to the left (favors decomposition of NH₃).*

2.  **For Endothermic Reactions (ΔH > 0):** Heat is considered a reactant. Increasing the temperature adds 'heat' to the system. The equilibrium will shift in the direction that consumes this added heat, i.e., the **forward direction**. This leads to an increase in the equilibrium constant (K).
    *Example: N₂(g) + O₂(g) + Heat ⇌ 2NO(g)*
    *Increasing temperature shifts equilibrium to the right (favors formation of NO).*

Conversely, decreasing the temperature will shift an exothermic reaction in the forward direction and an endothermic reaction in the backward direction.

### b) Effect of Pressure
Pressure changes primarily affect reactions involving gases. The effect of pressure depends on the change in the number of moles of gaseous substances (Δn_g) during the reaction.

1.  **If Δn_g ≠ 0 (Unequal moles of gaseous reactants and products):**
    *   **Increasing Pressure:** The system will try to reduce the pressure by shifting towards the side with **fewer moles of gas**. This reduces the total number of gas molecules, thereby reducing the pressure.
    *   **Decreasing Pressure:** The system will try to increase the pressure by shifting towards the side with **more moles of gas**.
    *Example: N₂(g) + 3H₂(g) ⇌ 2NH₃(g)*
    *(Reactants: 1+3=4 moles of gas; Products: 2 moles of gas. Δn_g = 2 - 4 = -2)*
    *Increasing pressure shifts equilibrium to the right (forward direction), favoring NH₃ formation.*

2.  **If Δn_g = 0 (Equal moles of gaseous reactants and products):**
    *   Changes in pressure have **no effect** on the position of equilibrium. The equilibrium constant (K) also remains unchanged.
    *Example: H₂(g) + I₂(g) ⇌ 2HI(g)*
    *(Reactants: 1+1=2 moles of gas; Products: 2 moles of gas. Δn_g = 2 - 2 = 0)*
    *Pressure changes do not affect this equilibrium.*

### c) Effect of Catalyst
A catalyst is a substance that increases the rate of a chemical reaction without being consumed in the reaction. It does this by providing an alternative reaction pathway with a lower activation energy.

*   **No Effect on Equilibrium Position:** A catalyst increases the rate of both the forward and reverse reactions equally. Therefore, it helps the system reach equilibrium faster, but it **does not change the position of equilibrium** (i.e., the relative amounts of reactants and products at equilibrium). The equilibrium constant (K) remains unchanged.
*   **Faster Attainment of Equilibrium:** Catalysts are useful in industrial processes to achieve equilibrium concentrations more quickly, thus increasing the rate of product formation.

In summary, temperature affects the value of K and the equilibrium position, pressure affects the equilibrium position only if Δn_g ≠ 0, and a catalyst affects only the rate of attainment of equilibrium, not its position or the value of K.

Question 8

Q8. What is the effect of adding:
a) H₂
b) I₂
c) an inert gas (at constant volume)
d) an inert gas (at constant pressure)
on the equilibrium of the reaction 2HI(g) ⇌ H₂(g) + I₂(g)

Accepted Answer

The reaction given is: 2HI(g) ⇌ H₂(g) + I₂(g)

Let's analyze the effect of each factor based on Le Chatelier's Principle:

### a) Effect of adding H₂
*   **Explanation:** H₂ is a product in the given equilibrium. Adding H₂ increases its concentration.
*   **Le Chatelier's Principle:** According to Le Chatelier's Principle, the system will try to counteract this increase by consuming the added H₂.
*   **Direction of Shift:** The equilibrium will shift in the **backward (reverse) direction**, favoring the formation of HI. This means more HI will be formed, and some I₂ will be consumed along with H₂.

### b) Effect of adding I₂
*   **Explanation:** I₂ is also a product in the given equilibrium. Adding I₂ increases its concentration.
*   **Le Chatelier's Principle:** The system will try to counteract this increase by consuming the added I₂.
*   **Direction of Shift:** The equilibrium will shift in the **backward (reverse) direction**, favoring the formation of HI. This means more HI will be formed, and some H₂ will be consumed along with I₂.

### c) Effect of adding an inert gas (at constant volume)
*   **Explanation:** An inert gas (like Argon or Helium) does not participate in the reaction. When an inert gas is added to a reaction mixture at constant volume, the total pressure of the system increases. However, the partial pressures (and thus concentrations) of the reacting gases (HI, H₂, I₂) remain unchanged because the volume is constant and their amounts are not changing.
*   **Le Chatelier's Principle:** Since the partial pressures/concentrations of the reactants and products are not affected, the reaction quotient (Q_c or Q_p) remains the same.
*   **Direction of Shift:** Therefore, adding an inert gas at constant volume has **no effect** on the position of equilibrium.

### d) Effect of adding an inert gas (at constant pressure)
*   **Explanation:** When an inert gas is added at constant pressure, the total volume of the system must increase to keep the pressure constant. This increase in volume effectively decreases the partial pressures (and thus concentrations) of all the reacting gases (HI, H₂, I₂).
*   **Le Chatelier's Principle:** The system will try to counteract this decrease in partial pressures by shifting towards the side with a greater number of moles of gas.
*   **Analysis of Moles of Gas:** For the given reaction, 2HI(g) ⇌ H₂(g) + I₂(g):
    *   Moles of gaseous reactants = 2
    *   Moles of gaseous products = 1 (H₂) + 1 (I₂) = 2
    *   Since the number of moles of gaseous reactants is equal to the number of moles of gaseous products (Δn_g = 0), a change in volume (due to adding inert gas at constant pressure) will have **no effect** on the position of equilibrium.
*   **Direction of Shift:** Therefore, adding an inert gas at constant pressure has **no effect** on the position of equilibrium for this specific reaction.

Question 9

Q9. The value of K_c for the reaction 3O_2(g) <=> 2O_3(g) is 2.0 × 10^(-50) at 25°C. If the equilibrium concentration of O_2 in a sealed vessel is 1.6 × 10^(-2) M, what is the concentr

Accepted Answer

The concentration of O_3 at equilibrium is 3.2 × 10^(-27) M.

Question 10

Q10. The reaction, CO(g) + 3H_2(g) <=> CH_4(g) + H_2O(g) is at equilibrium in a 1 L flask. The concentrations are: CO = 0.30 M, H_2 = 0.10 M, CH_4 = 0.02 M and H_2O = 0.02 M. Calculate

Accepted Answer

The value of K_c for the reaction is 13.33 M^(-2).

Question 11

Q11. What is the effect of:
a) addition of H_2
b) addition of CH_3OH
c) removal of CO
d) removal of CH_3OH
on the equilibrium of the reaction: 2H_2(g) + CO(g) <=> CH_3OH(g)

Accepted Answer

The effect of various changes on the equilibrium of the reaction 2H_2(g) + CO(g) <=> CH_3OH(g) can be predicted using Le Chatelier's Principle.

Question 12

Q12. The value of $K_c$ for the reaction $2A ightleftharpoons B + C$ is $2 	imes 10^{-3}$. At a given time, the composition of reaction mixture is $[A] = [B] = [C] = 3 	imes 10^{-4}

Accepted Answer

To determine the direction of the reaction, we need to calculate the reaction quotient ($Q_c$) and compare it with the equilibrium constant ($K_c$).

The given reaction is: $2A ightleftharpoons B + C$

The expression for the reaction quotient ($Q_c$) is:
$Q_c = \frac{[B][C]}{[A]^2}$

Given concentrations at a specific time:
$[A] = 3 	imes 10^{-4} M$
$[B] = 3 	imes 10^{-4} M$
$[C] = 3 	imes 10^{-4} M$

Substitute these values into the $Q_c$ expression:
$Q_c = \frac{(3 	imes 10^{-4} M)(3 	imes 10^{-4} M)}{(3 	imes 10^{-4} M)^2}$
$Q_c = \frac{9 	imes 10^{-8} M^2}{9 	imes 10^{-8} M^2}$
$Q_c = 1$

Now, compare $Q_c$ with the given $K_c$ value:
$K_c = 2 	imes 10^{-3}$
$Q_c = 1$

Since $Q_c (1) > K_c (2 	imes 10^{-3})$, the reaction is not at equilibrium. For the reaction to reach equilibrium, the concentrations of products must decrease and the concentrations of reactants must increase. Therefore, the reaction will proceed in the reverse direction (from right to left) to reach equilibrium.

Question 13

Q13. Suggest in which of the following reactions the concentration of reactants and products present at equilibrium will be practically same:
(a) $2CO(g) + O_2(g) ightleftharpoons 2C

Accepted Answer

The equilibrium constant ($K_c$) provides information about the relative amounts of reactants and products at equilibrium. Its magnitude indicates the extent to which a reaction proceeds to completion.

1.  **If $K_c$ is very large ($K_c \gg 1$):** The reaction proceeds almost to completion. At equilibrium, the concentration of products will be significantly higher than the concentration of reactants.
2.  **If $K_c$ is very small ($K_c \ll 1$):** The reaction proceeds to a very small extent. At equilibrium, the concentration of reactants will be significantly higher than the concentration of products.
3.  **If $K_c$ is close to 1 (typically between $0.01$ and $100$):** The concentrations of reactants and products at equilibrium will be comparable, or practically same.

Let's analyze each reaction:

**(a) $2CO(g) + O_2(g) ightleftharpoons 2CO_2(g)$; $K_c = 5 	imes 10^{10}$**
Here, $K_c$ is very large ($5 	imes 10^{10} \gg 1$). This indicates that at equilibrium, the reaction will have proceeded almost entirely to the right, meaning the concentration of products ($CO_2$) will be much higher than the concentrations of reactants ($CO$ and $O_2$). So, reactant and product concentrations will NOT be practically same.

**(b) $H_2(g) + I_2(g) ightleftharpoons 2HI(g)$; $K_c = 2.0$**
Here, $K_c$ is $2.0$, which is close to 1. This suggests that at equilibrium, neither reactants nor products are overwhelmingly favored. The concentrations of reactants ($H_2$, $I_2$) and products ($HI$) will be comparable or practically same.

**(c) $N_2(g) + 3H_2(g) ightleftharpoons 2NH_3(g)$; $K_c = 3.7 	imes 10^{-3}$**
Here, $K_c$ is very small ($3.7 	imes 10^{-3} \ll 1$). This indicates that at equilibrium, the reaction will have proceeded only to a small extent, meaning the concentrations of reactants ($N_2$, $H_2$) will be much higher than the concentration of the product ($NH_3$). So, reactant and product concentrations will NOT be practically same.

Therefore, the concentration of reactants and products present at equilibrium will be practically same for reaction **(b) $H_2(g) + I_2(g) ightleftharpoons 2HI(g)$**.

Question 14

Q14. One of the reactions that takes place in producing steel from iron ore is the reduction of iron(II) oxide by carbon monoxide to give iron metal and $CO_2$.
$FeO(s) + CO(g) ightl

Accepted Answer

The given reaction is: $FeO(s) + CO(g) ightleftharpoons Fe(s) + CO_2(g)$
Equilibrium constant $K_p = 0.265$ at 1050 K.
Initial partial pressures: $P_{CO} = 1.4 	ext{ atm}$, $P_{CO_2} = 0.80 	ext{ atm}$.

**Step 1: Write the expression for $K_p$**
Since $FeO(s)$ and $Fe(s)$ are pure solids, their partial pressures (or concentrations) are considered constant and are not included in the equilibrium constant expression.
$K_p = \frac{P_{CO_2}}{P_{CO}}$

**Step 2: Calculate the initial reaction quotient ($Q_p$)**
$Q_p = \frac{P_{CO_2, initial}}{P_{CO, initial}} = \frac{0.80 	ext{ atm}}{1.4 	ext{ atm}} = 0.571$

**Step 3: Determine the direction of the reaction**
Compare $Q_p$ with $K_p$:
$Q_p = 0.571$
$K_p = 0.265$
Since $Q_p > K_p$, the reaction will proceed in the reverse direction (from right to left) to reach equilibrium. This means $CO_2$ will decrease and $CO$ will increase.

**Step 4: Set up an ICE (Initial, Change, Equilibrium) table for partial pressures**
Let $x$ be the change in partial pressure.

| Species | Initial (atm) | Change (atm) | Equilibrium (atm) |
| :------ | :------------ | :----------- | :---------------- |
| $CO(g)$ | $1.4$         | $+x$         | $1.4 + x$         |
| $CO_2(g)$ | $0.80$        | $-x$         | $0.80 - x$        |

**Step 5: Substitute equilibrium partial pressures into the $K_p$ expression**
$K_p = \frac{P_{CO_2, eq}}{P_{CO, eq}}$
$0.265 = \frac{0.80 - x}{1.4 + x}$

**Step 6: Solve for $x$**
$0.265 (1.4 + x) = 0.80 - x$
$0.265 	imes 1.4 + 0.265x = 0.80 - x$
$0.371 + 0.265x = 0.80 - x$
$0.265x + x = 0.80 - 0.371$
$1.265x = 0.429$
$x = \frac{0.429}{1.265}$
$x \approx 0.339 	ext{ atm}$

**Step 7: Calculate the equilibrium partial pressures**
$P_{CO, eq} = 1.4 + x = 1.4 + 0.339 = 1.739 	ext{ atm}$
$P_{CO_2, eq} = 0.80 - x = 0.80 - 0.339 = 0.461 	ext{ atm}$

**Final Answer:**
At equilibrium, the partial pressure of $CO$ is $1.739 	ext{ atm}$ and the partial pressure of $CO_2$ is $0.461 	ext{ atm}$.

Question 15

Q15. The value of K_c for the reaction 3O₂(g) ⇌ 2O₃(g) is 2.0 × 10⁻⁵⁰ at 25°C. If the equilibrium concentration of O₂ in a sealed vessel is 1.6 × 10⁻² M, what is the concentration of O₃

Accepted Answer

To find the concentration of O₃ at equilibrium, we first write the equilibrium constant expression (K_c) for the given reaction. Then, we substitute the given values of K_c and the equilibrium concentration of O₂ into the expression and solve for the concentration of O₃.

Question 16

Q16. The reaction, CO(g) + 3H₂(g) ⇌ CH₄(g) + H₂O(g) is at equilibrium in a 1 L flask. The concentrations are CO = 0.30 mol L⁻¹, H₂ = 0.10 mol L⁻¹, CH₄ = 0.02 mol L⁻¹ and H₂O = 0.01 mol

Accepted Answer

To calculate the equilibrium constant (K_c) for the given reaction, we first write the K_c expression based on the stoichiometry of the balanced chemical equation. Then, we substitute the given equilibrium concentrations of reactants and products into the expression and perform the calculation.

Question 17

Q17. What is the effect of:
(a) addition of H₂
(b) addition of CH₃OH
(c) removal of CO
(d) removal of CH₃OH
on the equilibrium of the reaction: 2H₂(g) + CO(g) ⇌ CH₃OH(g)

Accepted Answer

The effect of changes in concentration on the equilibrium position of a reversible reaction is governed by Le Chatelier's Principle. According to this principle, if a change of condition is applied to a system in equilibrium, the system will shift in a direction that relieves the stress.

Question 18

Q18. The value of K_c for the reaction 2A ⇌ B + C is 2 × 10^{-3} at a given time, the composition of reaction mixture is [A] = [B] = [C] = 3 × 10^{-4} M. In which direction the reaction

Accepted Answer

To determine the direction of the reaction, we need to calculate the reaction quotient (Q_c) and compare it with the equilibrium constant (K_c).

The given reaction is: 2A(g) ⇌ B(g) + C(g)
Given K_c = 2 × 10^{-3}
Given concentrations: [A] = 3 × 10^{-4} M, [B] = 3 × 10^{-4} M, [C] = 3 × 10^{-4} M

1.  **Write the expression for the reaction quotient (Q_c):**
    Q_c = ([B][C]) / [A]^2

2.  **Substitute the given concentrations into the Q_c expression:**
    Q_c = (3 × 10^{-4} M × 3 × 10^{-4} M) / (3 × 10^{-4} M)^2
    Q_c = (9 × 10^{-8}) / (9 × 10^{-8})
    Q_c = 1

3.  **Compare Q_c with K_c:**
    We have Q_c = 1 and K_c = 2 × 10^{-3}.
    Clearly, Q_c (1) > K_c (0.002).

4.  **Determine the direction of the reaction:**
    Since Q_c > K_c, the reaction mixture has a higher concentration of products and/or lower concentration of reactants than it would have at equilibrium. To reach equilibrium, the reaction must proceed in the reverse direction (from products to reactants) to decrease the concentration of products and increase the concentration of reactants.

Therefore, the reaction will proceed in the **reverse direction**.

Question 19

Q19. What is the effect of:
(a) addition of H_2
(b) addition of CH_3OH
(c) removal of CO
(d) removal of CH_3OH
on the equilibrium of the reaction:
2H_2(g) + CO(g) ⇌ CH_3OH(g)

Accepted Answer

The given reaction is: 2H_2(g) + CO(g) ⇌ CH_3OH(g)
According to Le Chatelier's Principle, if a change of condition is applied to a system in equilibrium, the system will shift in a direction that relieves the stress.

(a) **Addition of H_2:**
    H_2 is a reactant. Increasing the concentration of a reactant will shift the equilibrium to the right (forward direction) to consume the added H_2 and produce more CH_3OH.

(b) **Addition of CH_3OH:**
    CH_3OH is a product. Increasing the concentration of a product will shift the equilibrium to the left (reverse direction) to consume the added CH_3OH and produce more H_2 and CO.

(c) **Removal of CO:**
    CO is a reactant. Decreasing the concentration of a reactant will shift the equilibrium to the left (reverse direction) to replenish the removed CO, consuming CH_3OH and producing more H_2.

(d) **Removal of CH_3OH:**
    CH_3OH is a product. Decreasing the concentration of a product will shift the equilibrium to the right (forward direction) to produce more CH_3OH, consuming H_2 and CO.

Question 20

Q20. At 473 K, equilibrium constant K_c for decomposition of phosphorus pentachloride, PCl_5 is 8.3 × 10^{-3}. If decomposition is depicted as:
PCl_5(g) ⇌ PCl_3(g) + Cl_2(g) Δ_r H^° =

Accepted Answer

The given reaction is: PCl_5(g) ⇌ PCl_3(g) + Cl_2(g)
Given K_c = 8.3 × 10^{-3} at 473 K.
Given Δ_r H^° = 124.0 kJ mol^{-1} (positive, so the reaction is endothermic).

(a) **Expression for K_c for the reaction:**
    For the reaction PCl_5(g) ⇌ PCl_3(g) + Cl_2(g), the equilibrium constant K_c is given by the ratio of the product of molar concentrations of products to the product of molar concentrations of reactants, each raised to the power of their stoichiometric coefficients.
    K_c = ([PCl_3][Cl_2]) / [PCl_5]

(b) **Value of K_c for the reverse reaction at the same temperature:**
    The reverse reaction is: PCl_3(g) + Cl_2(g) ⇌ PCl_5(g)
    The equilibrium constant for the reverse reaction (K_c') is the reciprocal of the equilibrium constant for the forward reaction (K_c).
    K_c' = 1 / K_c
    K_c' = 1 / (8.3 × 10^{-3})
    K_c' = 1 / 0.0083
    K_c' ≈ 120.48

(c) **Effect on K_c if:**
    The equilibrium constant K_c is a constant for a given reaction at a specific temperature. It does not change with changes in concentration, pressure, or the addition of a catalyst. It only changes with temperature.

(i) **More PCl_5 is added:**
        Adding more PCl_5 (a reactant) will shift the equilibrium to the right (forward direction) to consume the added PCl_5 and produce more PCl_3 and Cl_2. However, the value of K_c **remains unchanged** because K_c is independent of initial concentrations or the amount of reactants/products added.

(ii) **Pressure is increased:**
        Increasing the pressure will shift the equilibrium in the direction that has fewer moles of gas. In the given reaction, PCl_5(g) (1 mole) ⇌ PCl_3(g) + Cl_2(g) (2 moles). So, increasing pressure will shift the equilibrium to the left (reverse direction). However, the value of K_c **remains unchanged** because K_c is independent of pressure changes.

(iii) **The temperature is increased:**
        The reaction PCl_5(g) ⇌ PCl_3(g) + Cl_2(g) has Δ_r H^° = 124.0 kJ mol^{-1}, which means it is an endothermic reaction. For an endothermic reaction, increasing the temperature favors the forward reaction (product formation). This leads to an increase in the concentration of products and a decrease in the concentration of reactants at the new equilibrium. Therefore, the value of K_c **will increase** when the temperature is increased.

Question 21

Q21. Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following reaction:
CH_4(g) + H_2O(g) ⇌ CO(g) + 3H_2(g)
(a) Write an expression for K_p for the

Accepted Answer

The given reaction is: CH_4(g) + H_2O(g) ⇌ CO(g) + 3H_2(g)

(a) **Expression for K_p for the reaction:**
    For a gaseous reaction, the equilibrium constant K_p is expressed in terms of partial pressures of the gaseous reactants and products, each raised to the power of their stoichiometric coefficients.
    K_p = (P_{CO} × P_{H_2}^3) / (P_{CH_4} × P_{H_2O})

(b) **How will the values of K_p and composition of equilibrium mixture be affected by:**

(i) **Increasing the pressure:**
        *   **Effect on K_p:** The value of K_p is a constant for a given reaction at a specific temperature. It **remains unchanged** with a change in pressure.
        *   **Effect on equilibrium composition:** According to Le Chatelier's Principle, increasing the pressure on a gaseous equilibrium system will shift the equilibrium in the direction that has fewer moles of gas. Let's count the moles of gas on each side:
            Reactant side: 1 mole CH_4 + 1 mole H_2O = 2 moles of gas
            Product side: 1 mole CO + 3 moles H_2 = 4 moles of gas
            Since the product side has more moles of gas (4 moles) than the reactant side (2 moles), increasing the pressure will shift the equilibrium to the **left (reverse direction)**. This means the equilibrium mixture will have a higher concentration of CH_4 and H_2O, and a lower concentration of CO and H_2.

(ii) **Increasing the temperature:**
        To determine the effect of temperature, we need to know if the reaction is endothermic or exothermic. This information is not explicitly given, but the reaction is a steam reforming reaction, which is typically highly endothermic. Let's assume it's an endothermic reaction (ΔH > 0), which is standard for this process.
        *   **Effect on K_p:** For an endothermic reaction, increasing the temperature favors the forward reaction. This leads to an increase in the concentration (and partial pressure) of products and a decrease in the concentration (and partial pressure) of reactants at the new equilibrium. Therefore, the value of K_p **will increase**.
        *   **Effect on equilibrium composition:** Since increasing the temperature favors the forward (endothermic) reaction, the equilibrium will shift to the **right (forward direction)**. This means the equilibrium mixture will have a higher concentration of CO and H_2, and a lower concentration of CH_4 and H_2O.

(iii) **Using a catalyst:**
        *   **Effect on K_p:** A catalyst speeds up both the forward and reverse reactions equally. It helps the system reach equilibrium faster but does not change the position of equilibrium or the relative amounts of reactants and products at equilibrium. Therefore, the value of K_p **remains unchanged**.
        *   **Effect on equilibrium composition:** A catalyst does not affect the equilibrium composition. It only **reduces the time** required to reach equilibrium. The final concentrations of reactants and products at equilibrium will be the same as without the catalyst, assuming enough time is given.

Question 22

Q22. Calculate the pH of a 0.1 M ammonia solution. Calculate the pH after 50.0 mL of this solution is treated with 25.0 mL of 0.1 M HCl. (Dissociation constant of NH₃ = 1.77 × 10⁻⁵)

Accepted Answer

The pH of 0.1 M ammonia solution is 11.12.
The pH after treating 50.0 mL of this solution with 25.0 mL of 0.1 M HCl is 9.25.

Question 23

Q23. Calculate the pH of the following solutions:
(a) 0.01 M HCl
(b) 0.01 M NaOH
(c) 0.01 M H₂SO₄
(d) 0.01 M Ca(OH)₂

Accepted Answer

The pH values for the given solutions are:
(a) For 0.01 M HCl, pH = 2
(b) For 0.01 M NaOH, pH = 12
(c) For 0.01 M H₂SO₄, pH = 1.70
(d) For 0.01 M Ca(OH)₂, pH = 12.30

Question 24

Q24. The pH of a solution is 3.12. Calculate the concentration of H⁺ ions.

Accepted Answer

The concentration of H⁺ ions in the solution is 7.58 × 10⁻⁴ M.

Question 25

Q25. Calculate the degree of ionization and pH of 0.05 M of a weak base, BOH, having a K_b = 1.77 × 10⁻⁵.

Accepted Answer

For a 0.05 M solution of the weak base BOH with K_b = 1.77 × 10⁻⁵:
Degree of ionization (α) = 0.0188 or 1.88%
pH = 11.97

Question 26

Q26. Calculate the pH of a 0.1 M ammonia solution. Calculate the pH after 50.0 mL of this solution is treated with 25.0 mL of 0.1 M HCl. The dissociation constant of ammonia, K_b = 1.77

Accepted Answer

The pH of 0.1 M ammonia solution is 11.12.
The pH of the solution after treating 50.0 mL of 0.1 M ammonia with 25.0 mL of 0.1 M HCl is 9.25.

Question 27

Q27. What is the effect of:
(a) addition of H₂
(b) addition of CH₃OH
(c) removal of CO
(d) removal of CH₃OH
on the equilibrium of the reaction:
2H₂(g) + CO(g) ⇌ CH₃OH(g)

Accepted Answer

According to Le Chatelier's Principle, if a change of condition is applied to a system in equilibrium, the system will shift in a direction that relieves the stress. The given reaction is: 2H₂(g) + CO(g) ⇌ CH₃OH(g)

(a) **Addition of H₂**: H₂ is a reactant. Increasing the concentration of a reactant will cause the equilibrium to shift in the forward direction (towards products) to consume the added H₂.

(b) **Addition of CH₃OH**: CH₃OH is a product. Increasing the concentration of a product will cause the equilibrium to shift in the backward direction (towards reactants) to consume the added CH₃OH.

(c) **Removal of CO**: CO is a reactant. Decreasing the concentration of a reactant will cause the equilibrium to shift in the backward direction (towards reactants) to replenish the removed CO.

(d) **Removal of CH₃OH**: CH₃OH is a product. Decreasing the concentration of a product will cause the equilibrium to shift in the forward direction (towards products) to produce more CH₃OH.

Question 28

Q28. At 473 K, equilibrium constant K_c for decomposition of phosphorus pentachloride, PCl₅ is 8.3 × 10⁻³ . If decomposition is depicted as:
PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) Δ_r H° = 124.0 k

Accepted Answer

(a) The expression for K_c for the reaction PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) is K_c = [PCl₃][Cl₂] / [PCl₅].
(b) The value of K_c for the reverse reaction at the same temperature is approximately 120.48.
(c) (i) Increasing pressure has no effect on the value of K_c. (ii) Increasing temperature will increase the value of K_c because the reaction is endothermic.

Question 29

Q29. The pH of 0.1 M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.

Accepted Answer

Given:
Concentration of HCNO (C) = 0.1 M
pH of the solution = 2.34

**1. Calculate [H⁺] concentration:**
pH = -log[H⁺]
2.34 = -log[H⁺]
[H⁺] = 10⁻².³⁴ M
[H⁺] = 4.57 × 10⁻³ M

**2. Write the dissociation equilibrium for HCNO:**
HCNO(aq) ⇌ H⁺(aq) + CNO⁻(aq)

**3. Set up an ICE table:**
Initial:       0.1 M        0         0
Change:       -x           +x        +x
Equilibrium: (0.1 - x) M    x M       x M

From the pH calculation, we know that at equilibrium, [H⁺] = x = 4.57 × 10⁻³ M.

**4. Calculate equilibrium concentrations:**
[H⁺] = 4.57 × 10⁻³ M
[CNO⁻] = 4.57 × 10⁻³ M
[HCNO] = 0.1 - 4.57 × 10⁻³ = 0.1 - 0.00457 = 0.09543 M

**5. Calculate the ionization constant (K_a):**
K_a = ([H⁺][CNO⁻]) / [HCNO]
K_a = (4.57 × 10⁻³ M) * (4.57 × 10⁻³ M) / (0.09543 M)
K_a = (20.8849 × 10⁻⁶) / 0.09543
K_a = 2.188 × 10⁻⁴

**6. Calculate the degree of ionization (α):**
Degree of ionization (α) = (Amount dissociated) / (Initial amount)
α = [H⁺] / C
α = (4.57 × 10⁻³ M) / (0.1 M)
α = 0.0457

Alternatively, α can be expressed as a percentage:
α = 0.0457 × 100% = 4.57%

**Final Answer:**
The ionization constant (K_a) of cyanic acid is **2.19 × 10⁻⁴**.
The degree of ionization (α) in the solution is **0.0457** (or 4.57%).

Question 30

Q30. Calculate the pH of the following solutions: (a) 0.01 M HCl (b) 0.01 M NaOH (c) 0.01 M H₂SO₄ (d) 0.01 M Ca(OH)₂

Accepted Answer

The pH of the given solutions are:
(a) 0.01 M HCl: pH = 2
(b) 0.01 M NaOH: pH = 12
(c) 0.01 M H₂SO₄: pH = 1.7
(d) 0.01 M Ca(OH)₂: pH = 12.3

Question 31

Q31. Calculate the pH of 0.1 M solution of NH₄Cl. K_b for NH₄OH is 1.77 × 10⁻⁵.

Accepted Answer

The pH of a 0.1 M NH₄Cl solution is 5.12.

Question 32

Q32. If 50 mL of 0.1 M NaOH is added to 50 mL of 0.1 M acetic acid solution, the pH of the resulting solution is (Given K_a of acetic acid = 1.75 × 10⁻⁵).

Accepted Answer

The pH of the resulting solution is 8.73.

Question 33

Q33. What is the solubility of Al(OH)₃ in a buffer solution of pH = 4.0? K_sp for Al(OH)₃ = 1.3 × 10⁻³³.

Accepted Answer

The solubility of Al(OH)₃ in a buffer solution of pH = 4.0 is 1.3 × 10⁻³ M.

Question 34

Q34. The pH of 0.1 M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.

Accepted Answer

Given:
Initial concentration of HCNO (C) = 0.1 M
pH of the solution = 2.34

**Part 1: Calculation of Ionization Constant (Ka)**
1.  **Determine [H⁺] from pH:**
    pH = -log[H⁺]
    2.34 = -log[H⁺]
    [H⁺] = 10⁻².³⁴ M
    [H⁺] = 4.57 × 10⁻³ M

2.  **Set up the equilibrium expression for HCNO:**
    HCNO(aq) ⇌ H⁺(aq) + CNO⁻(aq)

Initial:      0.1 M          0           0
    Change:      -x             +x          +x
    Equilibrium: (0.1 - x) M     x M         x M

From the pH, we know that at equilibrium, [H⁺] = x = 4.57 × 10⁻³ M.

3.  **Calculate equilibrium concentrations:**
    [H⁺] = 4.57 × 10⁻³ M
    [CNO⁻] = 4.57 × 10⁻³ M
    [HCNO] = 0.1 - 4.57 × 10⁻³ = 0.1 - 0.00457 = 0.09543 M

4.  **Calculate Ka:**
    Ka = ([H⁺][CNO⁻]) / [HCNO]
    Ka = (4.57 × 10⁻³)(4.57 × 10⁻³) / (0.09543)
    Ka = (20.8849 × 10⁻⁶) / (0.09543)
    Ka = 2.188 × 10⁻⁴

**Part 2: Calculation of Degree of Ionization (α)**
Degree of ionization (α) = (concentration of ionized acid) / (initial concentration of acid)
α = [H⁺] / C
α = (4.57 × 10⁻³) / 0.1
α = 0.0457

Alternatively, α = √(Ka/C) is an approximation valid for very weak acids where α is small. Here, we have the actual ionized concentration.

Therefore, the ionization constant of cyanic acid is 2.19 × 10⁻⁴ and its degree of ionization is 0.0457.

Question 35

Q35. The ionization constant of phenol is 1.0 × 10⁻¹⁰. What is the concentration of phenolate ion in 0.05 M solution of phenol? What will be its degree of ionization if the solution is

Accepted Answer

Given:
Ionization constant of phenol (Ka) = 1.0 × 10⁻¹⁰
Initial concentration of phenol (C₆H₅OH) = 0.05 M

**Part 1: Concentration of phenolate ion in 0.05 M phenol solution (without common ion)**
Phenol (C₆H₅OH) is a weak acid that dissociates as follows:
C₆H₅OH(aq) ⇌ H⁺(aq) + C₆H₅O⁻(aq)

Initial:      0.05 M          0           0
Change:      -x             +x          +x
Equilibrium: (0.05 - x) M     x M         x M

Ka = ([H⁺][C₆H₅O⁻]) / [C₆H₅OH]
1.0 × 10⁻¹⁰ = (x)(x) / (0.05 - x)

Since Ka is very small (1.0 × 10⁻¹⁰), we can assume that x is much smaller than 0.05, so (0.05 - x) ≈ 0.05.
1.0 × 10⁻¹⁰ = x² / 0.05
x² = 1.0 × 10⁻¹⁰ × 0.05
x² = 5.0 × 10⁻¹²
x = √(5.0 × 10⁻¹²) = 2.236 × 10⁻⁶ M

Therefore, the concentration of phenolate ion, [C₆H₅O⁻] = x = 2.236 × 10⁻⁶ M.

**Part 2: Degree of ionization if the solution is also 0.01 M in sodium phenolate**
Sodium phenolate (C₆H₅ONa) is a strong electrolyte and dissociates completely to provide C₆H₅O⁻ ions.
C₆H₅ONa(aq) → Na⁺(aq) + C₆H₅O⁻(aq)
Initial [C₆H₅O⁻] from sodium phenolate = 0.01 M

Now, consider the dissociation of phenol in the presence of 0.01 M C₆H₅O⁻ (common ion effect):
C₆H₅OH(aq) ⇌ H⁺(aq) + C₆H₅O⁻(aq)

Initial:      0.05 M          0           0.01 M
Change:      -y             +y          +y
Equilibrium: (0.05 - y) M     y M         (0.01 + y) M

Ka = ([H⁺][C₆H₅O⁻]) / [C₆H₅OH]
1.0 × 10⁻¹⁰ = (y)(0.01 + y) / (0.05 - y)

Since Ka is very small and there is a common ion, y will be even smaller. We can make the approximations:
(0.05 - y) ≈ 0.05
(0.01 + y) ≈ 0.01

1.0 × 10⁻¹⁰ = (y)(0.01) / (0.05)
1.0 × 10⁻¹⁰ = y / 5
y = 5.0 × 10⁻¹⁰ M

Here, y represents [H⁺] at equilibrium, and also the concentration of phenol that has dissociated.

Degree of ionization (α) = (concentration of ionized phenol) / (initial concentration of phenol)
α = y / 0.05
α = (5.0 × 10⁻¹⁰) / 0.05
α = 1.0 × 10⁻⁸

Therefore, the concentration of phenolate ion in 0.05 M phenol solution is 2.24 × 10⁻⁶ M. If the solution is also 0.01 M in sodium phenolate, the degree of ionization of phenol is 1.0 × 10⁻⁸.

Question 36

Q36. Calculate the pH of a 0.1 M ammonia solution. Calculate the pH after 50.0 mL of this solution is treated with 25.0 mL of 0.1 M HCl. The dissociation constant of ammonia, K_b = 1.77

Accepted Answer

The pH of 0.1 M ammonia solution is 11.12.
The pH of the solution after treating 50.0 mL of 0.1 M ammonia with 25.0 mL of 0.1 M HCl is 9.25.

Question 37

Q37. The K_a for acetic acid is 1.74 × 10⁻⁵. Calculate the pH of 0.05 M acetic acid. Calculate the concentration of acetate ion in the solution. Calculate the concentration of acetate i

Accepted Answer

The pH of 0.05 M acetic acid is 3.03.
The concentration of acetate ion in 0.05 M acetic acid is 9.33 × 10⁻⁴ M.
The concentration of acetate ion in a mixture with 0.1 M HCl is 8.7 × 10⁻⁶ M.

Question 38

Q38. Ionic product of water at 310 K is 2.7 × 10⁻¹⁴. What is the pH of neutral water at this temperature?

Accepted Answer

The pH of neutral water at 310 K is 6.78.

Question 39

Q39. The pH of 0.1 M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.

Accepted Answer

The ionization constant (Ka) of cyanic acid (HCNO) is 2.0 × 10⁻⁴ and its degree of ionization (α) in 0.1 M solution is 0.0457.

Question 40

Q40. The ionization constant of phenol is 1.0 × 10⁻¹⁰. What is the concentration of phenolate ion in 0.05 M solution of phenol? What will be its degree of ionization if the solution is

Accepted Answer

In 0.05 M solution of pure phenol, the concentration of phenolate ion is 2.24 × 10⁻⁶ M.
If the solution is also 0.01 M with sodium phenolate, the degree of ionization of phenol is 1.0 × 10⁻⁸, and the pH of the solution is 9.0.

Question 41

Q41. The first ionization constant of H₂S is 9.1 × 10⁻⁸. Calculate the percentage of H₂S dissociated in 0.1 M solution. How much will the degree of dissociation change if the solution i

Accepted Answer

In 0.1 M H₂S solution:
Percentage of H₂S dissociated = 0.0954%
Concentration of S²⁻ = 1.2 × 10⁻¹³ M

In 0.1 M H₂S solution also 0.1 M in HCl:
Degree of dissociation of H₂S = 9.1 × 10⁻⁸
Concentration of S²⁻ = 1.092 × 10⁻²⁰ M

Question 42

Q42. Calculate the pH of the following solutions:
a) 0.01 M HBr
b) 0.01 M KOH

Accepted Answer

The pH of the solutions are:
a) For 0.01 M HBr, pH = 2
b) For 0.01 M KOH, pH = 12

Question 43

Q43. The pH of 0.1 M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.

Accepted Answer

The ionization constant (Ka) of cyanic acid (HCNO) is 2.0 x 10⁻⁴ and its degree of ionization (α) in a 0.1 M solution is 0.0457.

Question 44

Q44. The ionization constant of phenol is 1.0 × 10⁻¹⁰. What is the concentration of phenolate ion in 0.05 M solution of phenol? What will be its degree of ionization in this solution? I

Accepted Answer

In a 0.05 M solution of phenol:
1.  Concentration of phenolate ion ([C₆H₅O⁻]) = 2.236 x 10⁻⁶ M
2.  Degree of ionization (α) = 4.47 x 10⁻⁵

In the presence of 0.01 M sodium phenolate:
3.  Degree of ionization (α) = 1.0 x 10⁻⁸

Question 45

Q45. Calculate the pH of the following solutions:
(a) 0.01 M HCl
(b) 0.01 M NaOH
(c) 0.01 M H₂SO₄
(d) 0.01 M Ca(OH)₂

Accepted Answer

The pH of the given solutions are calculated as follows:
(a) For 0.01 M HCl, pH = 2
(b) For 0.01 M NaOH, pH = 12
(c) For 0.01 M H₂SO₄, pH = 1.7
(d) For 0.01 M Ca(OH)₂, pH = 12.3

Question 46

Q46. The pH of 0.005 M codeine (C₁₈H₂₁NO₃) solution is 9.95. Calculate its ionization constant and pK_b.

Accepted Answer

The ionization constant (K_b) for codeine is 1.62 × 10⁻⁶ and its pK_b is 5.79.

Question 47

Q47. The pK_a of acetic acid and pK_b of ammonium hydroxide are 4.76 and 4.75 respectively. Calculate the pH of ammonium acetate solution.

Accepted Answer

The pH of ammonium acetate solution is 7.005.

Question 48

Q48. The pH of 0.1 M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.

Accepted Answer

The ionization constant (Ka) of cyanic acid (HCNO) is 2.0 x 10⁻⁴, and its degree of ionization (α) in a 0.1 M solution is 0.0457.

Question 49

Q49. The ionization constant of phenol is 1.0 × 10⁻¹⁰. What is the concentration of phenolate ion in 0.05 M solution of phenol? What will be its degree of ionization in this solution? I

Accepted Answer

For a 0.05 M solution of phenol:
1.  Concentration of phenolate ion ([C₆H₅O⁻]) = 2.236 × 10⁻⁶ M
2.  Degree of ionization (α) = 4.472 × 10⁻⁵

In the presence of 0.01 M sodium phenolate:
3.  Degree of ionization (α) = 1.0 × 10⁻⁸

Question 50

Q50. The solubility product constant of Ag2CrO4 and AgBr are 1.1 × 10–12 and 5.0 × 10–13 respectively. Calculate the ratio of the solubilities of Ag2CrO4 to AgBr in pure water.

Accepted Answer

The ratio of the solubilities of Ag2CrO4 to AgBr in pure water is approximately 2.2.

Question 51

Q51. The pH of 0.1 M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.

Accepted Answer

The ionization constant (Ka) of cyanic acid is approximately 3.8 × 10^-4, and its degree of ionization (α) in the 0.1 M solution is approximately 0.0457 or 4.57%.

Question 52

Q52. The pH of 0.1 M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.

Accepted Answer

The ionization constant (Kₐ) of cyanic acid (HCNO) is 2.0 × 10⁻⁴, and its degree of ionization (α) in a 0.1 M solution is 0.0457.

Question 53

Q53. The solubility product constant of Ag2CrO4 and AgBr are 1.1 × 10–12 and 5.0 × 10–13 respectively. Calculate the ratio of the solubilities of Ag2CrO4 to AgBr in water.

Accepted Answer

The ratio of the solubilities of Ag2CrO4 to AgBr in water is approximately 91.94.

Question 54

Q54. The pH of 0.1 M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.

Accepted Answer

The ionization constant (Ka) of cyanic acid (HCNO) is 2.0 × 10⁻⁴ and its degree of ionization (α) in 0.1 M solution is 0.0457.

Question 55

Q55. The ionization constant of phenol is 1.0 × 10⁻¹⁰. What is the concentration of phenoxide ion in 0.05 M solution of phenol? What will be its degree of ionization in the solution? If

Accepted Answer

In 0.05 M solution of phenol:
Concentration of phenoxide ion = 2.24 × 10⁻⁶ M
Degree of ionization = 4.47 × 10⁻⁵

In the presence of 0.01 M sodium phenoxide:
Degree of ionization = 1.0 × 10⁻⁸

Question 56

Q56. Calculate the pH of 0.1 M solution of ammonia (NH₃), K_b = 1.77 × 10⁻⁵. Also calculate the pH after 50.0 mL of this solution is treated with 25.0 mL of 0.1 M HCl. The dissociation

Accepted Answer

The pH of 0.1 M ammonia solution is 11.12.
After treating 50.0 mL of 0.1 M ammonia solution with 25.0 mL of 0.1 M HCl, the pH of the resulting solution is 9.25.

Question 57

Q57. The K_sp of Ag₂CrO₄, AgCl, AgBr and AgI are respectively 1.1 × 10⁻¹², 1.8 × 10⁻¹⁰, 5.0 × 10⁻¹³ and 8.3 × 10⁻¹⁷. Which one of the above salts will precipitate last if AgNO₃ solution

Accepted Answer

To determine which salt precipitates last, we need to find the concentration of Ag⁺ ions required to initiate the precipitation of each salt. The salt requiring the highest [Ag⁺] will precipitate last.

Given K_sp values:
Ag₂CrO₄: K_sp = 1.1 × 10⁻¹²
AgCl: K_sp = 1.8 × 10⁻¹⁰
AgBr: K_sp = 5.0 × 10⁻¹³
AgI: K_sp = 8.3 × 10⁻¹⁷

Let the initial concentration of each anion be 'C' (since equal moles are present in the same volume, their concentrations will be equal). For simplicity, we can assume C = 1 M for comparison, or just use 'C' and see how [Ag⁺] depends on K_sp and C.

1.  **For AgCl:**
    AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)
    K_sp = [Ag⁺][Cl⁻]
    [Ag⁺] = K_sp / [Cl⁻] = 1.8 × 10⁻¹⁰ / C

2.  **For AgBr:**
    AgBr(s) ⇌ Ag⁺(aq) + Br⁻(aq)
    K_sp = [Ag⁺][Br⁻]
    [Ag⁺] = K_sp / [Br⁻] = 5.0 × 10⁻¹³ / C

3.  **For AgI:**
    AgI(s) ⇌ Ag⁺(aq) + I⁻(aq)
    K_sp = [Ag⁺][I⁻]
    [Ag⁺] = K_sp / [I⁻] = 8.3 × 10⁻¹⁷ / C

4.  **For Ag₂CrO₄:**
    Ag₂CrO₄(s) ⇌ 2Ag⁺(aq) + CrO₄²⁻(aq)
    K_sp = [Ag⁺]²[CrO₄²⁻]
    [Ag⁺]² = K_sp / [CrO₄²⁻]
    [Ag⁺] = √(K_sp / [CrO₄²⁻]) = √(1.1 × 10⁻¹² / C)

Now, let's calculate the required [Ag⁺] for each salt to start precipitating, assuming C = 1 M for comparison (the relative order will remain the same regardless of C, as long as C is the same for all anions):

1.  **For AgCl:** [Ag⁺] = 1.8 × 10⁻¹⁰ / 1 = 1.8 × 10⁻¹⁰ M
2.  **For AgBr:** [Ag⁺] = 5.0 × 10⁻¹³ / 1 = 5.0 × 10⁻¹³ M
3.  **For AgI:** [Ag⁺] = 8.3 × 10⁻¹⁷ / 1 = 8.3 × 10⁻¹⁷ M
4.  **For Ag₂CrO₄:** [Ag⁺] = √(1.1 × 10⁻¹² / 1) = √(110 × 10⁻¹⁴) ≈ 10.49 × 10⁻⁷ = 1.049 × 10⁻⁶ M

Comparing the required [Ag⁺] values:
AgI: 8.3 × 10⁻¹⁷ M (lowest [Ag⁺] required, precipitates first)
AgBr: 5.0 × 10⁻¹³ M
AgCl: 1.8 × 10⁻¹⁰ M
Ag₂CrO₄: 1.049 × 10⁻⁶ M (highest [Ag⁺] required, precipitates last)

Therefore, **Ag₂CrO₄** will precipitate last.

Question 58

Q58. The solubility product constant of Ag2CrO4 and AgBr are 1.1 × 10–12 and 5.0 × 10–13 respectively. Calculate the ratio of the molarities of their saturated solutions.

Accepted Answer

The ratio of the molarities of the saturated solutions of Ag2CrO4 and AgBr is approximately 1.05.

Question 59

Q59. The pH of 0.1 M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.

Accepted Answer

The ionization constant (Ka) of cyanic acid is 2.19 x 10⁻⁴ and its degree of ionization (α) in the 0.1 M solution is 0.0457.

Question 60

Q60. Calculate the solubility of AgCl in 0.1 M AgNO₃ solution. Ksp for AgCl is 1.8 × 10⁻¹⁰.

Accepted Answer

The solubility of AgCl in 0.1 M AgNO₃ solution is 1.8 × 10⁻⁹ M. This is calculated using the concept of the common ion effect, where the presence of Ag⁺ ions from AgNO₃ suppresses the dissolution of AgCl.

The equilibrium for the dissolution of AgCl is:
AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)

The solubility product constant expression is:
Ksp = [Ag⁺][Cl⁻]

Let 's' be the molar solubility of AgCl in the 0.1 M AgNO₃ solution. The dissolution of 's' moles of AgCl will produce 's' moles of Ag⁺ and 's' moles of Cl⁻.

However, the solution already contains 0.1 M Ag⁺ from the complete dissociation of AgNO₃. Therefore, the total concentration of Ag⁺ at equilibrium will be (0.1 + s) M.

At equilibrium:
[Ag⁺] = (0.1 + s) M
[Cl⁻] = s M

Substituting these values into the Ksp expression:
Ksp = (0.1 + s)(s)
1.8 × 10⁻¹⁰ = (0.1 + s)(s)

Since Ksp is very small, the solubility 's' will be much smaller than 0.1 M. Thus, we can make the approximation:
(0.1 + s) ≈ 0.1

So, the expression simplifies to:
1.8 × 10⁻¹⁰ ≈ (0.1)(s)
s = (1.8 × 10⁻¹⁰) / 0.1
s = 1.8 × 10⁻⁹ M

Thus, the solubility of AgCl in 0.1 M AgNO₃ is 1.8 × 10⁻⁹ mol/L.

Question 61

Q61. Predict if a precipitate will form when 50 mL of 0.01 M BaCl₂ solution is mixed with 50 mL of 0.002 M Na₂SO₄ solution. Ksp for BaSO₄ is 1.1 × 10⁻¹⁰.

Accepted Answer

Yes, a precipitate of BaSO₄ will form.

To determine if a precipitate will form, we must calculate the ionic product (Qsp) for BaSO₄ in the mixed solution and compare it with its solubility product (Ksp).

The reaction for potential precipitation is:
Ba²⁺(aq) + SO₄²⁻(aq) ⇌ BaSO₄(s)

First, we calculate the concentrations of Ba²⁺ and SO₄²⁻ ions in the final mixture.
When 50 mL of BaCl₂ is mixed with 50 mL of Na₂SO₄, the total volume becomes 50 + 50 = 100 mL. The concentration of each ion is halved due to dilution.

New concentration of Ba²⁺:
[Ba²⁺] = (Initial Molarity × Initial Volume) / Total Volume
[Ba²⁺] = (0.01 M × 50 mL) / 100 mL = 0.005 M = 5 × 10⁻³ M

New concentration of SO₄²⁻:
[SO₄²⁻] = (Initial Molarity × Initial Volume) / Total Volume
[SO₄²⁻] = (0.002 M × 50 mL) / 100 mL = 0.001 M = 1 × 10⁻³ M

Next, we calculate the ionic product, Qsp:
Qsp = [Ba²⁺][SO₄²⁻]
Qsp = (5 × 10⁻³ M) × (1 × 10⁻³ M)
Qsp = 5 × 10⁻⁶

Finally, we compare Qsp with Ksp:
Given Ksp for BaSO₄ = 1.1 × 10⁻¹⁰.
Qsp = 5 × 10⁻⁶

Since Qsp (5 × 10⁻⁶) > Ksp (1.1 × 10⁻¹⁰), the solution is supersaturated, and a precipitate of BaSO₄ will form until the ionic product equals the solubility product.

Chapter 7: Equilibrium