Ionic Equilibrium

To determine the solubility of Ag₂CrO₄ in a 0.1 M AgNO₃ solution, let’s consider the solubility equilibrium and how it is affected by the common ion effect.

First, let’s establish the dissociation equation for Ag₂CrO₄ :

$Ag_2CrO_4(s) \rightleftharpoons 2Ag^+ (aq) + CrO_4^{2-} (aq)$

The solubility product constant, K_sp, is given by :

$K_{sp} = [Ag^+]^2 [CrO_4^{2-}]$

Given that K_sp = 1.1 $\times$ 10^-12.

In a 0.1 M AgNO₃ solution, the concentration of Ag⁺ ions from AgNO₃ alone is already 0.1 M. Let’s denote the additional solubility of Ag₂CrO₄ in this solution as $s$.

Therefore, the total concentration of Ag⁺ ions will be (0.1 + 2s) M, and the concentration of CrO₄^2- ions will be $s$ M.

Substituting these into the K_sp expression :

$K_{sp} = (0.1 + 2s)^2 s$

Given that K_sp is a very small number, we can assume that $2s$ is much smaller than 0.1, therefore $0.1 + 2s \approx 0.1$. This simplifies our equation to :

$K_{sp} \approx (0.1)^2 s$

Substitute $K_{sp} = 1.1 \times 10^{-12}$ into the equation :

$1.1 \times 10^{-12} \approx (0.1)^2 s$

$1.1 \times 10^{-12} \approx 0.01 s$

Solving for $s$ :

$s = \frac{1.1 \times 10^{-12}}{0.01} = 1.1 \times 10^{-10}$

Hence, the solubility of Ag₂CrO₄ in a 0.1 M AgNO₃ solution is 1.1 $\times$ 10^-10 mol/L.

The correct option is B: 1.1 $\times$ 10^-10.

Let's break down this problem step-by-step:

Understanding the Reaction:

The hydrolysis of methyl acetate (MeCOOCH₃) is catalyzed by both strong acids (like HX) and weak acids (like HA). The reaction proceeds as follows:

$ MeCOOCH_3 + H_2O \xrightarrow[or \, HA]{H^+ \, or \, HX} MeCOOH + CH_3OH $

Rate of Reaction and Acid Strength:

The rate of acid-catalyzed hydrolysis depends on the concentration of H⁺ ions.

• Strong acids ionize completely. So, a 1M solution of HX provides 1M H⁺ ions.


• Weak acids ionize partially. So, a 1M solution of HA will provide significantly less than 1M H⁺ ions.

Information from the Problem:

We are told that the initial rate of hydrolysis with HA is 1/100^th of that with HX. Since the rate is directly proportional to [H⁺], this means the [H⁺] from HA is 1/100^th of that from HX.

Calculating K_a:

1. [H⁺] from HX: Since HX is a strong acid and fully ionizes, [H⁺] = 1M.


2. [H⁺] from HA: This is 1/100^th the [H⁺] from HX, so [H⁺] = (1/100) \* 1M = 10^-2 M.


3. Setting up the K_a expression:

$ K_a = \frac{[H^+][A^-]}{[HA]} $

Since the initial concentration of HA is 1M and it ionizes to a small extent, we can approximate:

• [HA] ≈ 1 M


• [H⁺] = [A^-] = 10^-2 M

1. Calculating K_a:

$ K_a = \frac{(10^{-2})(10^{-2})}{1} = 1 \times 10^{-4} $

Answer:

Therefore, the K_a of the weak acid HA is 1 × 10^-4 (Option A).