pH of saturated solution of silver salt of monobasic acid HA is found to be 9. Find the `K_(sp)` of sparingly soluble salt Ag A(s). Given `K_(a) (HA)=10^(-10)`

Options

A. `1.1 xx 10^(-9)`
B. `1.1 xx 10^(-10)`
C. `10^(-12)`
D. None of these

Answer

To find the solubility product \( K_{sp} \) of the sparingly soluble salt \( \text{AgA}(s) \), we will follow these steps:

### Step 1: Determine the concentration of \( \text{H}^+ \) ions from the pH
Given that the pH of the saturated solution is 9, we can calculate the concentration of hydrogen ions \( [\text{H}^+] \) using the formula:

\[ [\text{H}^+] = 10^{-\text{pH}} = 10^{-9} \, \text{M} \]

### Step 2: Write the dissociation reaction of the acid \( \text{HA} \)
The dissociation of the weak acid \( \text{HA} \) can be represented as:

\[ \text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- \]

The acid dissociation constant \( K_a \) is given by:

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

### Step 3: Relate the concentrations using \( K_a \)
Given \( K_a = 10^{-10} \) and \( [\text{H}^+] = 10^{-9} \), we can substitute these values into the expression for \( K_a \):

\[ 10^{-10} = \frac{(10^{-9})[\text{A}^-]}{[\text{HA}]} \]

Rearranging gives:

\[ [\text{A}^-] = \frac{10^{-10} [\text{HA}]}{10^{-9}} = 10^{-1} [\text{HA}] \]

### Step 4: Express the concentrations of \( \text{Ag}^+ \) and \( \text{A}^- \)
The sparingly soluble salt \( \text{AgA} \) dissociates as follows:

\[ \text{AgA}(s) \rightleftharpoons \text{Ag}^+ + \text{A}^- \]

Let the solubility of \( \text{AgA} \) be \( s \). Therefore, at equilibrium:

\[ [\text{Ag}^+] = s \quad \text{and} \quad [\text{A}^-] = s + 10^{-1} [\text{HA}] \]

### Step 5: Substitute \( [\text{A}^-] \) into the \( K_{sp} \) expression
The solubility product \( K_{sp} \) is given by:

\[ K_{sp} = [\text{Ag}^+][\text{A}^-] \]

Substituting the values:

\[ K_{sp} = s \cdot (s + 10^{-1} [\text{HA}]) \]

### Step 6: Determine \( [\text{HA}] \)
Since \( [\text{HA}] \) is the equilibrium concentration of the undissociated acid, we can assume that \( [\text{HA}] \approx s \) (because the dissociation is small). Thus:

\[ K_{sp} \approx s \cdot (s + 10^{-1} s) = s \cdot (1.1s) = 1.1s^2 \]

### Step 7: Calculate \( s \) using \( K_w \)
Using the ion product of water \( K_w = 10^{-14} \):

\[ K_w = [\text{H}^+][\text{OH}^-] \]

From the concentration of \( \text{H}^+ \):

\[ [\text{OH}^-] = \frac{K_w}{[\text{H}^+]} = \frac{10^{-14}}{10^{-9}} = 10^{-5} \, \text{M} \]

### Step 8: Equate \( [\text{A}^-] \) and \( [\text{Ag}^+] \)
Since \( [\text{A}^-] \) from the acid dissociation is \( 10^{-1} s \) and \( [\text{Ag}^+] = s \):

\[ s = 10^{-5} \]

### Step 9: Calculate \( K_{sp} \)
Substituting \( s \) back into the \( K_{sp} \) equation:

\[ K_{sp} = 1.1(10^{-5})^2 = 1.1 \times 10^{-10} \]

### Final Answer
Thus, the solubility product \( K_{sp} \) of \( \text{AgA}(s) \) is:

\[ K_{sp} = 1.1 \times 10^{-10} \]