If a saturated solution prepared by dissolving `Ag_(2)CO_(3)` in water has `[Ag^(+)] = 2.5 xx 10^(-)`. What is the value of `K_(sp)` for `Ag_(2)CO_(3)` ?

Options

A. `78.1 xx 10^(-12)`
B. `7.81 xx 10^(-12)`
C. `93.8 xx 10^(-12)`
D. `9.38 xx 10^(-12)`

Answer

To find the solubility product constant \( K_{sp} \) for \( Ag_2CO_3 \), we can follow these steps:

### Step 1: Write the dissociation equation
The dissociation of silver carbonate (\( Ag_2CO_3 \)) in water can be represented as:
\[ Ag_2CO_3 (s) \rightleftharpoons 2 Ag^+ (aq) + CO_3^{2-} (aq) \]

### Step 2: Identify the concentrations of ions
From the problem, we know that the concentration of \( Ag^+ \) ions is given as:
\[ [Ag^+] = 2.5 \times 10^{-4} \, M \]
Since each formula unit of \( Ag_2CO_3 \) produces 2 moles of \( Ag^+ \), the concentration of \( CO_3^{2-} \) ions will be half of that:
\[ [CO_3^{2-}] = \frac{[Ag^+]}{2} = \frac{2.5 \times 10^{-4}}{2} = 1.25 \times 10^{-4} \, M \]

### Step 3: Write the expression for \( K_{sp} \)
The solubility product constant \( K_{sp} \) can be expressed in terms of the concentrations of the ions at equilibrium:
\[ K_{sp} = [Ag^+]^2 \cdot [CO_3^{2-}] \]

### Step 4: Substitute the concentrations into the \( K_{sp} \) expression
Now, substituting the values we found:
\[ K_{sp} = (2.5 \times 10^{-4})^2 \cdot (1.25 \times 10^{-4}) \]

### Step 5: Calculate \( K_{sp} \)
Calculating \( (2.5 \times 10^{-4})^2 \):
\[ (2.5 \times 10^{-4})^2 = 6.25 \times 10^{-8} \]
Now, substituting this back into the \( K_{sp} \) expression:
\[ K_{sp} = 6.25 \times 10^{-8} \cdot 1.25 \times 10^{-4} \]
Calculating this gives:
\[ K_{sp} = 7.8125 \times 10^{-12} \]

### Step 6: Round the answer
Rounding this to three significant figures, we get:
\[ K_{sp} \approx 7.81 \times 10^{-12} \]

### Final Answer
The value of \( K_{sp} \) for \( Ag_2CO_3 \) is:
\[ K_{sp} = 7.81 \times 10^{-12} \]

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