Solubilities of `Ni(OH)_(2)` and `AgCN` are `S_(1)` and `S_(2)`. If `K_(sp)[Ni(OH)_(2)]=2xx10^(-15)`
`K_(sp)[AgCN]=6xx10^(-17)`, then

Options

A. `S_(1)gtS_(2)`
B. `S_(1)ltS_(2)`
C. `S_(1)=S_(2)`
D. Data is insufficient

Explanation

To solve the problem, we need to calculate the solubilities \( S_1 \) and \( S_2 \) for \( Ni(OH)_2 \) and \( AgCN \) respectively, using their solubility product constants \( K_{sp} \).

### Step 1: Calculate \( S_1 \) for \( Ni(OH)_2 \)

1. **Write the dissociation equation:**
\[ Ni(OH)_2 (s) \rightleftharpoons Ni^{2+} (aq) + 2OH^{-} (aq) \]

2. **Define solubility:**
Let the solubility of \( Ni(OH)_2 \) be \( S_1 \). Then, at equilibrium:
- \( [Ni^{2+}] = S_1 \)
- \( [OH^{-}] = 2S_1 \)

3. **Write the expression for \( K_{sp} \):**
\[ K_{sp} = [Ni^{2+}][OH^{-}]^2 = S_1 (2S_1)^2 = S_1 \cdot 4S_1^2 = 4S_1^3 \]

4. **Substitute the given \( K_{sp} \):**
\[ 4S_1^3 = 2 \times 10^{-15} \]

5. **Solve for \( S_1 \):**
\[ S_1^3 = \frac{2 \times 10^{-15}}{4} = 0.5 \times 10^{-15} = 5 \times 10^{-16} \]
\[ S_1 = (5 \times 10^{-16})^{1/3} \approx 0.79 \times 10^{-5} \text{ M} \]

### Step 2: Calculate \( S_2 \) for \( AgCN \)

1. **Write the dissociation equation:**
\[ AgCN (s) \rightleftharpoons Ag^{+} (aq) + CN^{-} (aq) \]

2. **Define solubility:**
Let the solubility of \( AgCN \) be \( S_2 \). Then, at equilibrium:
- \( [Ag^{+}] = S_2 \)
- \( [CN^{-}] = S_2 \)

3. **Write the expression for \( K_{sp} \):**
\[ K_{sp} = [Ag^{+}][CN^{-}] = S_2 \cdot S_2 = S_2^2 \]

4. **Substitute the given \( K_{sp} \):**
\[ S_2^2 = 6 \times 10^{-17} \]

5. **Solve for \( S_2 \):**
\[ S_2 = (6 \times 10^{-17})^{1/2} \approx 7.75 \times 10^{-9} \text{ M} \]

### Step 3: Compare \( S_1 \) and \( S_2 \)

Now we have:
- \( S_1 \approx 0.79 \times 10^{-5} \)
- \( S_2 \approx 7.75 \times 10^{-9} \)

Since \( 0.79 \times 10^{-5} > 7.75 \times 10^{-9} \), we can conclude that:
\[ S_1 > S_2 \]

### Final Answer:
The solubility of \( Ni(OH)_2 \) is greater than that of \( AgCN \), i.e., \( S_1 > S_2 \).

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