The solubility product of `CuS, Ag_2S and HgS` are `10^(-31),10^(-44),10^(-54)` respectively. The solubilities of these sulphides are in the order

Options

A. `HgS gt Ag_2S gt CuS`
B. `CuS gt Ag_2S gt HgS`
C. `Ag_2S gt CuS gt HgS`
D. `AgS gt HgS gt CuS`

Answer

To determine the solubility of the sulfides \( CuS \), \( Ag_2S \), and \( HgS \) based on their solubility products (\( K_{sp} \)), we will follow these steps:

### Step 1: Write the dissociation equations and expressions for \( K_{sp} \)

1. **For \( CuS \)**:
\[ CuS \rightleftharpoons Cu^{2+} + S^{2-} \]
The solubility product expression is:
\[ K_{sp} = [Cu^{2+}][S^{2-}] = S \cdot S = S^2 \]
Given \( K_{sp} = 10^{-31} \):
\[ S^2 = 10^{-31} \implies S = \sqrt{10^{-31}} = 10^{-15.5} \approx 3.16 \times 10^{-16} \]

2. **For \( Ag_2S \)**:
\[ Ag_2S \rightleftharpoons 2Ag^{+} + S^{2-} \]
The solubility product expression is:
\[ K_{sp} = [Ag^{+}]^2[S^{2-}] = (2S)^2 \cdot S = 4S^3 \]
Given \( K_{sp} = 10^{-44} \):
\[ 4S^3 = 10^{-44} \implies S^3 = \frac{10^{-44}}{4} = 2.5 \times 10^{-45} \]
Therefore:
\[ S = \left(2.5 \times 10^{-45}\right)^{1/3} \approx 1.4 \times 10^{-15} \]

3. **For \( HgS \)**:
\[ HgS \rightleftharpoons Hg^{2+} + S^{2-} \]
The solubility product expression is:
\[ K_{sp} = [Hg^{2+}][S^{2-}] = S \cdot S = S^2 \]
Given \( K_{sp} = 10^{-54} \):
\[ S^2 = 10^{-54} \implies S = \sqrt{10^{-54}} = 10^{-27} \]

### Step 2: Compare the solubilities

Now we have the solubilities:
- \( CuS: S \approx 3.16 \times 10^{-16} \)
- \( Ag_2S: S \approx 1.4 \times 10^{-15} \)
- \( HgS: S = 10^{-27} \)

### Step 3: Order the solubilities

To order the solubilities from highest to lowest:
- \( Ag_2S \) has the highest solubility (\( 1.4 \times 10^{-15} \))
- \( CuS \) is next (\( 3.16 \times 10^{-16} \))
- \( HgS \) has the lowest solubility (\( 10^{-27} \))

Thus, the order of solubility is:
\[ Ag_2S > CuS > HgS \]