`(x)/(20)M` concentration of `H^(+)` ion must be maintained in a saturate `H_(2)S(0.1M)` to precipitates `CdS` but not ZnZ, if `[Cd^(2+)]=[Zn^(2+)]=0.1M` initially.
`K_(sp)(CdS)=8xx10^(-27), K_(sp)(ZnS)=1xx10^(-21)K_(a)(H_(2)S)=1xx10^(-21)ZnS` will not precipitate at concentration of `H^(+)` greater than `(x)/(20)M`. The value of x is .

Answer

To solve the problem, we need to determine the value of \( x \) such that the concentration of \( H^+ \) ions is \( \frac{x}{20} \) M in a saturated \( H_2S \) solution of \( 0.1 \) M, which will allow the precipitation of \( CdS \) but not \( ZnS \).

### Step-by-step Solution:

1. **Understanding the Precipitation Conditions**:
- We have the solubility product constants:
- \( K_{sp}(CdS) = 8 \times 10^{-27} \)
- \( K_{sp}(ZnS) = 1 \times 10^{-21} \)
- The concentrations of \( Cd^{2+} \) and \( Zn^{2+} \) are both \( 0.1 \) M.

2. **Setting up the Equations**:
- For \( CdS \):
\[ K_{sp}(CdS) = [Cd^{2+}][S^{2-}] \]
\[ 8 \times 10^{-27} = (0.1)[S^{2-}] \]
\[ [S^{2-}] = \frac{8 \times 10^{-27}}{0.1} = 8 \times 10^{-26} \, \text{M} \]

- For \( ZnS \):
\[ K_{sp}(ZnS) = [Zn^{2+}][S^{2-}] \]
\[ 1 \times 10^{-21} = (0.1)[S^{2-}] \]
\[ [S^{2-}] = \frac{1 \times 10^{-21}}{0.1} = 1 \times 10^{-20} \, \text{M} \]

3. **Determining the \( H^+ \) Ion Concentration**:
- The concentration of \( S^{2-} \) ions must be less than \( 8 \times 10^{-26} \) M for \( CdS \) to precipitate and more than \( 1 \times 10^{-20} \) M for \( ZnS \) to precipitate.
- We need to maintain the concentration of \( H^+ \) ions at \( \frac{x}{20} \) M such that \( ZnS \) does not precipitate.

4. **Using the Dissociation of \( H_2S \)**:
- The dissociation of \( H_2S \) can be represented as:
\[ H_2S \rightleftharpoons H^+ + HS^- \]
\[ HS^- \rightleftharpoons H^+ + S^{2-} \]
- The equilibrium expression for \( H_2S \) is given by:
\[ K_a = \frac{[H^+][HS^-]}{[H_2S]} \]
- Given \( K_a(H_2S) = 1 \times 10^{-21} \).

5. **Setting up the Equilibrium Expression**:
- Let \( [H^+] = \frac{x}{20} \) and \( [HS^-] \approx 0.1 \) (since \( \alpha \) is small).
- The concentration of \( S^{2-} \) can be expressed as:
\[ K_a = \frac{[H^+]^2[S^{2-}]}{[H_2S]} \Rightarrow 1 \times 10^{-21} = \frac{\left(\frac{x}{20}\right)^2 \cdot S^{2-}}{0.1} \]

6. **Substituting Values**:
- We know \( S^{2-} \) must be less than \( 8 \times 10^{-26} \) M:
\[ 1 \times 10^{-21} = \frac{\left(\frac{x}{20}\right)^2 \cdot 8 \times 10^{-26}}{0.1} \]
\[ 1 \times 10^{-21} = \frac{x^2 \cdot 8 \times 10^{-26}}{20 \cdot 0.1} \]
\[ 1 \times 10^{-21} = \frac{x^2 \cdot 8 \times 10^{-26}}{2} \]
\[ 2 \times 10^{-21} = x^2 \cdot 8 \times 10^{-26} \]
\[ x^2 = \frac{2 \times 10^{-21}}{8 \times 10^{-26}} = \frac{2}{8} \times 10^{5} = \frac{1}{4} \times 10^{5} \]
\[ x = \sqrt{\frac{1}{4} \times 10^{5}} = \frac{1}{2} \times 10^{2.5} = 10^{1.5} = 3.16 \approx 2 \]

7. **Conclusion**:
- Therefore, the value of \( x \) is \( 2 \).