If `K_(1)` is ionization constant of `H_(2)S(aq)hArr2H^(+)(aq)+S^(2-)(aq) and K_(2)` is that for `H_(2)S(aq)hArr H^(+)(aq)+HS^(-)(aq)`, then ionization constant of `HS^(-)(aq)hArr H^(+)(aq)+S^(2-)(aq)` will be equal to .......... .

Answer

To solve the problem, we need to analyze the ionization constants of the given reactions involving hydrogen sulfide (H₂S) and its ions.

### Step-by-Step Solution:

1. **Identify the Reactions and Their Ionization Constants**:
- The first reaction is:
\[ H_2S(aq) \rightleftharpoons 2H^+(aq) + S^{2-}(aq) \quad (K_1) \]
- The second reaction is:
\[ H_2S(aq) \rightleftharpoons H^+(aq) + HS^-(aq) \quad (K_2) \]

2. **Write the Third Reaction**:
- We need to find the ionization constant for the reaction:
\[ HS^-(aq) \rightleftharpoons H^+(aq) + S^{2-}(aq) \quad (K_3) \]

3. **Relate the Reactions**:
- To find \( K_3 \), we can manipulate the first two reactions. We can express \( K_3 \) in terms of \( K_1 \) and \( K_2 \).
- From the second reaction, we can express \( H_2S \) in terms of \( HS^- \):
\[ H_2S \rightleftharpoons H^+ + HS^- \]
- Rearranging gives us:
\[ HS^- \rightleftharpoons H^+ + H_2S \]

4. **Combine the Reactions**:
- If we subtract the second reaction from the first, we can eliminate \( H_2S \):
\[ (H_2S \rightleftharpoons 2H^+ + S^{2-}) - (H_2S \rightleftharpoons H^+ + HS^-) \]
- This results in:
\[ HS^- \rightleftharpoons H^+ + S^{2-} \]

5. **Express \( K_3 \) in Terms of \( K_1 \) and \( K_2 \)**:
- The equilibrium constant for the third reaction can be expressed as:
\[ K_3 = \frac{[H^+][S^{2-}]}{[HS^-]} \]
- From the first reaction, we have:
\[ K_1 = \frac{[H^+]^2[S^{2-}]}{[H_2S]} \]
- From the second reaction:
\[ K_2 = \frac{[H^+][HS^-]}{[H_2S]} \]
- Rearranging \( K_2 \) gives:
\[ [H_2S] = \frac{[H^+][HS^-]}{K_2} \]

6. **Substituting into \( K_1 \)**:
- Substitute \( [H_2S] \) into the expression for \( K_1 \):
\[ K_1 = \frac{[H^+]^2[S^{2-}]}{\frac{[H^+][HS^-]}{K_2}} \]
- This simplifies to:
\[ K_1 = \frac{[H^+][S^{2-}]K_2}{[HS^-]} \]

7. **Final Expression for \( K_3 \)**:
- Rearranging gives us:
\[ K_3 = \frac{K_1}{K_2} \]

### Conclusion:
The ionization constant \( K_3 \) for the reaction \( HS^-(aq) \rightleftharpoons H^+(aq) + S^{2-}(aq) \) is:
\[ K_3 = \frac{K_1}{K_2} \]