What is the `pH`of `0.01M H_(2)S` solution ? Also determine `[HS^(-)]` & `[S^(2-)]` Given: For `H_(2)S ,K_(a_(1))=9xx10^(-8),K_(a_(2))=1.2xx10^(-13)`

Explanation

To find the pH of a 0.01 M solution of \( H_2S \) and to determine the concentrations of \( [HS^-] \) and \( [S^{2-}] \), we will follow these steps:

### Step 1: Write the dissociation reactions
The dissociation of \( H_2S \) can be represented as follows:
1. \( H_2S \rightleftharpoons H^+ + HS^- \) (with \( K_{a1} = 9 \times 10^{-8} \))
2. \( HS^- \rightleftharpoons H^+ + S^{2-} \) (with \( K_{a2} = 1.2 \times 10^{-13} \))

### Step 2: Set up the equilibrium expressions
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