Calculate the maximum possible concentration of `Mn^(2+)` in water that is saturated with `H_(2)S` (which is `0.1M` at `300K`) and maintained at `pH = 3` with `HCI`. The equilibrium constant (s) for dissociation of `H_(s)S` are:
`H_(2)S hArr H^(oplus)+HS^(Theta),K_(1)=9 xx 10^(-8)`
`HS^(Theta) hArr H^(oplus)+S^(2-),K_(2)=1 xx 10^(-12)` and
`K_(sp)` of `MnS = 3 xx 10^(-22)`

Explanation

To solve the problem of calculating the maximum possible concentration of \( \text{Mn}^{2+} \) in water saturated with \( \text{H}_2\text{S} \) at a pH of 3, we can follow these steps:

### Step 1: Determine the concentration of \( \text{H}^+ \) ions
Given that the pH of the solution is 3, we can calculate the concentration of \( \text{H}^+ \) ions using the formula:
\[ [\text{H}^+] = 10^{-\text{pH}} = 10^{-3} \, \text{M} \]

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