A lead salts is dissolved in `HC1` which si `94%` ionised. It is found to have `0.1M Pb^(2+)` and `0.28M H^(o+)` ions. The solution is satured with `H_(2)S(g)`. Calculate the amount of `Pb^(2+)` ions that remains unprecipitated.
`K_(sp)` of `PbS = 4 xx 10^(-29)`,
`K_(sp)` of `H_(2)S = 1.1 xx 10^(-22)`

Explanation

To solve the problem, we need to calculate the amount of \( \text{Pb}^{2+} \) ions that remains unprecipitated when the solution is saturated with \( \text{H}_2\text{S} \). We will follow these steps:

### Step 1: Determine the concentration of \( \text{H}^+ \) ions
Given that the concentration of \( \text{H}^+ \) ions is \( 0.28 \, \text{M} \) and that \( \text{HCl} \) is \( 94\% \) ionized, we can calculate the effective concentration of \( \text{H}^+ \) ions.

\[ \text{Concentration of } \text{H}^+ = 0.28 \times \frac{94}{100} = 0.2632 \, \text{M} \]

### Step 2: Calculate the concentration of \( \text{S}^{2-} \) ions
Using the \( K_{sp} \) expression for \( \text{H}_2\text{S} \):

\[ K_{sp} = [\text{H}^+]^2 [\text{S}^{2-}] \]

We know \( K_{sp} \) for \( \text{H}_2\text{S} \) is \( 1.1 \times 10^{-22} \). Rearranging the equation to find \( [\text{S}^{2-}] \):

\[ [\text{S}^{2-}] = \frac{K_{sp}}{[\text{H}^+]^2} \]

Substituting the values:

\[ [\text{S}^{2-}] = \frac{1.1 \times 10^{-22}}{(0.2632)^2} = \frac{1.1 \times 10^{-22}}{0.0695} \approx 1.59 \times 10^{-21} \, \text{M} \]

### Step 3: Write the \( K_{sp} \) expression for \( \text{PbS} \)
The dissociation of \( \text{PbS} \) can be represented as:

\[ \text{PbS} \rightleftharpoons \text{Pb}^{2+} + \text{S}^{2-} \]

The \( K_{sp} \) expression is:

\[ K_{sp} = [\text{Pb}^{2+}][\text{S}^{2-}] \]

Given \( K_{sp} \) for \( \text{PbS} \) is \( 4 \times 10^{-29} \).

### Step 4: Calculate the concentration of \( \text{Pb}^{2+} \) ions
Rearranging the \( K_{sp} \) expression to find \( [\text{Pb}^{2+}] \):

\[ [\text{Pb}^{2+}] = \frac{K_{sp}}{[\text{S}^{2-}]} \]

Substituting the values:

\[ [\text{Pb}^{2+}] = \frac{4 \times 10^{-29}}{1.59 \times 10^{-21}} \approx 2.52 \times 10^{-8} \, \text{M} \]

### Step 5: Determine the amount of \( \text{Pb}^{2+} \) that remains unprecipitated
Initially, the concentration of \( \text{Pb}^{2+} \) ions was \( 0.1 \, \text{M} \). After saturation with \( \text{H}_2\text{S} \), the concentration of \( \text{Pb}^{2+} \) will be reduced to \( 2.52 \times 10^{-8} \, \text{M} \).

Thus, the amount of \( \text{Pb}^{2+} \) that remains unprecipitated is:

\[ \text{Unprecipitated } \text{Pb}^{2+} = 0.1 \, \text{M} - 2.52 \times 10^{-8} \, \text{M} \approx 0.1 \, \text{M} \]

### Final Answer:
The amount of \( \text{Pb}^{2+} \) ions that remains unprecipitated is approximately \( 0.1 \, \text{M} \).

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