Concept:Note: The question wording is slightly flawed. Based on the numerical options, it implies that \( 4.0 \times 10^{-28} \) is actually the Solubility Product (\( K_{sp} \)), and it is asking for the molar solubility (\( s \)), which equals the ionic concentration.Formula:$$ \text{PbS}_{(s)} \rightleftharpoons \text{Pb}^{2+} + \text{S}^{2-} \implies K_{sp} = (s)(s) = s^2 $$
Solution:- Assuming \( K_{sp} = 4.0 \times 10^{-28} \).
- Set up the equation: \( s^2 = 4.0 \times 10^{-28} \).
- Take the square root of both sides to find solubility (\( s \)): \( s = \sqrt{4.0 \times 10^{-28}} \).
- The square root of 4.0 is 2. The square root of \( 10^{-28} \) (divide exponent by 2) is \( 10^{-14} \).
- Therefore, the molar solubility (and thus individual ionic concentration) is \( 2 \times 10^{-14} \text{ M} \).
Why other options are incorrect:Option A forgot to square root the base number 4. Options C and D are mathematically incorrect square roots.
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