A saturated 0.01 M H2S solution is buffered at pH 3 with exactly sufficient Pb(NO3)2 to not precipitate PbS. Ka of H2S = 10^23, Ksp of PbS = 10^28. The concentration of Pb^+2 in the solution is 10^x. What is x?
Question
A saturated 0.01 M H2S solution is buffered at pH 3 with exactly sufficient Pb(NO3)2 to not precipitate PbS. Ka of H2S = 10^23, Ksp of PbS = 10^28. The concentration of Pb^+2 in the solution is 10^x. What is x?
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Answer
To find the value of ( x ) in this problem, we need to calculate the concentration of ( \text{Pb}^{2+} ) in a buffered ( \text{H}_2\text{S} ) solution.
Given data:
 pH = 3
 ( [\text{H}_2\text{S}] = 0.01 , M )
 ( K_a ) of ( \text{H}_2\text{S} = 10^{23} )
 ( K_{sp} ) of ( \text{PbS} = 10^{28} )
Steps to solve:

Determine the ([ \text{H}^+ ]) concentration:
 Since pH = 3, $$ [ \text{H}^+ ] = 10^{3} , \text{M} $$

Use the dissociation equation of ( \text{H}_2\text{S} ):
 The dissociation of ( \text{H}_2\text{S} ) is given by: $$ \text{H}_2\text{S} \rightleftharpoons 2\text{H}^+ + \text{S}^{2} $$

Apply the acid dissociation constant ( K_a ):
 The expression for ( K_a ) is: $$ K_a = \frac{[\text{H}^+]^2 [\text{S}^{2}]}{[\text{H}_2\text{S}]} $$
 Plugging in the values: $$ 10^{23} = \frac{(10^{3})^2 [\text{S}^{2}]}{0.01} $$
 Simplify: $$ 10^{23} = \frac{10^{6} [\text{S}^{2}]}{10^{2}} $$
 Solve for ([ \text{S}^{2} ]): $$ 10^{23} = 10^{4} [\text{S}^{2}] $$ $$ [\text{S}^{2}] = 10^{19} , \text{M} $$

Consider the solubility product ( K_{sp} ) of ( \text{PbS} ):
 The dissociation of ( \text{PbS} ) is: $$ \text{PbS} \rightleftharpoons \text{Pb}^{2+} + \text{S}^{2} $$
 The expression for ( K_{sp} ) is: $$ K_{sp} = [\text{Pb}^{2+}] [\text{S}^{2}] $$
 Given: $$ 10^{28} = [\text{Pb}^{2+}] (10^{19}) $$
 Solve for ([ \text{Pb}^{2+} ]): $$ [\text{Pb}^{2+}] = \frac{10^{28}}{10^{19}} = 10^{9} , \text{M} $$
Therefore, ( \text{Pb}^{2+} ) concentration is ( 10^{9} , \text{M} ), and the value of ( x ) is:
$$ \boxed{9} $$
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