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:

  1. Determine the ([ \text{H}^+ ]) concentration:

    • Since pH = 3, $$ [ \text{H}^+ ] = 10^{-3} , \text{M} $$
  2. Use the dissociation equation of ( \text{H}_2\text{S} ):

    • The dissociation of ( \text{H}_2\text{S} ) is given by: $$ \text{H}_2\text{S} \rightleftharpoons 2\text{H}^+ + \text{S}^{2-} $$
  3. 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-} ]): $$ 10^{-23} = 10^{-4} [\text{S}^{2-}] $$ $$ [\text{S}^{2-}] = 10^{-19} , \text{M} $$
  4. 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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