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?

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, $$[{\text{H}}^{+}]=10^{-3},\text{M}$$$$\text{H}^+$$ = 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:

$$\boxed{9}$$