Question

If $K_{sp}\left(PbS{O}_4\right)=1.8\times {10}^{-8}$ and $K_a\left(HS{O}_4^{⊝}\right)=1.0\times {10}^{-2}$ for the following reaction, then the equilibrium constant for the reaction is
$PbS{O}_4\left(s\right)+H^{\oplus }\left(aq\right)\rightleftharpoons HS{O}_4^{⊝}\left(aq\right)+P{b}^{2+}\left(aq\right)$

• A. $1.8\times {10}^{-6}$
• B. $1.8\times {10}^{-10}$
• C. $2.8\times {10}^{-10}$
• D. $1.0\times {10}^{-2}$

Answer 1

The correct option is A $1.8\times {10}^{-6}$

$K_{sp}=\left[P{b}^{+2}\right[S{O}_4^{-2}]$

$K_{a\left(HS{O}_{\$}^{-}\right)}=\frac{\left[H^{+}\right]\left[S{O}_4^{2-}\right]}{\left[HS{O}_4^{-}\right]}$

$PbS{O}_4\left(s\right)+H^{\oplus }\left(aq\right)\rightleftharpoons HS{O}_4^{⊝}\left(aq\right)+P{b}^{2+}\left(aq\right)$

So, the equilibrium constant for the reaction is

$K_{eq}=\frac{K_{sp}}{K_a}=\frac{1.8\times {10}^{-8}}{{10}^{-2}}=1.8\times {10}^{-6}$