Equilibrium Question 852
Question: If $ K {sp}(PbSO_4)=1.8\times {10^{-8}} $ and $ K{a}(HSO_4^{-})=1.0\times {10^{-2}} $ the equilibrium constant for the reaction. $ PbSO_4(s)+{H^{+}}(aq)\rightarrow HSO_4^{-}(aq)+P{b^{2+}}(aq) $ is
Options:
A) $ 1.8\times {10^{-6}} $
B) $ 1.8\times {10^{-10}} $
C) $ 2.8\times {10^{-10}} $
D) $ 1.0\times {10^{-2}} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ PbSO_4(s)\rightarrow $
$ P{b^{2+}}(aq)+SO_4^{2-}(aq)(K _{sp}) $ ….(i) $ HSO_4^{-}(aq)\rightarrow $
$ {H^{+}}( aq )+SO_4^{2-}( aq )(K_{a}) $ ….(ii) Subtracting equation (ii) from (i), then $ PbSO_4(s)+{H^{+}}(aq)\rightarrow $
$ P{b^{2+}}(aq)+HSO_4^{-}(aq)(K_{eq}) $
$ \therefore K_{eq}=\frac{K {sp}}{K{a}} $
$ =\frac{1.8\times {10^{-8}}}{1.0\times {10^{-2}}}=1.8\times {10^{-6}} $
