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}} $



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