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1. |
If \[{{K}_{sp}}(PbS{{O}_{4}})=1.8\times {{10}^{-8}}\] and \[{{K}_{a}}(HSO_{4}^{-})=1.0\times {{10}^{-2}}\] the equilibrium constant for the reaction. \[PbS{{O}_{4}}(s)+{{H}^{+}}(aq)\rightleftharpoons HSO_{4}^{-}(aq)+P{{b}^{2+}}(aq)\] is |
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» B. \[1.8\times {{10}^{-10}}\] | |