A) \[3.3750\times {{10}^{-12}}~\]
B) \[1.6875\times {{10}^{-10}}\]
C) \[1.6875\times {{10}^{-12}}\]
D) \[1.6875\times {{10}^{-11}}\]
Correct Answer: C
Solution :
Given, \[2[A{{g}^{+}}]=1.5\times {{10}^{-4}}mol/L\] \[[A{{g}^{+}}]=0.75\times {{10}^{-4}}\] \[[A{{g}^{+}}]=[CrO_{4}^{2-}]\] \[=0.75\times {{10}^{-4}}mol/L\] \[A{{g}_{2}}Cr{{O}_{4}}\rightleftharpoons 2A{{g}^{+}}+CrO_{4}^{2-}\] \[{{K}_{sp}}={{[A{{g}^{+}}]}^{2}}.[CrO_{4}^{2-}]\] \[{{K}_{sp}}={{(1.5\times {{10}^{-4}})}^{2}}\times (0.75\times {{10}^{-4}})\] \[{{K}_{sp}}=2.25\times {{10}^{-8}}\times 0.75\times {{10}^{-4}}\] \[{{K}_{sp}}=1.6875\times {{10}^{-12}}\]You need to login to perform this action.
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