A) \[3.3\times {{10}^{-7}}\]
B) \[5.0\times {{10}^{-7}}\]
C) \[4.0\times {{10}^{-6}}\]
D) \[5.0\times {{10}^{-6}}\]
Correct Answer: B
Solution :
Given, \[pH\]of\[Ba{{(OH)}_{2}}=12\] \[\therefore \] \[pOH=14-pH\] \[=14-12=2\] We know that, \[pOH=-\log [O{{H}^{-}}]\] \[2=-\log [O{{H}^{-}}]\] \[[O{{H}^{-}}]=\text{antilog}\,\text{(-2)}\] \[[O{{H}^{-}}]=1\times {{10}^{-2}}\] \[Ba{{(OH)}_{2}}\]dissolves in water as \[\underset{s\,mol\,{{L}^{-1}}}{\mathop{Ba{{(OH)}_{2}}}}\,(s)\underset{s\,}{\mathop{B{{a}^{2+}}}}\,+\underset{2s\,}{\mathop{2O{{H}^{-}}}}\,\] \[\therefore \] \[[O{{H}^{-}}]=2s=1\times {{10}^{-2}}\] \[[B{{a}^{2+}}]=\frac{[O{{H}^{-}}]}{2}=\frac{1\times {{10}^{-2}}}{2}\] \[{{K}_{sp}}=[B{{a}^{2+}}]{{[O{{H}^{-}}]}^{2}}\] \[=\left( \frac{1\times {{10}^{-2}}}{2} \right){{(1\times {{10}^{-2}})}^{2}}\] \[=0.5\times {{10}^{-6}}=5\times {{10}^{-7}}\]
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