A) \[\text{4}\text{.00 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-6}}}{{\text{M}}^{\text{3}}}\]
B) \[\text{4}\text{.00 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-7}}}{{\text{M}}^{\text{3}}}\]
C) \[\text{5}\text{.00 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-6}}}{{\text{M}}^{\text{3}}}\]
D) \[\text{5}\text{.00 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-7}}}{{\text{M}}^{\text{3}}}\]
Correct Answer: D
Solution :
Given, pH of \[Ba{{(OH)}_{2}}=12\] \[\therefore \] \[[{{H}^{+}}]=[1\times {{10}^{-12}}]\] and\[[O{{H}^{-}}]=\frac{1\times {{10}^{-14}}}{1\times {{10}^{-12}}}[\because [{{H}^{+}}][O{{H}^{-}}]=1\times {{10}^{-14}}]\] \[=1\times {{10}^{-2}}mol/L\] \[Ba{{(OH)}_{2}}\xrightarrow[{}]{{}}\underset{s}{\mathop{B{{a}^{2+}}}}\,+\underset{2s}{\mathop{2O{{H}^{-}}}}\,\] \[{{K}_{sp}}=[B{{a}^{2+}}{{[O{{H}^{-}}]}^{2}}\] \[=[s]{{[2s]}^{2}}\] \[=\left[ \frac{1\times {{10}^{-2}}}{2} \right]{{(1\times {{10}^{-2}})}^{2}}\] \[=0.5\times {{10}^{-6}}=5.0\times {{10}^{-7}}{{M}^{3}}\]
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