A) 10.53
B) 8.47
C) 6.94
D) 3.47
Correct Answer: A
Solution :
\[Mg{{(OH)}_{2}}\rightleftharpoons \underset{S}{\mathop{M{{g}^{2+}}}}\,+\underset{2S}{\mathop{2O{{H}^{-}}}}\,\] \[{{K}_{sp}}Mg{{(OH)}_{2}}=[M{{g}^{2+}}]{{[O{{H}^{-}}]}^{2}}\] \[\Rightarrow \] \[{{K}_{sp}}Mg{{(OH)}_{2}}=4{{S}^{3}}\] \[1.96\times {{10}^{-11}}=4{{S}^{3}}\] or \[S={{\left[ \frac{1.96\times {{10}^{-11}}}{4} \right]}^{1/3}}\] or \[S={{(4.9\times {{10}^{-12}})}^{1/3}}\] \[\therefore \] \[S=1.69\times {{10}^{-4}}\] So, concentration of \[[O{{H}^{-}}]=2S\] \[\therefore \] \[[O{{H}^{-}}]=3.38\times {{10}^{-4}}\] \[\Rightarrow \] \[pOH=-\log [O{{H}^{-}}]\] \[=-\log [3.38\times {{10}^{-4}}]\] \[\therefore \] \[pOH=3.471\] \[pH=14-pOH\] \[=14-3.471\] \[\therefore \] \[pH=10.529\]
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