A) 5
B) 9
C) 11.5
D) 8.5
Correct Answer: B
Solution :
\[M{{(OH)}_{2}}\rightleftharpoons \underset{s}{\mathop{{{M}^{+}}}}\,+\underset{2s}{\mathop{2O{{H}^{-}}}}\,\] \[{{K}_{sp}}=(s){{(2s)}^{2}}=4{{s}^{3}}=5\times {{10}^{-16}}\] \[\therefore \] \[s=\sqrt[3]{\frac{5\times {{10}^{-6}}}{4}}=5\times {{10}^{-6}}\] Cone. of \[O{{H}^{-}}=2\times 5\times {{10}^{-6}}={{10}^{-5}}\,\text{mol}\,\text{d}{{\text{m}}^{-3}}\] \[pOH=-\log [O{{H}^{-}}]\] \[=-\log {{10}^{-5}}=5\] \[pH=14-pOH=14-5=9.\]
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