A) \[0.158\times {{10}^{-14}}\]
B) \[0.3162\times {{10}^{-14}}\]
C) \[3.163\times {{10}^{-14}}\]
D) \[2.163\times {{10}^{-14}}\]
Correct Answer: C
Solution :
\[Sr{{(OH)}_{2}}(s)\xrightarrow{{}}S{{r}^{2+}}(aq)+2O{{H}^{-}}(aq)\] \[[S{{r}^{2+}}]=0.1581\,M\] \[[O{{H}^{-}}]=2\times 0.1581=0.3126\,M\] Now, \[{{K}_{w}}=[{{H}_{3}}{{O}^{+}}]\,[O{{H}^{-}}]\] So, \[[{{H}_{3}}{{O}^{+}}]=\frac{{{K}_{w}}}{[O{{H}^{-}}]}=\frac{{{10}^{-14}}}{0.3126}\] \[=3.2\times {{10}^{-14}}\] \[=3.163\times {{10}^{-14}}\]
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