A) \[5\times {{10}^{-10}}\,\text{N}\]
B) \[4\times {{10}^{-10}}\,\text{N}\]
C) \[2\times {{10}^{-5}}\,\text{N}\]
D) \[9\times {{10}^{-4}}\,\text{N}\]
Correct Answer: A
Solution :
\[{{[H]}^{+}}[O{{H}^{-}}]={{K}_{w}}\] \[2\times {{10}^{^{-5}}}\times [O{{H}^{-}}]=1\times {{10}^{-14}}\] \[[OH]=\frac{1\times {{10}^{-14}}}{2\times {{10}^{-5}}}\] \[=0.5\times {{10}^{-9}}\] \[=5\times {{10}^{-10}}\]
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