A) \[{{10}^{-2}}\]
B) \[{{10}^{-12}}\]
C) \[{{10}^{-11}}\]
D) \[{{10}^{-10}}\]
Correct Answer: B
Solution :
\[[{{H}^{+}}]=C.\alpha =0.1\times 0.1=1\times {{10}^{-2}}\] \[\because \] \[[{{H}^{+}}]\,[O{{H}^{-}}]={{K}_{w}}\] \[\therefore \] \[[1\times {{10}^{-2}}]\,[O{{H}^{-}}]=1\times {{10}^{-14}}\] \[\,[O{{H}^{-}}]=\frac{1\times {{10}^{-14}}}{1\times {{10}^{-2}}}\] \[\,[O{{H}^{-}}]={{10}^{-12}}g/ion/L\]You need to login to perform this action.
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