A) \[11\]
B) \[3\]
C) \[14\]
D) \[{{10}^{-11}}\]
Correct Answer: A
Solution :
Given, \[{{K}_{w}}=1\times {{10}^{-14}}mo{{l}^{2}}{{L}^{-2}}\] Molarity of \[KOH=\] normality of \[KOH\] \[N=1\times {{10}^{-3}}\] or \[[O{{H}^{-}}]=1\times {{10}^{-3}}\] \[[{{H}^{+}}]=\frac{1\times {{10}^{-14}}}{[O{{H}^{-}}]}=\frac{1\times {{10}^{-14}}}{1\times {{10}^{-3}}}=1\times {{10}^{-11}}\] \[pH=-\log [{{H}^{+}}]\] \[pH=-\log \,{{10}^{-11}}\] \[pH=11\]
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