A) 3
B) 10-11
C) 2
D) 11
Correct Answer: B
Solution :
Given, \[[O]=\frac{1}{1000}={{10}^{-3}}N\] \[(\therefore KOH\xrightarrow{{}}{{K}^{+}}OH)\] \[pOH=-\log \,\,[O{{H}^{-}}]\] \[=-\log \,\,{{10}^{-3}}\] \[poH=3\] \[pH+pOH=14\] or \[pH=14-3=11\]
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