A) 20
B) 14
C) 12
D) 10
Correct Answer: C
Solution :
Key Idea: \[pOH=-\log [O{{H}^{-}}]\] \[pH+pOH=14\] Given: \[[NaOH]=0.01\,M={{10}^{-2}}M\] \[\because \] \[NaOH\] is strong base therefore it ionises completely. \[NaOH\to N{{a}^{+}}+O{{H}^{-}}\] \[{{10}^{-2}}\] \[{{10}^{-2}}\] \[{{10}^{-2}}\] \[\therefore \] \[[O{{H}^{-}}]={{10}^{-2}}\] \[pOH=-\log \,[O{{H}^{-}}]\] \[=-\log ({{10}^{-2}})\] = 2 \[pH+pOH=14\] \[\therefore \] \[pH=14-pOH\] \[=14-2=12\]
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