A) 6
B) 13
C) 8
D) 10
Correct Answer: B
Solution :
\[\underset{\begin{smallmatrix} 1\,mol \\ 0.04\,F \end{smallmatrix}}{\mathop{NaOH}}\,\xrightarrow[{}]{{}}\underset{1\,mol}{\mathop{N{{a}^{+}}}}\,+\underset{\begin{smallmatrix} 1\,mol \\ 0.04\,mol \end{smallmatrix}}{\mathop{O{{H}^{-}}}}\,\] \[\therefore \]Concentration of \[O{{H}^{-}}=\frac{0.04}{400}\times 1000={{10}^{-1}}\] \[pOH=\log [O{{H}^{-}}]=-\log {{10}^{-1}}\] \[=1\] \[pH=14-1=13\]You need to login to perform this action.
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