Given: 
A) \[-0.907V\]
B) \[0.907V\]
C) \[-0.313\]
D) \[0.313\]
Correct Answer: B
Solution :
[b]
\[(pH=9.0,\,pOH=14-9=5,\,\overset{\bigcirc -}{\mathop{O}}\,H={{10}^{-5}}M\] Activity of \[{{H}_{2}}O=1\]) 
\[-\frac{0.06}{6}\log \frac{5.0\times {{10}^{-3}}\times {{({{10}^{-5}}M)}^{6}}}{2.5\times {{10}^{-3}}M}\] \[=0.61V-0.01\,(\log 2\times {{10}^{-30}})\] \[=0.61V-0.01(0.3-30)\] \[(\log 2\,\,\approx \,\,0.3)\] \[=0.61\,V-0.01\times (-29.7)\] \[=0.61V+0.297\approx 0.907V\]
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