A) 0.81 V
B) 0.071 V
C) 0.0591 V
D) 1.182 V
Correct Answer: C
Solution :
[c] \[{{E}_{cell}}=\frac{0.0591}{1}\log \frac{{{[{{H}^{+}}]}_{RHS}}}{{{[{{H}^{+}}]}_{LHS}}}\] |
\[{{E}_{cell}}=0.0591[p{{H}_{LHS}}-p{{H}_{RHS}}]\] |
\[p{{H}_{LHS}}p{{K}_{a}}+\log \frac{[{{A}^{-}}]}{[HA]}\] |
\[=8+\log \frac{1}{0.1}=9\] \[p{{H}_{RHS}}=8\] |
\[{{E}_{cell}}=0.0591(9-8)=0.0591V\] |
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