A current of 0.193 amp is passed through 100 mL of 0.2M NaCl for an hour. Calculate pH of solution after electrolysis. Assume no volume change.
Answer & explanation
Correct answer: option 2
The correct answer is option 2. 12.86
Given,
$i=0.193A,V=0.1L,C=0.2M,t=3600s$
$Q = it=0.193×3600=694.8C$
$n_e=\frac{Q}{F}=\frac{694.8C}{96500}=0.0072$
$2H_2O+2e^-→H_2+2OH^-$
$n_{OH^-}=n_e$
$[OH^-]=\frac{n_{OH^-}}{V}=\frac{0.0072}{0.1L}mol$
$[OH^-]=0.072M$
$P_{OH}=-\log_{10}[OH^-]$
$P_{OH}=-\log_{10}(0.072)$
$P_{OH}=1.14$
$pH=14-P_{OH}$
$pH=14-1.14$
$pH=12.86$