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.

12.86

The correct answer is option 2. 12.86

Given,

$i=0.193A,V=0.1L,C=0.2M,t=3600s$

$Q = it=0.193×3600=694.8C$

$n_e=\frac{Q}{F}=\frac{694.8C}{96500}=0.0072$

$2H_2O+2e^-→H_2+2OH^-$

$n_{OH^-}=n_e$

$[OH^-]=\frac{n_{OH^-}}{V}=\frac{0.0072}{0.1L}mol$

$[OH^-]=0.072M$

$P_{OH}=-\log_{10}[OH^-]$

$P_{OH}=-\log_{10}(0.072)$

$P_{OH}=1.14$

$pH=14-P_{OH}$

$pH=14-1.14$

$pH=12.86$