Question

A current of 0.250 A is passed through 400 ml of a 2.0 M solution of NaCl for 32 minutes and 10 seconds. What will be the pH of the solution after the current is turned off?
Take log 2 = 0.3, log 3 = 0.5, log 5 = 0.7

Answer 1

After electrolysis, aqueous NaCl is converted into aqueous NaOH.
$\text{The quantity of electricity passed}=\frac{0.250\times (32\times 60+10)}{96500}=5\times {10}^{-3}F$
$\text{The number of equivalents of}{\text{OH}}^{-}\text{ions formed}=5\times {10}^{-3}$
$\therefore$
$\text{Molarity of}\text{NaOH}=\frac{5\times {10}^{-3}}{0.400}=125\times {10}^{-4}\text{M}\text{pOH}=-\log (125\times {10}^{-4})=4.00–3\log 5=1.9\therefore \text{pH}=14-1.9=12.1$