In an exper iment 0.04F was passed through 400mL of 1 M solution of NaCl. What would be the pH of the solution after electrolysis?

Options

A. 8
B. 10
C. 13
D. 6

Answer

C

Explanation

`NaCl+H_(2)Ooverset("Electrolysis")toNaOH+underset("At anode")((1)/(2)H_(2))+underset("At cathode")((1)/(2)Cl_(2))`
400 ml of 1 M NaCl solution contains NaCl
`=(1)/(1000)xx100mol=0.4mol`
As `Na^(+)+e^(-)toNa overset(H_(2)O)toNaOH+(1)/(2)H_(2)`
1 F produces 1 mol of NaOH
`therefore0.04F` produces NaOH=0.04 mol
Thus, 400 ml of the solution now contain 0.04 mol of NaOH
`therefore`Molar concentration of NaOH sol.
`=(0.04)/(400)xx1000=0.1M`
`[OH^(-)]=0.1M[H^(+)]=10^(-13)MthereforepH=13`