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?

Correct Answer - C
`NaCl+H_(2)Ooverset( "Electrolysis")rarrNaOH+(1)/(2)H_(2)+(1)/(2)Cl_(2)`
400mL of 1 M NaCl solution contain NaCl
`=(1)/(1000)xx400` mol =0.04mol
Since `Na^(+)+erarrNaoverset(H_(2)O)rarrNaOH+(1)/(2)H_(2)`
1F produces 1 mol NaOH.
`therefore 0.04` F will produce 0.04mol of NaOH from 0.04mol of NaCl
Thus 0.04mol of NaOH is present in 400mL of solution.
` [OH^(-)]=[NaOH ]=(0.04)/(400)xx1000`
`=0.1M=10^(-1)M`
`therefore[H^(+)]=(k_(H))/([OH^(-)])=(10^(-14))/(10^(-1))=10^(-13)`
`therefore pH=-"log"[H^(+)]=-"log" 10^(-13)=13`