Answer to Question #122966 in Inorganic Chemistry for Waisele

Question

A solution of NaCl is neutral, with an expected PH of 7. If electrolysis was carried out on 500 mL of an NaCl solution with a current of 0.500 A, how many seconds would it take for the PH of the solution to rise to a value of 9? (PH is related to H+ concentration by PH = -log [H+]).

Answer 1

NaCl + H₂O --> NaOH + Cl₂ + H₂;

2H₂O +2e -->H₂ + 2OH^-;

Before electrolysis:

pOH=14-pH=14-7=7;

[OH^-]=10^(-pOH)=10^-7 mol/L;

n(OH-)=10^(-7) * 0.500=5*10^(-8)

After:

pOH=14-9=5;

n(OH^-)=10^(-5) * 0.500=5*10^(-6);

Amount of OH^- formed by electrolysis = 5*10^(-6) - 5*10^(-8)=4.95 * 10^-6 mol;

t= F*n(OH-)/I=96500*4.95*10^(-6)/0.5=0.95535 s