Question

Calculate the pH of 0.015M NaOCl. No K value is given.

Answer 1

Ka of HClO = 3*10^-8

NaOCl(aq) -------------> Na^+ (aq) + OCl^- (aq)

0.015M ------------------- 0.015M ------- 0.015M

Kb = Kw/Ka

      = 1*10^-14/(3*10^-8)   = 3.34*10^-7

------------- OCl^- (aq) + H2O(l) -------------> HClO(aq) + OH^- (aq)

I ------------- 0.015 -----------------------------------0 ---------------- 0

C-------------- -x ------------------------------------ +x ----------------- +x

E----------- 0.015-x---------------------------------+x ----------------+x

            Kb   = [HClO][OH^-]/[OCl^-]

           3.34*10^-7 = x*x/(0.015-x)

          3.34*10^-7(0.015-x) = x^2

          x   = 7.06*10^-5

[OH^-]   = x   = 7.06*10^-5M

POH   = -log[OH^-]

          = -log(7.06*10^-5)

          = 4.15

PH   = 14-POH

        = 14-4.15

         = 9.85 >>>>answer