Question

A.Neglecting activity coefficients, calculate the pH of an aqueous solution of 5.0 x 10-7 M KOH.

B. What fraction of the total [H+ ] is derived from the autoionization of water?

Answer 1

KOH is a strong base, so it will disassociate completely. Water will disassociate in turn. Therefore, you set up an ICE table for water.

Complete ionization of the strong acid:

KOH ----> K+ + OH-
I.5E-7........ ..0...........0
C.-5E-7... .+5E-7...+5E-7
E..0........... .5E-7.....5E7

a) Since KOH will disassociate completely, the H+ concentration is already 5E-7 M.

..H2O <----> H+ + OH-
I........... ...5E-7.....0...
C.......... ...+x.......+x..
E.......... ...5E-7+x..+x..

Kw = [H+][OH-]

1E-14 = (5E-7+x)x

x²+ 5x10^-7x - 1x 10^-14

You can either solve this quadratic, (I useD a calculator).

x = [H+] = 5.19 x 10 ^-7

Water will disassociate to form 5.19 x 10 ^-7 M H+. Add this to the concentration given.

5.19 x 10 ^-7 + 5E-7 = 10.19 x10^-7

pH = -log[H+] = 5.99 0r 6

b) Since you know that water disassociates to form 5.19 x 10 ^-7 M H+, you simply divide this number by the total H+ concentration.

5.19 x 10 ^-7 / 10.19 x10^-7 = 0.509 * 100% = 51%

51% of the H+ comes from the disassociation of water.