Question

The table of diprotic acids below shows how the acidity of the 1st proton is always much higher than the acidity of the second proton.Thus, the pH of the solution is dominated by the equilibrium for the loss of the 1st proton.

Compound	Ka1	Ka2
H2CO3	4.3 X 10-7	5.6 X10-11
H2S	8.9 X 10-8	1.0 X 10-19
H2Te	2.3 X 10-3	1.6 X 10-11
H2SO3	1.7 X 10-2	6.4 X 10-8

What is the pH for a 1.12 M solution of Li2S?

Answer 1

Concentration of Li2S = 1.12M

Li2S ------------------ 2 Li+ + S-2

S-2 + H2O ---------------- HS- + OH-

1.12                                      0        0

-x                                        +x          +x

1.12-x                                 +x          +x

Ka = 8.9x10^-8

Ka x Kb = Kw                     where Kw = 1.0x10^-14

Kb = Kw/Ka = 1.0x10^-14/8.9x10^-8 = 1.12 x10^-7

Kb= 1.12x10^-7

Kb = [HS-][OH-]/[S-2]

1.12x10^-7 = x*x/(1.12-x)

for solving the equation

x= 0.000354

[OH-] = 0.000354M

-log[OH-] = -log(0.000354)

POH= 3.45

PH+POH= 14

PH = 14-POH

PH= 14-3.45

PH= 10.55.