Question

Write the equilibrium equations, identify the equilibrium variables such as x +/- y, and calculate the pH of a 0.600 liter solution made up as 1.14 M diprotic base B(aq). Kb1=6.87x10^-6; Kb2=7.15x10^-8

Answer 1

0.600 liter solution made up as 1.14 M diprotic base B

B + H2O = BH(+) + OH-

Initial:    B = 1.14 M, BH+ = 0, OH- = 0  
Change:    B = -X M, BH+ = +X, OH- = +X  
Equilibrium:    B = 1.14-X M, BH+ = X, OH- = X  

Kb1 = [BH(+)]*[OH-]/[B] = 6.87E-6
X^2/(1.14-X) = 6.87E-6
Assuming X << 1.14
X = (1.14*6.87E-6)^0.5 = 2.8E-3
[OH-] = X = 2.8E-3 M
pOH = -Log(2.8E-3) = 2.55
pH = 14 - pOH = 14-2.55=11.45

The 2nd equilibrium constant Kb2=7.15E-8 is very low and the pH will not change much
in the 2nd equilibria.

BH(+) + H2O = BH2(2+) + OH-

Initial:    BH+ = 2.8E-3 M,    BH2(2+) = 0, OH- = 2.8E-3 M  
Change:    BH+ = -X M,        BH2(2+) = +X, OH- = +X  
Equilibrium:    BH+ = 2.8E-3-X M, BH2(2+) = X, OH- = (2.8E-3+X)  

Kb2 = [BH2(2+)]*[OH-]/[BH+] = 7.15E-8
X*(2.8E-3+X)/(2.8E-3-X) = 7.15E-8
Assuming X << 2.8E-3
X = 7.15E-8
[OH-] = (2.8E-3+X) = 2.8E-3 M
pOH = -Log(2.8E-3) = 2.55
pH = 14 - pOH = 14-2.55=11.45