B is a weak base with Kb= 2.6 x 10^-6. Its salt is BHCl.?

Question

Calculate the pH of the solution by mixing 40.0 mL of 0.075M BHCl with 60.0mL of 0.050M B.

Answer 1

B + H2O <==> BH+ + OH-
Kb = 2.6x10^(-6) = [BH+]*[OH-] /[B]
BHCl is a salt thus is completely dissociated to BH+ and Cl-.
By mixing 40.0 mL of 0.075M BHCl with 60.0mL of 0.050M B, the initial concentration of BHCl reduces to 0.40*0.075 = 0.030(M), and the initial concentration of B reduces to 0.60*0.050 = 0.030M, too. Let [OH-] = X, we have:
B.... + ..H2O <==> BH+... + ..OH-
0.03-X....................0.03+X....X
Kb = 2.6x10^(-6) = [BH+]*[OH-] /[B] = (0.03+X)*X/(0.03-X)
X^2 + 0.03X + 2.6x10^(-6)*X - 2.6x10^(-6)*0.03 = 0
X^2 + 0.03X - 7.8x10^(-8) = 0
Please solve X = [OH-] and use pH = 14 + log [OH-].