Question

Calculate the pH of the following salt solutions

0.427 M NaC₂H₃O₂

  

0.289 M CH₃NH₃Cl

Answer 1

1)CH3NH2 + H2O ---> CH3NH3^+ + OH^-
Kb = 3.7x10^-4 = [CH3NH3^+][OH^-]/[CH3NH2]
3.7x10^-4 = (x)(x)/0.289 -x and assuming x is small relative to 0.289 , it can be omitted
3.7x10^-4 = x^2/0.289

x = 0.03578

[OH-] = 0.03578
pOH = -log 0.03578 = 1.45
pH = 14 - pOH = 12.55

1)C2H3O2^- + H2O <===> HC2H3O2 + Oh^-

shows NaC2H3O2 acting as a base.

We need to know the Kb for the acetate ion:

Kw = K1 times Kb

1.0 x 10^-14 = (1.8 x 10^-5) (Kb)

Kb = 5.56 x 10^-10

Next, we calculate the hydroxide ion concentration:

5.56 x 10^-10 = [(x) (x)] / 0.427

x = 3.608x10^-5 M

pOH = - log 3.608x10^-5 = 4.44

pH = 9.56