Calculate the pH of the solution and the concentrations

Question

The 𝐾a of propanoic acid (C2H5COOH) is 1.34×10−5. Calculate the pH of the solution and the concentrations of C2H5COOH and C2H5COO− in a 0.249 M propanoic acid solution at equilibrium.

pH=

[C2H5COOH]= M

[C2H5COO−]= M

Answer 1

Let C2H5COOH be HA

Let C2H5COO- be A-

HA <—> H+ + A-

Ka = 1.34x10-5 = [H+][A-] / [HA]

1.34x-5 = (x)(x) / 0.249-x. (Assume x is small relative to 0.249)

1.34x-5 = (x)(x) / 0.249

x ^2 = 3.29x10-6

x = 1.81x10-3. (this is less than 1% of 0.249 so assumption was ok)

[H+] = 1.8x10-3

pH = - log 1.81x10-3

pH = 2.74

C2H5COOH] = 0.249 - .00181= 0.247M

[C2H5COO−] = 1.81x10-3 M