Question

10 mL of 0.0100 M HCl are added to 23 mL of 0.0100 M acetic acid. What is the pH of the resulting solution?

Answer 1

V total = 10+23 = 33 mL

mmol of H+ from HCl = MV = 10*0.01 = 0.1 mmol; [H+] = mmol/V = 0.1/33 = 0.0030

mmol of acetic acid initially = MV = 0.01*23 = 0.23 mmol

note that..

[H+]total = [H+]strong acid + [H+]weak acid

Ka for acetic acid = 1.8*10^-5

then

Ka = [H+][A-]/[HA]

initially

[H+] = 0.0030

[A-] = 0

[HA] = 0.23 /33 = 0.00697

in equilbirium

[H+] = 0.0030 + x

[A-] = 0 + x

[HA] = 0.23 /33 = 0.00697 - x

1.8*10^-5 = (x)(0.003+x) / (0.00697 - x)

(1.8*10^-5)(0.00697 -x) = x^2 + 0.003x

x^2 + (0.003+1.8*10^-5)* x - 1.254*10^-7 = 0

x = 4.09*10^-5

then

[H+] = 0.0030 + x

[H+] = 0.0030 + 4.09*10^-5 = 0.0030409

pH = -log(0.0030409) = 2.52