Question

A 30 mL sample of 0.015 M JOH is titrated woith 0.125 M HCL. Calculate the pH for the solution after the following additons of HCL: a) 0mL, b) 10mL, c) 25mL, d) 36mL, e) 45mL, f) 60mL.

Answer 1

a) 0 mL

KOH = 0.015 M

[OH-] = 0.015 M

pOH = -log [OH-]

pOH = 1.82

pH + pOH = 14

pH = 12.18

b) 10 mL

millimoles of KOH = 30 x 0.015 = 0.45

millimoles of HCl = 10 x 0.125 = 1.25

acid remains[H+] = 1.25 - 0.45 / (30 +10) = 0.02 M

pH = -log [H+]

pH = 1.70

c) 25 mL

millmoles of HCl = 0.125 x 25 = 3.125

acid concentration [H+]= 3.125 - 0.45 / (55) = 0.0568

pH = 1.25

d) 36 mL

millmoles of HCl = 0.125 x 36 = 4.5

acid concentration [H+] = 4.5 - 0.45 / (30+36) = 0.061

pH = 1.21