Question

The titration of 50.00 mL of 0.160-M HCL with 0.160-M NaOH (the titrant) is carried out in a chemistry laboratory. Calculate the pH of the solution after these volumes of the titrant have been added:

i. 0.0 mL

ii. 25.0 mL

iii. 49.9 ml

Answer 1

HCl   + NaOH   ----------------->   NaCl   + H2O

millimoles of HCl = 50 x 0.160 = 8

i) initially only HCl   present

pH = - log [H⁺]

pH = - log [0.16]

pH = 0.79

ii) millimoles of NaOH added = 25 x 0.16 = 4.0

8 - 4 = 4 millimoles HCl left

[HCl] = 4 / 75 = 0.053 M

pH = - log [H⁺]

pH = - log [0.053]

pH = 1.27

iii) millimoles of NaOH added = 49.9 x 0.16 = 7.984

8 - 7.984 = 0.016

[HCl] = 0.016 / 99.9 = 0.00016 M

pH = - log [H⁺]

pH = - log [0.00016]

pH = 3.79