Question

A 40.0 mL sample of 0.100 M HCl is titrated with 0.100 M NaOH. Calculate the pH after the addition of each of the following volumes of NaOH: (a) 0.0 mL, (b) 10.0 mL, (c) 20.0 mL, (d) 40.0 mL, (e) 60.0 mL. A plot of the pH of the solution as a function of the volume of added titrant is known as a pH titration curve. Using the available data points, plot the pH titration curve for the above titration.

Answer 1

The reaction is

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

40x0.1 0 0 0 initial mmoles

a) initial pH of HCl before adding NaOH

pH = - log [H+]

= - log 0.1

= 1.0

b)

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

40x0.1 0 0 0 initial mmoles

- 10x0.1 - - change

3 0 10 - equilibrium mmoles

Thus [H+] = mmoles/ volume

= 3/50

and pH = - log [H+]

= --log 3/50

= 1.2218

c)

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

40x0.1 0 0 0 initial mmoles

- 20x0.1 - - change

2 0 2 - equilibrium mmoles

Thus [H+] = 2/60

pH = - log (2/60)

= 1.477

d)

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

40x0.1 0 0 0 initial mmoles

- 40x0.1 - - change

0 0 4 - equilibrium mmoles

Thus the solution has no acid , no base only salt.

Thus pH = 7.0

e)

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

40x0.1 0 0 0 initial mmoles

- 60x0.1 - - change

0 2 4 - equilibrium mmoles

Now the solution contains excess base

[base] = mmoles / total volume

= 2/100

and pOH = - log [OH-]

= -log (2/100)

= 1.699

and pH = 14-1.699

= 12.301