Question

Given a titration of 50.00 mL of 0.150 M KOH with 0.200 M HCl, show calculations to create a titration curve using Excel from the initial pH of the KOH ending with 5 mL of HCl past the end point.

Answer 1

At equivalence

mmoles of KOH = mmoles of HCl

50x 0.15 = V x0.2

thus volume of HCl at equivalence = 37.5mL

1) initial pH of KOH

molarity of KOH = 0.150M

thus [OH-] = 0.15 M

and pOH = -log [OH-]

= -log0.15

=0.8239

thus pH = 14-0.8239

= 13.1760

2) at half equivalence point , that after adiition 18.75 ml

KOH + HCl -------------> KCl + H2O

50x0.15=7.5 18.75x0.2=3.75 0 0 initial mmoles

3.75 0 3.75 3.75 after reaction

Thus [OH-] = mmoles /total volume

= 3.75/68.75

=0.05454 M

Thus pOH = -log 0.05454

= 1.26 32

and pH = 14-1.2632

= 12.736

3) 5 mL before equivalence

KOH + HCl -------------> KCl + H2O

50x0.15=7.5 32.5x0.2=6.5 0 0 initial mmoles

1.0 0 6.75 6.75 after reaction

Thus [OH-] = mmoles /total volume

=1.0/82.5

pOH = -log 1/82.5

= 1.9164

pH = 12.08

4) AT equivalence

As there is no base or acid remains, the pH of solution = 7.0 (the salt of strong acidand strong base ) is neutral.

5) 5 mL past equivalence

KOH + HCl -------------> KCl + H2O

50x0.15=7.5 42.5x0.2=8.5 0 0 initial mmoles

0 1.0 7.5 7.5 after reaction

The solution now has excess of acid

Thus [H+] = mmoles /total volume

= 1/92.5

and pH = -log 1/92.5

= 1.966