Custom Search

 Strong Acid Strong Base Titrations

The titration of 50.0mL of 0.150M HCl with 0.150M NaOH is carried out in a chemistry laboratory. Calculate the pH of the solution after these volumes of the titrant have been added. Use these results to plot the titration curve.

 pH Vol. NaOH 0.0mL 25.0mL 49.9mL 50.0mL 50.1mL 75.0ml

0.0ml of NaOH            50.0mL 0.150M HCl

pH=-log0.150M

pH= 0.82

25.0ml of NaOH to 50.0mL of HCl

Moles H+=50.0mL x 0.150M=7.50mmol

Moles OH-=25.0mL x 0.150M=3.75mmol

 H+  + OH- è  H2O I 7.50mmol 3.75mmol C -3.75mmol -3.75mmol E 3.75mmol 0

 [H+]= 3.75mmol =0.05M 25.0mL +50.0mL

pH=-log[H+]

pH=-log[0.05M]=0.82

49.9ml of NaOH to 50.0mL of HCl

Moles H+=50.0mL x 0.150M=7.50mmol

Moles OH-=49.9mL x 0.150M=7.485mmol

 H+  + OH- è  H2O I 7.50mmol 7.485mmol C -7.485mmol -7.485mmol E 0.015mmol 0

 [H+]= 0.015mmol =0.000 15M 49.9mL +50.0mL

pH=-log[H+]

pH=-log[0.000 15M]=3.82

50.0ml of NaOH to 50.0mL of HCl

Moles H+=50.0mL x 0.150M=7.50mmol

Moles OH-=50.0mL x 0.150M=7.50mmol

 H+  + OH- è  H2O I 7.50mmol 7.50mmol C -7.50mmol -7.50mmol E 0 0

neutralization  pH=7

50.1ml of NaOH to 50.0mL of HCl

Moles H+=50.0mL x 0.150M=7.50mmol

Moles OH-=50.1mL x 0.150M=7.515mmol

 H+  + OH- è  H2O I 7.50mmol 7.515mmol C -7.50mmol -7.50mmol E 0 0.015mmol

 [OH- ]= 0.015mmol =0.000 15M 50.1mL +50.0mL

pOH=-log[OH-]

pOH=-log[0.000 15M]=3.82

pH=14-3.82=10.18

75.0ml of NaOH to 50.0mL of HCl

Moles H+=50.0mL x 0.150M=7.50mmol

Moles OH-=75.0mL x 0.150M=11.25mmol

 H+  + OH- è  H2O I 7.50mmol 11.25mmol C -7.50mmol -7.50mmol E 0 3.75mmol

 [OH- ]= 3.75mmol =0.030M 75.0mL +50.0mL

pOH=-log[OH-]

pOH=-log[0.030M]=1.52

pH=14-1.52=12.48

 pH Vol. NaOH 0.82 0.0mL 1.30 25.0mL 3.82 49.9mL 7.00 50.0mL 10.18 50.1mL 12.48 75.0ml