Titration of H2SO4 and NaOH

April 26th, 2008

Titration that involve H2SO4 and NaOH is an examples of acid-base titration of a strong base acid and strong base, both the titrant and the analyte are completely ionized. The balanced reaction and its ionic reaction are:

H2SO4 + 2NaOH -> Na2SO4 + 2H2O
H+ + SO42- + 2 Na+ + 2OH- -> 2Na+ + SO42- + 2H2O
H+ + + 2OH- -> H2O

The H+ combined with OH- to form H2O, and other ions (Na+ and SO42-) remain unchanged. This is an example of neutralization reaction, the net result of this neutralization is conversion of H2SO4 into a neutral solution Na2SO4 in the equivalent point.

By constructing a titration curve we can easily explain how the equivalent point and the end point of the H2SO4 and NaOH titration can be detected. A titration curve is constructed by plotting the pH of the solution as a function of the volume of titrant added. The volume changes during titration must be employed for determining the concentrations of the species in the solutions (H+, OH- ). This is the example of the titration curve of 0.1 M H2SO4 100 mL versus 0.1 M of NaOH.

titration curve

How you can make the titration curve above?

We began with 0.1 mL H2SO4 100 mL in Erlenmeyer flask and then 0.1 mL of NaOH in the buret.

At the start of the titration there is only H2SO4 in the solution so the pH of this solution is:

[H2SO4] = 0.1 M
[H+] = 2 x 0.1 = 0.2 M
pH = - log [H+] = -log(0.2) = 0.699

as the titration begins, NaOH solution is added from the burette into the H2SO4 solution in the Erlenmeyer. The pH of the solution when the 20 mL NaOH 0.1 M is added to the solution is:

moles of H2SO4 = 0.1 M x 100 mL = 10 mmol
moles of NaOH = 0.1 M x 20 mL = 2 mmol

H2SO4

+

2NaOH

->

Na2SO4

+

2H2O

Initial

10 mmol

2 mmol

0 0
Reacted

1 mmol

2 mmol

2 mmol

1 mmol

Equilibrium 9 mmol

0

2 mmol

1 mmol

[H2SO4] = 9 mmol / (100 mL + 20 mL ) = 0.075 M
[H+] = 2 x 0.075 M = 0.150 M
pH = - log [H+] = -log(0.150) = 0.824

you can do with similar way to calculate the pH of the solution when the next 40, 60, 80,….260 mL of NaOH solution is added and here are the results

Volume NaOH

pH

0

0.699

20

0.824

40

0.942

60

1.058

80

1.176

100

1.301

120

1.439

140

1.602

160

1.813

180

2.146

200

7.000

220

12.574

240

12.615

260

12.647

Other related articles

Calculate the molarity of H2SO4
Calculate of volume H2SO4 that neutralize 200 grams of NaOH
Back titration of H2SO4 and NaOH

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Entry Filed under: Titration

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