Titration
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 + 2H2OH+ + SO42- + 2 Na+ + 2OH- -> 2Na+ + SO42- + 2H2OH+ + + 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.
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.
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 MpH = - log [H+] = -log(0.2) = 0.699
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 MpH = - 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 mmolmoles of NaOH = 0.1 M x 20 mL = 2 mmol
moles of H2SO4 = 0.1 M x 100 mL = 10 mmolmoles of NaOH = 0.1 M x 20 mL = 2 mmol
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