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The Titration Formula

This is the very simple formula for you to calculate the problems that involve the titration. Just always to remember that in titration “mole equivalent of titrant in the equivalent point is equal to the mole equivalent of analyte”, using this statement we can write:

N is the Normality
V is the volume of the solution

Because normality (N) is n x M, thus the formula above can be written as,

n is the total of H+ or OH- in acid base titration, or the total of electrons that release or gain in the redox titration.

Add comment April 26th, 2008

Titration of H2SO4 and NaOH

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⁺ + SO4^2- + 2 Na⁺ + 2OH^- -> 2Na⁺ + SO4^2- + 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.

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] = 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

Other related articles

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

Add comment April 26th, 2008

Calculate the molarity of H2SO4

Chemistry Tutor Problem

250 ml of H2SO4 unknown concentration with a methyl red indicator present is titrated with 0.140 M NaOH, the red solution turns into yellow after 22.6 ml of NaOH added, What is the Molarity of H2SO4?

Chemistry Tutor - Solution

The balanced reaction is:

H2SO4 + 2 NaOH –> Na2SO4 + 2H2O

You can solve this problem into two ways first using the stoichiometry calculation and second using titration formula.

1. using stoichiometry calculation

First calculate the moles of NaOH that has reacted with H2SO4

Moles NaOH = 22.6 ml x 0.140 M = 3.164 mmol

From the balanced reaction above we know that the moles ratio of H2SO4 and NaOH is 1:2 thus the moles of H2SO4 that reacted with 3.164 mmol of NaOH is

Moles H2SO4 = ½ x 3.164 mmol = 1.583 mmol

And the molarity of H2SO4 is

[H2SO4] = 1.583 mmol/250 mL = 0.006328 M

2. Using titration formula

Just remember the titration formula that you can use to solve all of the titration problems. This formula you can used for all of types of titration. Just simple “at the equivalent point the mol-equivalent of the titratnt is equal to the mol-equivalent of the analyte” thus

NV acid = NV base or
n x M x V acid = n x M x V

n is the number of H+ in acid or the number of OH- in base, substitute the value from the data above into the formula we get,

2 x M x 250 mL (H2SO4) = 1 x 22.6 ml x 0.140 M
M H2SO4 = (1 x 22.6 ml x 0.140 M) / (2 x 250 mL)
M H2SO4 = 0.006328 M

The second way is more simple than the first one, and you can always use this way when you have the problems that involve titration calculation.

Add comment April 26th, 2008

Calculate of Volume H2SO4 that Neutralize 200 grams of NaOH

Chemistry Tutor Problem
What volume of 2.50 N H2SO4 is needed to neutralize 200 grams of NaOH? How would I write the equation? Would it be 2H20 + Na2SO4?

Chemistry Tutor - Solution

Using the titration formula you can easily solve this problem. First you need to convert the 200 g of NaOH into its moles after that substituting the moles into the titration formula you will get the moles of H2SO4 that needed to neutralized the NaOH. To convert the moles of H2SO4 into volume you just need to use the molarity equation. See the way to solve this problem:

NV acid = NV base or
NV (H2SO4) = n x M x V NaOH
NV (H2SO4) = n x mol / V x V (NaOH)
V (H2SO4) = ( n x mol ) NaOH / N H2S)4

The moles of NaOH
= g / Molar Mass
= 200 / 39.998
= 5.000 mol

V (H2SO4) = ( n x mol ) NaOH / N H2S)4
V (H2SO4) = ( 1 x 5.000 ) / 2.50 N
V (H2SO4) = 2.000 L

Add comment April 26th, 2008