Question

A solution of sulfuric of sulfuric acid $H_{2}SO_{4}$ was titrated with a 0.100 M NaOH solution. If 20.76 mL of the NaOH solution is required to neutralize completely 10.54 mL of the sulfuric acid solution, what is the molarity of the sulfuric acid solution?

$H_{2}SO_{4}(aq)+2NaOH(aq)\rightarrow Na_{2}SO_{4}(aq)+2H_{2}O(I)$

Answer 1

Solution

Given data

Molarity of NaOH = 0.100 M

Volume of NaOH = (20.76 ml * 1 L )/ 1000 ml =0.02076 L

Volume of H2SO4 = (10.54 ml* 1 L) / 1000 ml = 0.01054 L

Molarity of H2SO4 = ?

Balanced reaction equation

H2SO4 + 2NaOH      ------ > Na2SO4   + 2H2O

We know the molarity and volume of the NaOH used for the titration therefore lets calculate the moles of the NaOH using its molarity and volume in liter

Formula to calculate moles using the molarity and volume is as follows

Moles = molarity * volume in liter

Moles of NaOH = 0.100 mol / L * 0.02076 L

                           = 0.002076 mol NaOH

Now using the mole ratio of the NaOH and H2SO4 lets calculate the moles of the H2SO4 titrated

Mole ratio of the NaOH to H2SO4 is 2 : 1

(0.002076 mol NaOH * 1 mol H2SO4) / 2 mol NaOH = 0.001038 mol H2SO4

Now lets calculate the molarity of the H2SO4 using its moles and volume

Molarity = moles / volume in liter

Molarity of H2SO4 = 0.001038 mol / 0.01054 L

                                  = 0.0985 M

Therefore concentration of the H2SO4 = 0.0985 M