When 25.0 mL of sulfuric acid solution was completely neutralized in a titration with 0.05 M NaOH solution, it took 18.3 mL of the NaOH (aq) to complete the job. The reaction is:

NaOH (aq) + H2SO4 (aq) → Na2SO4 (aq) + H2O (l)

What was the molarity of the sulfuric acid solution?

Question

Brent Andrews

Chemistry

3 July, 15:06

When 25.0 mL of sulfuric acid solution was completely neutralized in a titration with 0.05 M NaOH solution, it took 18.3 mL of the NaOH (aq) to complete the job. The reaction is:

NaOH (aq) + H2SO4 (aq) → Na2SO4 (aq) + H2O (l)

What was the molarity of the sulfuric acid solution?

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Answers (1)

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Taryn Decker · Answer 1

Balanced equation will be as follows

2NaOH + H2SO4=Na2SO4 + 2H2O

from MaVa/MbVb=Na/Nb where Ma is molarity of acid=?

Va is volume of acid=25.0 mL

Mb is molarity of base = 0.05 M

Vb is volume of base=18.3 mL

Nb number of moles of base = 2

Na is number of moles of acid = 1

Ma*25 mL / 18.3mL*0.05M = 1/2

Ma=18.3*0.05M / 50 = 0.0183M

The molarity of sulphuric acid = 0.0183M

When 25.0 mL of sulfuric acid solution was completely neutralized in a titration with 0.05 M NaOH solution, it took 18.3 mL of the NaOH (aq) to complete the job. The reaction is:NaOH (aq) + H2SO4 (aq) → Na2SO4 (aq) + H2O (l)What was the molarity of the sulfuric acid solution?

When 25.0 mL of sulfuric acid solution was completely neutralized in a titration with 0.05 M NaOH solution, it took 18.3 mL of the NaOH (aq) to complete the job. The reaction is:

NaOH (aq) + H2SO4 (aq) → Na2SO4 (aq) + H2O (l)

What was the molarity of the sulfuric acid solution?