Answer to Question #79350 in General Chemistry for Shahmeer Faisal

Question

Determine the concentration of 20.0 mL of the sulphuric aci when 1.00 mol/L sodium hydroxide is used as the titrant.

Observations:Initial volume of NaOH: ________________
Final volume of NaOH: _______________
Volume of NaOH added: ________________

Answer 1

reaction between sulfuric acid and sodium hydroxide is of an acid-base type, or is also known as a neutralization reaction. In this process, both compounds undergo a reaction to neutralize the acid and base properties. The products of this process are salt and water.

The balanced equation of this reaction is:

H2SO4 + 2NaOH → Na2SO4 + 2H2O

In this process, 2 moles of sodium hydroxide combine with one mole of sulfuric acid. This results in the formation of two moles of water and one mole of sodium sulfate.

For example, the amount of NaOH needed to react with sulfuric acid is 12 ml, 13 ml, and 12.5 ml, respectively. Thus, by taking their average, 12.5 ml of NaOH neutralized the acid with the unknown concentration.

Number of Moles (N) = Volume (V) x Molarity (M)

N = 0.0125L x 1mol/L = 0.0125 moles

Now, according to the equation, exactly half the number of moles of sulfuric acid take part in the reaction; i.e., the number of moles required for the neutralization process are:

N (for sulfuric acid) = 0.0125/2 = 0.00625 moles

The volume of sulfuric acid used in this experiment is 20 ml.

The concentration of H2SO4 is calculated:

Molarity (M) = Number of Moles (N)/Volume (V) M = 0.00625/0.02 = 0.3125 mol/L