Question

A 15.00 mL sample of a solution of H2SO4 of unknown concentration was titrated with 0.3200M NaOH. the titration required 21.30 mL of the base. Assuming complete neutralization of the acid,
1) What was the normality of the acid solution?
2) What was the molarity of the acid solution?

Answer 1

Answer:

a. 0.4544 N

b. $5.112 \times 10^{-5 M}$

Explanation:

For computing the normality and molarity of the acid solution first we need to do the following calculations

The balanced reaction

$H_2SO_4 + 2NaOH = Na_2SO_4 + 2H_2O$

$NaOH\ Mass = Normality \times equivalent\ weight \times\ volume$

$= 0.3200 \times 40 g \times 21.30 mL \times 1L/1000mL$

= 0.27264 g

$NaOH\ mass = \frac{mass}{molecular\ weight}$

$= \frac{0.27264\ g}{40g/mol}$

= 0.006816 mol

Now

Moles of $H_2SO_4$ needed  is

$= \frac{0.006816}{2}$

= 0.003408 mol

$Mass\ of\ H_2SO_4 = moles \times molecular\ weight$

$= 0.003408\ mol \times 98g/mol$

= 0.333984 g

Now based on the above calculation

a. Normality of acid is

$= \frac{acid\ mass}{equivalent\ weight \times volume}$

$= \frac{0.333984 g}{49 \times 0.015}$

= 0.4544 N

b. And, the acid solution molarity is

$= \frac{moles}{Volume}$

$= \frac{0.003408 mol}{15\ mL \times 1L/1000\ mL}$

= 0.00005112

=$5.112 \times 10^{-5 M}$

We simply applied the above formulas