Question

You are given 5.00 mL of an H2SO4 solution of unknown concentration. You divide the 5.00-mL sample into five 1.00-mL samples and titrate each separately with 0.1000 M NaOH. In each titration the H2SO4 is completely neutralized. The average volume of NaOH solution used to reach the endpoint is 12.6 mL. What was the concentration of H2SO4 in the 5.00-mL sample?

Answer 1

Equation $\mathrm{N}_1\mathrm{V}_1 = \mathrm{N}_2\mathrm{V}_2$ can help to solve this task. N is normality, Normality (N) is another ratio that relates the amount of solute to the total volume of solution.

It is defined as the number of equivalents per liter of solution:

Normality = number of equivalents / 1 L of solution

There is a very simple relationship between normality and molarity:

$\mathrm{N} = \mathrm{n} \times \mathrm{M}$ (where n is an integer)

n for $\mathrm{H}_2\mathrm{SO}_4$ is 2 (2H⁺)

So $\mathrm{N}_{\mathrm{acid}} = 2\mathrm{M}_{\mathrm{acid}}$

n for NaOH is 1 N = M

If $\mathrm{V}_{\mathrm{acid}} = 1\mathrm{ml}$

$\mathrm{V}_{\mathrm{base}} = 12.6\mathrm{ml}$

$\mathrm{N}_{\mathrm{base}} = 0.1000$

N of acid can be found as:

$\mathrm{N}_{\mathrm{acid}} = \mathrm{N}_{\mathrm{base}} \cdot \mathrm{V}_{\mathrm{base}} / \mathrm{V}_{\mathrm{acid}} = 0.1000 * 12.6 / 1 = 1.26$

Molarity is: $\mathrm{N} / 2 = 1.26 / 2 = \mathbf{0.63}\mathbf{M}$