Question

It takes 38 mL of 0.75 M NaOH solution to completely neutralize 155 mL of a sulfuric acid solution (H2SO4). What is the concentration of the H2SO4 solution?

Answer 1

Answer:

Molarity = 0.09M

Explanation:

no. of moles of NaOH = M x Volume(Litre)

                                     = 0.75 x 38 / 1000

                                     = 0.0285 moles

2 moles of NaOH = 1 mole of H2SO4

no. of moles of H2SO4 = 0.0285 / 2

                                       = 0.014 moles

Molarity = 0.014 / 0.155

              = 0.09 M

Answer 2

The concentration of sulfuric acid is 0.95 M.

Explanation:

Given:

Volume of $H_2SO_4$, $V_1 = 15mL$

The volume of NaOH, $V_2 = 38 mL$

Molarity of NaOH, $M_2 = 0.75 M$

To find: Molarity of $H_2SO_4 (M_1)$

Formula Used: $n_1M_1V_1 = n_2M_2V_2$ where $n_1$ and $n_2$are the number of proton/ hydroxides transferred during the reaction

Solution:

$n_1 = 2$ (since each molecule of $H_2SO_4$ can give 2 protons)

$n_2 = 1$ (since each molecule of NaOH can give 1 hydroxide ion)

Substituting the values in the formula, we obtain

$n_1M_1V_1 = n_2M_2V_2$

$2\times M_1 \times 15 mL = 1 \times 0.75 M \times 38 mL$

$M_1 = \frac{1 \times 0.75 M \times 38 mL}{2 \times 15 mL}$

$M_1$ = 0.95 M

Thus, the concentration of sulfuric acid is 0.95 M.