Question

A volume of 10cm3 of sulfuric acid of unknown concentration neutralized 25cm3 of sodium hydroxide solution of concentration 0.4 mol dm-3. The concentration of the acid is:
0.1 mol dm-3

0.25 mol dm-3

0.4 mol dm-3

0.5 mol dm-3

Answer 1

Answer:

0.5 mol dm⁻³

Explanation:

The reaction that takes place is:

• H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

First we convert the given volumes to dm³(1000cm³ = 1 dm³):

• 10 cm³ / 1000 = 0.010 dm³

• 25 cm³ / 1000 = 0.025 dm³

Now we calculate how many NaOH moles reacted, using the given volume and concentration:

• 0.4 mol/dm³ * 0.025 dm³ = 0.01 mol NaOH

Then we convert NaOH moles into H₂SO₄ moles, using the stoichiometric coefficients:

• 0.01 mol NaOH * $\frac{1molH_2SO_4}{2molNaOH}$ = 0.005 mol H₂SO₄

Finally we calculate the concentration of H₂SO₄:

• 0.005 mol H₂SO₄ / 0.010 dm³ = 0.5 mol/dm³