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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

1 Answer

3 votes

Answer:

0.5 mol dm⁻³

Step-by-step 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 *
    (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
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