Calculating a volume - higher tier - Quantitative chemistry - (CCEA) - GCSE Chemistry (Single Science) Revision - CCEA

Calculating a volume - higher tier

Worked example

25.0 cm³ of 0.3 mol/dm³ sodium hydroxide solution is exactly neutralised by 0.1 mol/dm³ sulfuric acid. Calculate the volume of sulfuric acid required.

Step 1: Calculate the number of moles of sodium hydroxide in moles

Volume of sodium hydroxide solution = 25.0 ÷ 1000 = 0.025 dm³

Number of moles of sodium hydroxide = concentration × volume

Number of moles of sodium hydroxide = 0.3 mol/dm³ × 0.025 dm³

= 0.0075 mol

Step 2: Find the number of moles of sulfuric acid in moles

The balanced symbol equation is:

2NaOH(aq) + H₂SO₄(aq) → Na₂SO₄(aq) + 2H₂O(l)

∴ the mole ratio NaOH:H₂SO₄ is 2:1.

Therefore 0.0075 mol of NaOH reacts with (0.0075 ÷ 2) = 0.00375 mol of H₂SO₄

Step 3: Calculate the volume of sulfuric acid

Rearrange:

Concentration in mol/dm³ = \(\frac{moles~of~solute}{volume~in~dm^3}\)

Volume in dm³ = \(\frac{moles~of~solute}{Concentration~in~mol/dm^3}\)

Volume in dm³ = \(\frac{0.00375}{0.1}\)

= 0.0375 dm³ (37.5 cm³)

Alternatively using volumes in cm³

moles of sodium hydroxide = \(\frac{volume (cm^3 ) × concentration}{1000} = \frac{25.0 × 0.3}{1000}\)= 0.0075 mol

moles of sulfuric acid = \(\frac{0.0075}{2}\)= 0.00375 mol (as 2:1 ratio of NaOH:H₂SO₄)

Rearranging the equation above gives:\(volume = \frac {moles × 1000}{concentration}\)

\(volume~of~sulfuric~acid = \frac {moles × 1000}{concentration} = \frac{0.00375 × 1000}{0.1}\)

= 37.5 cm³

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Quantitative chemistry - (CCEA)Calculating a volume - higher tier