Sodium hydroxide (NaOH) reacts with sulfuric acid (H2SO4) according to the equation: NaOH(aq) + H2SO4(aq) -> Na2SO4(aq) + 2H2O(l). In an experiment to make some sodium sulfate (Na2SO4), a chemist neutralizes 250 cm3 of 0.5 M (0.5 mol dm-3) H2SO4 with solid NaOH. The resulting solution is evaporated to dryness to recover the Na2SO4. Atomic masses: Na = 23, H = 1, S = 32, O = 16. – How many moles of H2SO4 are in 250 cm3 of 0.05 M sulfuric acid? – How many moles of NaOH does the chemist need to react with this amount of H2SO4? – What is the mass of NaOH required? – What is the number of moles of Na2SO4 produced? – What is the concentration of the Na2SO4 solution when the reaction is complete? – What mass of Na2SO4 would be produced after drying the product (assume 100% yield)? – A small experimental point: Why would the chemist need to be very careful in carrying out this reaction?
The Correct Answer and Explanation is:
Correct Answers:
- Moles of H₂SO₄ in 250 cm³ of 0.5 M solution:
Volume = 250 cm³ = 0.250 dm³
Moles = Molarity × Volume = 0.5 mol/dm³ × 0.250 dm³ = 0.125 mol - Moles of NaOH needed:
Balanced equation:
2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
From the ratio, 1 mole of H₂SO₄ reacts with 2 moles of NaOH
So, moles of NaOH = 2 × 0.125 = 0.25 mol - Mass of NaOH required:
Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol
Mass = moles × molar mass = 0.25 × 40 = 10 grams - Moles of Na₂SO₄ produced:
From the equation, 1 mole H₂SO₄ produces 1 mole Na₂SO₄
So, moles of Na₂SO₄ = 0.125 mol - Concentration of Na₂SO₄ solution:
Total volume = 250 cm³ = 0.250 dm³
Concentration = moles / volume = 0.125 mol / 0.250 dm³ = 0.5 mol/dm³ - Mass of Na₂SO₄ produced after drying:
Molar mass of Na₂SO₄ = (2 × 23) + 32 + (4 × 16) = 46 + 32 + 64 = 142 g/mol
Mass = 0.125 mol × 142 g/mol = 17.75 grams - Why the chemist needs to be careful:
The reaction between NaOH and H₂SO₄ is exothermic. This means it releases heat. If the solid NaOH is added too quickly to the acid, the solution can become hot rapidly and boil or splash, which may be dangerous. The chemist should add the solid slowly with stirring to control the reaction and avoid overheating. Proper lab safety such as goggles and gloves should be used to prevent burns from hot acid or alkali.
Explanation
This reaction is a classic acid–base neutralization between sulfuric acid (H₂SO₄) and sodium hydroxide (NaOH), forming a salt (sodium sulfate) and water. To determine the correct quantities for this reaction, we start by converting the volume of the acid solution into liters, since molarity (mol/dm³) relates to liters. For a 0.5 M solution and 0.250 dm³ volume, the moles of acid are 0.125.
The balanced chemical equation shows a 1:2 molar ratio between H₂SO₄ and NaOH, meaning two moles of NaOH are needed per mole of acid. Therefore, 0.25 moles of NaOH are required to fully react with 0.125 moles of H₂SO₄.
To find the mass of NaOH, we use its molar mass, which is the sum of sodium, oxygen, and hydrogen atomic masses. With 0.25 moles needed, we calculate a required mass of 10 grams.
Since each mole of H₂SO₄ produces one mole of Na₂SO₄, the amount of product formed is 0.125 moles. The concentration of the final sodium sulfate solution remains 0.5 mol/dm³ before evaporation, as no extra volume is added. Once the solution is evaporated, solid Na₂SO₄ remains. Using its molar mass, the dry mass of sodium sulfate is found to be 17.75 grams.
Safety during this reaction is vital because it is exothermic. If NaOH is added too quickly, the heat can cause splattering or burns. Handling chemicals cautiously and using personal protective equipment ensures the procedure remains safe and controlled.
