What is the percentage of sulfurous acid solution obtained by dissolving 1 liter of sulfur dioxide in 10 ml of water?

univerkov | June 10, 2021 | education

SO2 + H2O = H2SO3.
V (SO2) = 1 l.
V (H2O) = 10 ml = 10 g.
We find the amount of the substance of water and sulfur dioxide.
n = m / M = 10 g / 18 g / mol = 0.56 mol.
n (SO2) = V / Vm = 1 L / 22.4 L / mol = 0.045 mol.
This means that SO2 is in short supply, therefore, sulfuric acid is equal to 0.045 mol.
n (H2O) = 0.56 mol – 0.045 mol = 0.515 mol.
W in solution (H2SO3) = m (H2SO3) / m (solution) = m (H2SO3) / (m (H2SO3) + m (H2O) rest) =
= (n (H2SO3) × M (H2SO3)) / (n (H2SO3) × M (H2SO3) + n (H2O) rest × M (H2O) rest)) =
= (0.045 × 82g / mol) / (0.045 × 82 + 0.515 × 18) = 3.69 / (3.69 + 9.27) = 3.69 / 12.96 = 0.284; 28.4%.
Answer: W (H2SO3) = 28.4%.

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