# Complete neutralization of 110 g of sulfuric acid solution required 80 g of 10%

**Complete neutralization of 110 g of sulfuric acid solution required 80 g of 10% sodium hydroxide solution. Calculate the mass fraction of acid in the original solution.**

Find the mass of NaOH in solution.

W = m (substance): m (solution) × 100%,

hence m (substance) = (m (solution) × w): 100%.

m (substance) = (80 g × 10%): 100% = 8 g.

Find the amount of NaOH.

n = m: M.

M (NaOH) = 40 g / mol.

n = 8 g: 40 g / mol = 0.2 mol.

Let’s find the quantitative ratios of substances.

2NaOH + H2SO4 = Na2SO4 + 2H2O.

For 2 mol of NaOH, there is 1 mol of H2SO4.

Substances are in quantitative ratios of 2: 1.

The amount of the substance H2SO4 is 2 times less than that of NaOH.

n (H2SO4) = ½ n (NaOH) = 0.2: 2 = 0.1 mol.

Let’s find the mass of H2SO4.

M (H2SO4) = 98 g / mol.

m = n × M.

m = 98 g / mol × 0.1 mol = 9.8 g.

W = m (substance): m (solution) × 100%,

W = (9.8: 110) × 100% = 8.91%

Answer: 8.91%.