Calculate the volume of a 20% sodium hydroxide solution with a density of 1.22 g / m3

| May 22, 2021 | education

Calculate the volume of a 20% sodium hydroxide solution with a density of 1.22 g / m3 that will be required to neutralize 45 g of acetic acid.

Let’s find the amount of substance СН3СООН according to the formula:

n = m: M.

M (CH3COOH) = 60 g / mol.

n = 45 g: 60 g / mol = 0.75 mol.

Let’s compose the reaction equation, find the quantitative ratios of substances.

NaOH + CH3COOH → CH3COONa + H2O.

According to the reaction equation, there is 1 mol of NaOH for 1 mol of acid. Substances are in quantitative ratios 1: 1.

The amount of substance will be the same.

n (CH3COOH) = n (NaOH) = 0.75 mol.

Let’s find the mass of NaOH by the formula:

m = n × M,

M (NaOH) = 40 g / mol.

m = 0.75 mol × 40 g / mol = 30 g.

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

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

m (solution) = (30 g: 20%) × 100% = 150 g.

Find the volume of the NaOH solution.

V = m: p.

V = 150 g: 1.22 g / m3 = 122.95 m3.

Answer: 122.95 m3.



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