Determine the volume of sodium hydroxide solution with a mass fraction of alkali 18% (density 1.25 g / ml)

| June 15, 2021 | education

Determine the volume of sodium hydroxide solution with a mass fraction of alkali 18% (density 1.25 g / ml), which is required to neutralize aminoacetic acid with a mass of 22.5 g.

Find the amount of aminoacetic acid substance by the formula:

n = m: M.

M (CH2 (NH2) COOH) = 75 g / mol.

n = 22.5 g: 75 g / mol = 0.3 mol.

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

CH2 (NH2) COOH + NaOH = CH2 (NH2) COONa + H2O.

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

The amount of substance will be equal.

n (CH2 (NH2) COOH) = n (NaOH) = 0.3 mol.

Let’s find the mass of NaOH.

M (NaOH) = 40 g / mol.

m = n × M.

m = 40 g / mol × 0.3 mol = 12 g.

The volume of the solution is determined by the formula:

V = m: p.

Let’s find the mass of the solution.

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

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

m (solution) = (12g: 18%) × 100% = 66.67 g.

Find the volume of the solution.

V = 66.67 g: 1.25 g / ml = 53.34 ml.

Answer: 53.34 ml.



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