Making and diluting a standard solution

A-Level (AQA) · unreviewed

Standard solution made from a weighed solid, then a portion diluted.

Worked example

A standard solution was made by dissolving 2.380 g of Na₂CO₃ (M = 106.0 g mol⁻¹) in water and making it up to 250.0 cm³. 25.0 cm³ of it was transferred to another volumetric flask and made up to 100.0 cm³. What is the final concentration?

1. Moles from mass: solve for n(Na₂CO₃)
$n = \dfrac{m}{M}$
= 0.02245 mol
2. Concentration from moles and volume: solve for c of the standard solution
$c = \dfrac{n}{V}$
= 0.08981 mol dm⁻³
3. Dilution: solve for c after dilution
$c_2 = \dfrac{c_1V_1}{V_2}$
= 0.02245 mol dm⁻³

Answer: c(diluted) = 0.0225 mol dm⁻³

Method

n = m/M; c = n/V with V in dm³. Diluting a portion: c₂ = c₁V₁/V₂ (the moles in the portion are unchanged).

Common mistakes

That is the concentration in the flask. The portion was then diluted, which lowers it by V₁/V₂.

Formulas used

Moles from mass
$n = \dfrac{m}{M}$
Concentration from moles and volume
$c = \dfrac{n}{V}$
Dilution
$c_1V_1 = c_2V_2$

Practise this with fresh numbers

Making and diluting a standard solution

Worked example

Method

Common mistakes

Formulas used

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