Calculations in chemistry (Higher) - AQAVolumes of solutions in reactions

Calculating amounts from concentration and volume

The amount in moles of a solute in a given volume of solution can be calculated if the concentration of the solution is known. The mass of solute can then be calculated.

Rearranging the equation for concentration below:

Gives:

Amount of solute in mol = concentration in mol/dm³ × volume in dm³

Example

Calculate the amount of sodium hydroxide, NaOH, in 25.0 cm³ of solution of concentration 0.1 mol/dm³.

Converting the volume from cm³ to dm³, 25.0 cm³ = 25.0 ÷ 1000 = 0.025 dm³.

Amount of NaOH in mol = concentration in mol/dm³ × volume in dm³

= 0.1 mol/dm³ × 0.025 dm³

= 0.0025 mol

Calculating masses from concentration and volume

If the amount in mol of a solute in a given volume of solution is known, its mass can also be calculated.

More concentration calculations

If the volumes of two solutions that react completely are known, and the concentration of one of the solutions is known, then the concentration of the other solution can be calculated.

Example

25.0 cm³ of hydrochloric acid, HCl, of concentration 1.0 mol/dm³ reacts with 20.0 cm³ of sodium hydroxide, NaOH, solution. Calculate the concentration of the NaOH solution. The equation for the reaction is HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

Converting volumes from cm³ to dm³:

• volume of HCl = 25.0 cm³ = 25.0 ÷ 1000 = 0.025 dm³
• volume of NaOH = 20.0 cm³ = 20.0 ÷ 1000 = 0.020 dm³

Substituting into the equation to find the amount of HCl in the given volume:

Amount of HCl in mol = concentration in mol/dm³ × volume in dm³

= 1.0 mol/dm³ × 0.025 dm³

= 0.025 mol

The equation shows that 1 mol of HCl reacts with 1 mol of NaOH. So the amount of NaOH in 20.0 cm³ is 0.025 mol.

Substituting into the equation to find the concentration of NaOH:

= 1.25 mol/dm³