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Standard solutions is taught in the note for Volumetric Analysis. Read it below and look for the section on this topic.

Volumetric Analysis

12 min read

Standard solutions

Volumetric analysis finds how much of a substance is in a solution by measuring the volume of another solution that reacts with it exactly. A standard solution is a solution whose concentration is known accurately. A primary standard is a pure, stable solid (for example anhydrous sodium carbonate) that can be weighed accurately to make a standard solution.

Concentration c is the number of moles (n) dissolved in each dm³ (V) of solution. To change mol/dm³ to g/dm³, multiply by the molar mass.
Worked examplePreparing a standard solution

How many grams of anhydrous sodium carbonate (Na₂CO₃, M = 106) are needed to make 250 cm³ of a 0.1 mol/dm³ solution?

  1. Moles needed.
  2. Mass.

Answer: Weigh 2.65 g, dissolve it, and make it up to 250 cm³ in a volumetric flask.

The titration

In a titration a solution from a burette is added slowly to a measured volume of another solution in a conical flask, with an indicator, until the reaction is just complete. This point is the end point. Repeat until two results agree within 0.10 cm³ (concordant results) and take their mean.

Titration apparatusA retort stand holding a burette filled with solution. Below it a conical flask sits on a white tile and contains a measured volume of the other solution with a few drops of indicator. The tap on the burette controls the drops.burette (alkali)tapconical flask: acid + indicatorwhite tileretort stand
The pH changes slowly, then jumps at the end point, and the indicator changes colour.
Open the acid–base titration lab
Run a titration and find the unknown concentration.
TitrationSuitable indicator
Strong acid + strong alkaliMethyl orange or phenolphthalein
Strong acid + weak alkaliMethyl orange
Weak acid + strong alkaliPhenolphthalein

Volumetric calculations

c is concentration, V is volume and n is the number of moles of each substance in the balanced equation (for example HCl + NaOH: n_a = n_b = 1; H₂SO₄ + 2NaOH: n_a = 1 and n_b = 2).
Worked exampleConcentration of an acid

25.0 cm³ of sodium hydroxide of concentration 0.100 mol/dm³ needs 20.0 cm³ of hydrochloric acid to neutralise it. Find the concentration of the acid.

  1. HCl + NaOH → NaCl + H₂O, so the acid and alkali react 1 : 1.
  2. Moles of alkali.
  3. Same moles of acid in 20.0 cm³.

Answer: The acid is 0.125 mol/dm³.

Try itTitration calculation

Enter the alkali (in the flask) and the volume of acid needed. Choose how many moles of alkali react with one mole of acid.

Applications in real life and industry

ApplicationHow titration is used
FoodAcidity of vinegar and fruit juice; vitamin C in juice
MedicineStrength of antacid tablets and drugs
WaterHardness of water and chlorine in a swimming pool
IndustryChecking the concentration of acids and alkalis in the manufacture of soap and fertiliser
EnvironmentAcidity of rain water and soil
Check yourself The end point of a titration is when…
Check yourself 20.0 cm³ of 0.10 mol/dm³ NaOH is neutralised by 10.0 cm³ of HCl. What is the HCl concentration?
Check yourself Why are concordant results needed?

Key terms

Standard solution
A solution of accurately known concentration.
Titration
Adding one solution from a burette to another until they have just reacted.
End point
The point at which the reaction is just complete.
Concentration
Moles of solute per dm³ of solution.

Summary

  • A standard solution has a known concentration; c = n ÷ V.
  • In a titration one solution is added from a burette until the indicator shows the end point; use concordant results.
  • Use c_a V_a / n_a = c_b V_b / n_b with the balanced equation.
  • Titration is used in food, medicine, water testing and industry.
Volumetric calculations