5.5 - Titration Calculations
Concentration and volume from a titre
- A titration gives you one measured volume and one mean titre.
- To calculate the unknown concentration, you need:
- The concentration of the standard (known) solution.
- The balanced equation, which gives the reacting mole ratio.
- Every volume must be in cubic decimetres (dm3). Convert cm3 to dm3 by dividing by 1,000.
The three steps
Name which solution is the known one before you start - step 1 begins with it, and step 3 ends with the unknown.
- Find the moles of the known solution:
Moles = concentration × volume
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Find the moles of the other solution from the balanced equation. The big numbers in front of the formulae give the reacting ratio.
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Turn those moles into the answer:
concentration (or volume) = moles ÷ volume (or concentration)
Other answers from the same steps
- Grams per cubic decimetre - Work through to mol/dm3 first, then convert the result at the end.
- Identifying an unknown substance - Divide the dissolved mass by the moles found to give the relative formula mass of one mole. That number picks out the substance.
- Degree of hydration of a hydrated salt - Subtract the relative formula mass of the anhydrous salt; divide what's left by 18 to give the number of water molecules in the formula.
Worked example - Concentration of sulfuric acid
A student measured 25.0 cm3 of lithium hydroxide solution of concentration 0.200 mol/dm3 into a conical flask with a volumetric pipette and added a few drops of a suitable indicator. Sulfuric acid of unknown concentration was run in from a burette until the indicator changed colour, and the mean of the concordant titres was 22.0 cm3.
2LiOH(aq) + H2SO4(aq) → Li2SO4(aq) + 2H2O(l)
Calculate the concentration of the sulfuric acid in mol/dm3.
Step 1 - Moles of the known solution
The lithium hydroxide concentration was given, so start with it. Convert its volume to dm3 first.
Volume = 25.0 ÷ 1,000 = 0.0250 dm3
moles of LiOH = concentration × volume = 0.200 × 0.0250 = 0.00500 mol
Step 2 - Moles of the other solution from the equation
The equation shows 2 moles of LiOH reacting with 1 mole of H2SO4, so there were half as many moles of acid as alkali.
moles of H2SO4 = 0.00500 ÷ 2 = 0.00250 mol
Step 3 - Concentration of the sulfuric acid
Its volume was the 22.0 cm3 from the burette, which is 0.0220 dm3.
Concentration = moles ÷ volume = 0.00250 ÷ 0.0220 = 0.113636... mol/dm3
Rounded to three significant figures, matching the precision of the data given, the sulfuric acid has a concentration of 0.114 mol/dm3.
Step 4 - Sense-check
The acid supplied half as many moles as the alkali, packed into a slightly smaller volume, so it should come out a little above half of 0.200 mol/dm3. A result of 0.114 mol/dm3 fits; a value of 0.455 mol/dm3 - more concentrated than the alkali itself - would mean the ratio had been applied upside down.