Mr Cole Chemistry

Titration Calculation Questions

Titration Calculations

Amount of Substance Worksheet

Easy Titration Calculations

20.0cm3 of 0.5moldm-3 hydrochloric acid was titrated against sodium hydroxide. 13cm3 of sodium hydroxide was required to neutralise the hydrochloric acid. What was the concentration of the sodium hydroxide?
Working Out
HCl + NaOH → NaCl + H2O
Step 1: Calculate moles of HCl n = c × v
n(HCl) = 0.5 × 0.0200 = 0.01 mol
Step 2: Determine Ratio and Moles of NaOH Ratio is 1:1
n(NaOH) = 0.01 mol
Step 3: Calculate Concentration c = n / v
c(NaOH) = 0.01 / 0.013 = 0.7692…
Answer = 0.77 mol dm-3
25.0cm3 of sulfuric acid was titrated against 2moldm-3 potassium hydroxide. 21.3cm3 of potassium hydroxide was required to neutralise the sulfuric acid. What is the concentration of the acid?
Working Out
H2SO4 + 2KOH → K2SO4 + 2H2O
Step 1: Calculate moles of KOH n(KOH) = 2 × 0.0213 = 0.0426 mol
Step 2: Ratio Ratio Acid:Base is 1:2
n(H2SO4) = 0.0426 / 2 = 0.0213 mol
Step 3: Calculate Concentration c = 0.0213 / 0.0250 = 0.852
Answer = 0.852 mol dm-3
30cm3 of 0.4moldm-3 sodium hydrogen carbonate were titrated against 0.5moldm-3 ethanoic acid. What volume of ethanoic acid is the expected titre?
Working Out
CH3COOH + NaHCO3 → CH3COONa + H2O + CO2
Step 1: Calculate moles of Base n(NaHCO3) = 0.4 × 0.030 = 0.012 mol
Step 2: Ratio Ratio is 1:1
n(Acid) = 0.012 mol
Step 3: Calculate Volume v = n / c = 0.012 / 0.5 = 0.024 dm3
Answer = 24.0 cm3

Medium Difficulty Titration Calculations

10.2g of sodium carbonate was dissolved in 250cm3 of water. 25.0cm3 samples of the sodium carbonate were titrated against a monoprotic acid of unknown concentration. The table below is a record of the results.

Trial Rough 1 2 3 4
Starting volume / cm3 100 82.2 65.6 48.7 32.4
Ending volume / cm3 82.2 65.6 48.7 32.4 16
Change in volume / cm3 17.8 16.6 16.9 16.3 16.4
a) Outline the method for obtaining a solution of known concentration of sodium carbonate.
  • Place weighing boat / weighing bottle on a balance and add the required mass of sodium carbonate.
  • Record the mass of the boat and the solid.
  • Pour the sodium carbonate into a beaker and then reweigh the weighing boat.
  • The difference tells you the mass of sodium carbonate in the beaker.
  • It is possible to simply put the beaker on the balance and add the required mass of sodium carbonate rather than using a boat.
  • Next, add enough water to dissolve the solid and do so using a stirring rod.
  • Next, pour the solution into a volumetric flask using a flask, and then rinse the beaker, stirring rod and funnel with distilled water and pour the rinsings into the flask as well.
  • Add water up to the line on the volumetric flask and invert several times to mix.
b) Calculate the concentration of the acid.
Mean Titre Average of concordant results (Trials 3 and 4).
(16.3 + 16.4) / 2 = 16.35 cm3
Step 1: Moles of Na2CO3 (Total) Mr(Na2CO3) = (23 × 2) + 12 + (16 × 3) = 106.0
n = 10.2 / 106.0 = 0.0962 mol (in 250 cm3)
Step 2: Moles in 25 cm3 sample n(sample) = 0.0962 / 10 = 0.00962 mol
Step 3: Stoichiometry (Monoprotic Acid HA) Na2CO3 + 2HA → 2NaA + H2O + CO2
Ratio Carbonate:Acid is 1:2.
n(Acid) = 0.00962 × 2 = 0.01924 mol
Step 4: Concentration c = 0.01924 / 0.01635 = 1.177…
Answer = 1.18 mol dm-3
100g Arsenic acid hemi hydrate is dissolved in 1.00dm3 of water. 25.0cm3 samples of this solution are titrated against potassium hydrogen carbonate solution. The table below is a record of the results. What is the concentration of the potassium hydrogen carbonate?
Working Out
Moles of arsenic acid: 100 / 150.9 = 0.663 moles
Moles of arsenic acid in 25cm3: 0.663/40 = 0.0166
Volume of KHCO3: 40.3cm3
Moles of KHCO3: 0.0166 x 3 = 0.0497
Conc of KHCO3: 0.0497 / 0.0403 = 1.23 mol dm-3
14.1g of an unlabelled solid acid are dissolved in 250cm3 of water. A 25cm3 sample of the acid is titrated against 1moldm-3 NaOH, requiring 22.45cm3 to neutralise. Suggest an identity of the acid.
Working Out
Note: 14.1g of solid gives you an answer of 62.8, which is HNO3
Moles of NaOH: 1 x 0.02245 = 0.02245
Moles of acid (if monoprotic): 0.02245 moles of acid in 25cm3
Moles of acid (if monoprotic): 0.2245 moles of acid in 250cm3
Mr of acid (if monoprotic): 14.1 / 0.2245 = 62.8 = 63 = close enough to nitric acid
A 5.00 g mixture of propan-1-ol and propan-2-ol was reacted with excess potassium dichromate and 5cm3 of 2moldm-3 of sulfuric acid under reflux. The propanoic acid produced was titrated against 2moldm-3 NaOH where it required 36.4cm3 to fully neutralise the propanoic acid. What proportion of the mixture, by mass, was propan-1-ol.
Working Out

Note: The NaOH neutralises BOTH the propanoic acid produced AND the sulfuric acid added at the start.

Step 1: Moles of H2SO4 added initially n = c × v
n(H2SO4) = 2 × 0.005 = 0.01 mol
Step 2: Moles of NaOH neutralising H2SO4 Ratio H2SO4:NaOH is 1:2
n(NaOH for H2SO4) = 0.01 × 2 = 0.02 mol
Step 3: Total Moles of NaOH used n(Total NaOH) = 2 × 0.0364 = 0.0728 mol
Step 4: Moles of NaOH reacting with Propanoic Acid n(NaOH for Acid) = 0.0728 – 0.02 = 0.0528 mol
Step 5: Mass of Propan-1-ol Ratio NaOH:Propanoic Acid:Propan-1-ol is 1:1:1
n(Propan-1-ol) = 0.0528 mol
Mr(C3H8O) = 60.0
Mass = 0.0528 × 60.0 = 3.168 g
Step 6: Percentage by Mass % = (3.168 / 5.00) × 100 = 63.36%
Answer = 63.4%
An unknown solid was tested to determine its identity. It was found that it reacted with hydrochloric acid, effervescing. The gas produced was tested but produced inconclusive results due to the small amounts produced. The solid was added to water but it did not react. The conductivity of the solid was tested and it was found to not conduct electricity. A 5.50g sample of the solid was titrated against 2.00moldm-3 hydrochloric acid, requiring 39.9cm3 of acid to neutralise the solid. Suggest an identity of the solid.
Working Out
Step 1: Moles of HCl n(HCl) = 2.00 × 0.0399 = 0.0798 mol
Step 2: Determine Formula Mass based on Stoichiometry Effervescence suggests a carbonate (CO2) or metal (H2).
Assuming a divalent metal carbonate (MCO3) or metal (M), reaction ratio with HCl is 1:2.
n(solid) = 0.0798 / 2 = 0.0399 mol
Step 3: Calculate Molar Mass Mr = 5.50 / 0.0399 = 137.8 g mol-1
Answer Potassium Carbonate (K2CO3)
Mr = 138.2. This fits the calculation (137.8) almost perfectly. Being an ionic solid, it does not conduct electricity, fitting the description better than Barium metal.
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