
Equipment used to measure volumes precisely in titrations

The steps in a titration
Example 1: Calculations from titration resultsIn a titration, 25.00 cm3 of 0.05 mol dm-3 hydrochloric acid was neutralised by 8.50 cm3 of sodium hydroxide solution. Calculate the concentration of the sodium hydroxide solution.
Answer
Step 1: Find the number of moles of acid
moles of acid = concentration x volume in dm3
moles of acid = 0.05 x 25/1000 = 1.25 x 10-3 mol
Step 2: Deduce the number of moles of alkali
The equation for the reaction shows the mole ratio is 1:1
HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)
∴ moles of alkali = 1.25 x 10-3 mol
Step 3: Work out the concentration of the alkali
concentration = moles/volume in dm3
concentration = 1.25 x 10-3/0.0085 = 0.15 mol dm-3
Example 2: Calculating the pH in a strong acid-strong base titration50.0 cm3 of 0.10 mol dm3 NaOH is gradually added to 25.0 cm3 of 0.15 mol dm3 hydrochloric acid. Determine the pH after 45 cm3 of NaOH has been added. (Kw = 1 x 10-14 mol2 dm-6 at 298 K)
Answer
Step 1: Find the number of moles of acid
moles of acid = concentration x volume in dm3
moles of acid = 0.15 x 25/1000 = 3.75 x 10-3 mol
Step 2: Deduce the number of moles of alkali added
The equation for the reaction shows the mole ratio is 1:1
HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)
moles of alkali added = 0.10 x 45/1000 = 4.50 x 10-3 mol
∴ moles of alkali in excess = (4.50 x 10-3- 3.75 x 10-3) = 7.5 x 10-4 mol
Step 3: Work out the concentration of the alkali
concentration = moles/volume in dm3
concentration = 7.5 x 10-4/0.070 = 0.0107 mol dm-3
Step 4: Use Kw to find the concentration of H+
Kw = [H+][OH-]
[H+] = Kw /[OH-] = 1.00 x 10-14/0.0107 = 9.35 x 10-13
Step 5: Find the pH
-log[H+] = -log(9.35 x 10-13)
pH = 12.03
You will be asked to perform calculations only on monoprotic acids. A monoprotic acid has only one acidic hydrogen, like HCl. Sulfuric acid, H2SO4, is a diprotic acid.
转载自savemyexams
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