WJEC Chemistry for AS Level Student Book: 2nd Edition (Draft)

 1.3 Chemical calculations

Study point

Worked example A 25.0cm 3 sample of aqueous sodium hydroxide was exactly neutralised by 21.0cm 3 of 0.150mol dm −3 sulfuric acid. Calculate the concentration of the sodium hydroxide solution in (a) mol dm −3 (b) gdm −3 . H 2 SO 4 + 2NaOH →  Na 2 SO 4 + 2H 2 O Step 1 Amount in moles of H 2 SO 4  = 0.150 × 0.021 = 3.15 × 10 −3  mol (Divide 21.0cm 3 by 1000 to change it into dm 3 ) Step 2 From the equation, 1mol H 2 SO 4 requires 2mol NaOH 3.15 × 10 −3  mol H 2 SO 4 require 6.30 × 10 −3  mol NaOH Step 3 (a) Concentration of NaOH = 6.30 × 10 −3 0.025  = 0.252mol dm −3 (Divide 25.0cm 3 by 1000 to change it into dm 3 ) (b) M r NaOH = 40.0 Concentration NaOH = 40.0 × 0.252 = 10.1gdm −3 Back-titration In a back-titration, the amount of excess reactant (e.g. acid) unused at the end of a reaction is found and so the amount used can be calculated. Using the stoichiometric ratios of the acid and base allows us to calculate the exact amount of the second reactant (e.g. base) that has reacted. Worked example A sample containing ammonium sulfate was warmed with 100cm 3 of 1.00moldm −3 sodium hydroxide solution. After all the ammonia had been evolved, the excess of sodium hydroxide solution was neutralised by 50.0cm 3 of 0.850moldm −3 hydrochloric acid. What mass of ammonium sulfate did the sample contain? The two reactions taking place are (i) The reaction between the ammonium sulfate and sodium hydroxide (NH 4 ) 2 SO 4 (s) + 2NaOH(aq) →  2NH 3 (g) + Na 2 SO 4 (aq) + 2H 2 O(l) (ii) The neutralisation of the sodium hydroxide NaOH(aq) + HCl(aq) →  NaCl(aq) + H 2 O(l) Step 1 Calculate the amount in moles of HCl used in the neutralisation, this will give the amount in moles of NaOH unused in reaction (i) Amount in moles of HCl = 0.850 × 50.0 1000  = 0.0425mol From equation (ii) mole ratio of HCl :NaOH is 1:1 Therefore amount of moles NaOH unused is 0.0425 Step 2 Calculate the amount in moles of NaOH that reacted with the (NH 4 ) 2 SO 4 Initial amount of NaOH = 1.00 × 100 1000  = 0.100mol Amount in moles of NaOH used in reaction (i) = 0.100 − 0.0425 = 0.0575mol Step 3 Calculate the amount in moles of (NH 4 ) 2 SO 4 used From equation (i) mole ratio of NaOH: (NH 4 ) 2 SO 4 is 2:1 Therefore 0.0575 moles NaOH react with 0.0288 moles (NH 4 ) 2 SO 4 Step 4 Calculate the mass of (NH 4 ) 2 SO 4 in the sample Molar mass of (NH 4 ) 2 SO 4 is 132gmol −1 Mass of (NH 4 ) 2 SO 4  = 0.0288 × 132 = 3.80g

In acid–base titrations there are five things you need to know: 1 The volume of the acid solution. 2 The concentration of the acid solution. 3 The volume of the base solution. 4 The concentration of the base solution. 5 The equation for the reaction. If you know four of these, you can calculate the fifth.

Knowledge check 25.0cm 3 of hydrochloric acid are neutralised by 18.5cm 3 of a 0.200mol dm –3 solution of sodium hydroxide. Calculate the concentration of the acid.

14

Knowledge check 22.60 cm 3 of a 0.0956 mol dm –3 solution of nitric acid react with 24.00 cm 3 of barium hydroxide solution to form barium nitrate and water. Calculate the concentration of the barium hydroxide. 15

43 DRAFT