Chemistry OCR F321 past paper questions and answers on acids
- Created by: KatieL
- Created on: 15-12-12 19:45
Acids 1 (i)
1. A student carries out experiments using acids, bases and salts.
Calcium nitrate, Ca(NO3)2, is an example of a salt.
The student prepares a solution of calcium nitrate by reacting dilute nitric acid, HNO3, with the base calcium hydroxide, Ca(OH)2.
(i) Why is calcium nitrate an example of a salt?
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[1]
Acids 1 (i) Mark scheme
1. (i) The H+ ion in an (nitric) acid has been replaced by a metal ion
OR by a Ca2+ ion
DO NOT ALLOW it has been produced by the reaction of an acid and a base as this is stated in the question.
IGNORE references to replacement by NH4+ ions or positive ions.
ALLOW H OR Hydrogen for H+;
DO NOT ALLOW Hydrogen atoms
ALLOW Ca OR Calcium for Ca2+.
DO NOT ALLOW Calcium atoms
ALLOW ‘metal’ for ‘metal ion
1
Acids 1 (ii)
(ii) Write the equation for the reaction between dilute nitric acid and calcium hydroxide. Include state symbols.
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[2]
Acids 1 (ii) Mark scheme
(ii) 2HNO3(aq) + Ca(OH)2(aq) → Ca(NO3) 2 (aq)+ 2H2O(l)
Formulae
Balance AND states
ALLOW multiples
ALLOW (aq) OR (s) for Ca(OH) 2
2
Acids 1 (iii)
(iii) Explain how the hydroxide ion in aqueous calcium hydroxide acts as a base when it neutralises dilute nitric acid.
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[1]
Acids 1 (iii) Mark scheme
(iii) Accepts a proton OR accepts H+
ALLOW H+ + OH–→ H2O
ALLOW OH– reacts with H+ OR OH– takes H+
ALLOW OH– ‘attracts’ H+ if ‘to form water’ is seen
DO NOT ALLOW OH– neutralises H+ (‘neutralises’ is in the question)
1
Acids 2 (a) (i)
(a) A student carries out a titration to find the concentration of some sulfuric acid.
The student finds that 25.00 cm3 of 0.0880 mol dm–3 aqueous sodium hydroxide, NaOH, is neutralised by 17.60 cm3 of dilute sulfuric acid, H2SO4.
H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)
(i) Calculate the amount, in moles, of NaOH used.
answer = ................................... mol
[1]
Acids 2 (a) (i) Mark scheme
(a) (i) Calculate correctly (0.0880 x 25)/1000 = 2.20 x 10–3
OR 0.00220 mol
ALLOW 0.0022 OR 2.2 × 10–3 mol
Acids 2 (a) (ii)
(ii) Determine the amount, in moles, of H2SO4 used.
answer = ................................... mol
[1]
Acids 2 (a) (ii) Mark scheme
(ii) Calculates correctly 0.00220/2= 1.10 × 10–3 mol
OR 0.00110 mol
ALLOW 0.0011 OR 1.1 × 10–3 mol
ALLOW ECF for answer (i)/2 as calculator value or correct rounding to 2 significant figures or more but ignore trailing zeroes
Acids 2 (a) (iii)
(iii) Calculate the concentration, in mol dm–3, of the sulfuric acid.
answer = ................................... mol dm–3
Acids 2 (a) (iii) Mark scheme
(iii) (0.00110 x 1000)/ 17.60 = 0.062 mol dm–3
OR 6.25 × 10–2 mol dm–3
ALLOW 0.063 OR 6.3 × 10–2 mol dm–3
ALLOW ECF for answer (ii) × 1000/17.60
OR
ECF from (i) for answer (i)/2 × 1000/17.60 as calculator value or correct rounding to 2 significant figures or more but ignore trailing zeroes
Acids 2 (b) (i)
(b) After carrying out the titration in (a), the student left the resulting solution to crystallise. White crystals were formed, with a formula of Na2SO4•x H2O and a molar mass of 322.1 g mol–1.
(i) What term is given to the ‘•x H2O’ part of the formula?
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Acids 2 (b) (i) Mark scheme
(b) (i) (The number of) Water(s) of crystallisation
IGNORE hydrated OR hydrous
Acids 2 (b) (ii)
(ii) Using the molar mass of the crystals, calculate the value of x.
answer = ...................................
[2]
Acids 2 (b) (ii) Mark scheme
(ii) 142.1
ALLOW 142
ALLOW Mr expressed as a sum
ALLOW ECF from incorrect Mr and x is calculated correctly
x = (322.1 - 142.1)/18.0 = 10
ALLOW ECF values of x from nearest whole number to calculator value
ALLOW 2 marks if final answer is 10 without any working
2
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