Chemistry Nov/Dec 2013-ALT

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Question 1
In a titration experiment, 22.50 cm3 of an acid solution A containing 10.6 g of NaHSO4 per dm3 reacted with 25.0 cm3 of solution B containing X g of NaOH per dm3. The equation for the reaction is: (a) From the information given above, calculate the: (i) concentrate of A in moldm– 3; (ii) concentrate of B in moldm– 3; (iii) value of X; (iv) mass of Na2SO4 formed during the reaction. [ H = 1.00; 0 = 16.0; Na = 23.0; S = 32.0] (16 marks) (b) (i) Name a suitable indicator for the titration experiment. (ii) State the apparatus used to measure the volume of solution: I. A; II B; (3 marks) In a titration experiment, 22.50 cm3 of an acid solution A containing 10.6 g of NaHSO4 per dm3 reacted with 25.0 cm3 of solution B containing X g of NaOH per dm3. The equation for the reaction is: NaHSO4(aq) + NaOH(aq) Na2SO4(aq) + H2(l) (a) From the information given above, calculate the: (i) concentrate of A in moldm– 3; (ii) concentrate of B in moldm– 3; (iii) value of X; (iv) mass of Na2SO4 formed during the reaction. [ H = 1.00; 0 = 16.0; Na = 23.0; S = 32.0] (16 marks) (b) (i) Name a suitable indicator for the titration experiment. (ii) State the apparatus used to measure the volume of solution: I. A; II B; (3 marks)

Question 1

In a titration experiment, 22.50 cm3 of an acid solution A containing 10.6 g of NaHSO4 per dm3 reacted with 25.0 cm3 of solution B containing X g of NaOH per dm3. The equation for the reaction is:

(a) From the information given above, calculate the:
(i) concentrate of A in moldm– 3;
(ii) concentrate of B in moldm– 3;
(iii) value of X;
(iv) mass of Na2SO4 formed during the reaction. [ H = 1.00; 0 = 16.0; Na = 23.0; S = 32.0] (16 marks)

(b) (i) Name a suitable indicator for the titration experiment.
(ii) State the apparatus used to measure the volume of solution: I. A; II B; (3 marks)

In a titration experiment, 22.50 cm3 of an acid solution A containing 10.6 g of NaHSO4 per dm3 reacted with 25.0 cm3 of solution B containing X g of NaOH per dm3. The equation for the reaction is: NaHSO4(aq) + NaOH(aq) Na2SO4(aq) + H2(l)

(a) From the information given above, calculate the:

(i) concentrate of A in moldm– 3;
(ii) concentrate of B in moldm– 3;
(iii) value of X;
(iv) mass of Na2SO4 formed during the reaction. [ H = 1.00; 0 = 16.0; Na = 23.0; S = 32.0] (16 marks)

(b)

(i) Name a suitable indicator for the titration experiment.
(ii) State the apparatus used to measure the volume of solution: I. A; II B; (3 marks)

ANSWER

In (a), majority of the candidates lost mark for writing the unit of molar mass in “g” instead of in “g mol”. Although majority of the candidates were able to calculate the concentration of both the acid and base in moldm– 3, a lot of candidate lost marks for not expressing their answers to 3 significant figures.

In (b), candidates were able to name suitable indicators for the titration and they were also able to name the apparatus used to measure the volume of the solution of the acid and base.

The required answers include: (a) (i) Molar mass NaHSO4 = [23.0 + 1.00 + 32.0 + 64.0] = 120gmol-1 (No score for wrong Unit)
Conc. of A = 10.6/120
= 0.0883mol dm-3 (correct evaluation to 3sig.figure) (No score for wrong Unit).

(ii) Reaction: NaHSO4(aq) + NaOH(aq) ? Na2SO4(aq) + H2O(l)
From the reaction equation CAVA / CBVB = 1 /1
CB = CAVA / VB
CB = (0.0883 X 22.5) / 25.0= 0.0795mol dm-3(No score for wrong Unit)

iii) Molar mass of NaOH = (23.0 + 16.0 + 1.00) = 40gmol-1 (No score for wrong Unit)
Mass of NaOH gdm-3 = 40 x 0.0795
= 3.18gdm-3
Value of X = 3.18

(iv) Amount of NaOH in 25.0cm3 of B
= (25.0 x 0.0795)/1000
= 0.00199 mol
From the reaction equation
1 mol NaOH gives 1 mol Na2SO4
0.00199 mol NaOH = 0.00199 mol Na2SO4

Molar mass of Na2SO4 = (2 x 23.0) + 32.0 + (4 x 16.0)
= 142gmol-1
Mass of Na2SO4 = 142 x 0.00199
= 0.283g

Alternative Method 2 to (iv) Mass of NaOH in 25.0cm3 of 0.0795moldm-3
m = MCV
= 40 x 0.0795 x 25 / 1000
= 0.0795g
Molar Mass of Na2SO4 = 142gmol-1
From the balanced equation:
40g NaOH = 142g of Na2SO4
0.0795g NaOH = Xg Na2SO4
mass of Na2SO4 = 0.0795 x 142 g / 40
= 0.282g

(b) (i) Phenolphtalein (Accept methyl orange)
(ii) I – Burette
II – Pipette

Question 2

D is an inorganic salt. E is a solution of an inorganic salt. The tests recorded in the table below were performed on D and E.

Copy and complete the table.

	Test	Observation	Inference
(a)	D + distilled water	Did not dissolve	……………………..
(b)	D + dil. HCI + heat	Dissolved to form blue or greenish-blue solution	…………………….
(c)	Solution of D + NaOH(aq) in drops then in excess	Blue gelatinous precipitate …………………………	……………………..
(d)	Solution of D + NH3(aq) in drops then in excess	………………………….. ……………………………	Cu2+ present Cu2+ confirmed
(e)	(i) E(aq) + NaOH(aq) in drops and in excess	No precipitate	………………….
(e)	(ii)Mixture in 2(e)(i) warmed	…………………………… ………………………….. …………………………..	NH3 gas evolved, therefore NH4+ present
(f)	E(aq) +BaCl2(aq)	White chalky precipitate. No gas evolved, precipitate did not dissolve	……………………. ……………………. ……………………

(a) From the information given above, calculate the:

(i) concentrate of A in moldm– 3;
(ii) concentrate of B in moldm– 3;
(iii) value of X;
(iv) mass of Na2SO4 formed during the reaction. [ H = 1.00; 0 = 16.0; Na = 23.0; S = 32.0] (16 marks)

(b)

(i) Name a suitable indicator for the titration experiment.
(ii) State the apparatus used to measure the volume of solution: I. A; II B; (3 marks)

ANSWER

The expected answers include:

	Test	Observation	Inference
(a)	D + distilled water	Did not dissolve	D is an insoluble salt
(b)	D + dil. HCl + heat	Dissolved to form blue or greenish blue solution	Cu2+ may be present
(c)	Solution of D + NaOH(aq) in drops then in excess	Blue gelatinous precipitate. Precipitate insoluble	Cu2+ present
(d)	Solution of D + NH3(aq) in drops then in excess	Blue gelatinous Precipitate Precipitate dissolves forming deep blue solution	Cu2+ present Cu2+ present confirmed
(e)(i)	E(aq) + NaOH(aq) in drops and in excess	No precipitate	NH4+may be present
(ii)	Mixture in 2(e)(i) warmed	Colourless gas evolved Pungent/choking/irritating smell turns red litmus to blue/forms dense white fume with conc. HCl	NH3 gas evolved NH4+ present
(f)	E(aq) + BaCl2(aq) + excess HCl(aq)	White chalky precipitate. No gas evolved Precipitate did not dissolve	CO32-SO32- SO42- SO42- Present (SO42- must have been mentioned above)

The expected answers include:

	Test	Observation	Inference
(a)	D + distilled water	Did not dissolve	D is an insoluble salt
(b)	D + dil. HCl + heat	Dissolved to form blue or greenish blue solution	Cu2+ may be present
(c)	Solution of D + NaOH(aq) in drops then in excess	Blue gelatinous precipitate. Precipitate insoluble	Cu2+ present
(d)	Solution of D + NH3(aq) in drops then in excess	Blue gelatinous Precipitate Precipitate dissolves forming deep blue solution	Cu2+ present Cu2+ present confirmed
(e)(i)	E(aq) + NaOH(aq) in drops and in excess	No precipitate	NH4+may be present
(ii)	Mixture in 2(e)(i) warmed	Colourless gas evolved Pungent/choking/irritating smell turns red litmus to blue/forms dense white fume with conc. HCl	NH3 gas evolved NH4+ present
(f)	E(aq) + BaCl2(aq) + excess HCl(aq)	White chalky precipitate. No gas evolved Precipitate did not dissolve	CO32-SO32- SO42- SO42- Present (SO42- must have been mentioned above)

ANSWER

The expected answers include:

(a) (i)

(ii) Electrolytic cell

(iii) I- Tetraoxosulphate (VI) acid – sodium chloride solution
II – Glucose solution
– Sugar solution
– Petrol/Kerosene

(b) – Evaporating dish/Beaker
– Tripod stand and wire gauze
– Bunsen burner
– Burette/Pipette
– Glass rod / stirrer
– Thermometer
– Weighing balance/Scale

(c) (i) Ammonia solution
(ii) Ca2+ – No precipitate with NH3(aq)
Pb2+ – White chalky precipitate insoluble in excess
Zn2+ – White gelatinous precipitate soluble in excess
Al3+ – White gelatinous precipitate insoluble in excess

(d) End point is a stage/point in a titration where a given amount of reagent in a solution (acid oxidizing agent) completely react with another reagent (base reducing agent). It is usually indicated by colour change.

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Chemistry Nov/Dec 2013-ALT

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