Questions On Acids And Bases

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13.3 Acids and Bases Assessed Homework Mark Scheme 2007

1. B
2. D
3. C
4. C
5. D
6. A
7. B

8. Penalise pH given to 1 dp first time it would have scored only

(a) (i) Kw = [H+] [OH–] (1)

(ii) pH = –log [H+] (1) or in words or below unless contradiction

(iii) Calculation: [H+] = [pic] (1) = 2.34 × 10–7 \ pH = 6.63 or 6.64 (1)

Explanation: pure water \ [H+] = [OH–] (1) 5

(b) (i) [OH–] = 0.150 \ [H+] = 10–14/0.15 = 6.66 × 10–14 or pOH = 0.82

\ pH = 13.18 (1) or pH= 13.17

(ii) moles OH– = (35 × 10–3) × 0.150 = 5.25 × 10–3 (1)a

moles H+ = (40 × 10–3) × 0.120 = 4.8(0) × 10–3 (1)b

\ excess moles of OH– = 4.5 × 10–4 (1)c

\ [OH–] = (4.5(0) × 10–4) × 1000/75d (1)e

[H+] = [pic] = 1.66 × 10–12 or pOH = 2.22

\ pH = 11.78 (1)f or 11.77 6

(c) (i) Ka = [pic] (1)

(ii) [H+] = 1.80 × 10–2 × 0.150 = 2.70 × 10–3 (1)

Ka = [pic] = 4.86 × 10–5 (1) mol dm–3 (1) 5 or [pic] = 4.95 × 10–5

Notes

(a) If Kw includes H2O allow 6.63 if seen otherwise no marks likely

(b) (ii) If no vol, max 4 for a, b, c, f answr = 10.65 If wrong volume max 5 for a, b, c, e, f If no substraction max 3 for a, b, d If missing 1000 max 5 for a, b, c, d, f answer = 8.78 If uses excess as acid, max 4 for a, b, d, f answer = 2.22 If uses excess as acid and no volume, max 2 for a, b answer = 3.35

(c) If wrong Ka in (i) max 2 in part (ii) for [H+] (1) and conseq units (1) but mark on fully from minor errors eg no [ ] or charges missing
[18]

9. (a) Ka = [pic] 1 (All three sets of square brackets needed, penalise missing brackets or missing charge once in the question) (Don’t penalise extra [H+]2/[HA])

(b) Ka = [pic] or [H+] = [A–] 1 [H+] = [pic] = 6.02 × 10–3 1 pH = 2.22 1 (must be to 2dp) (allow 4th mark consequential on their [H+])

(c) (i) pH (almost) unchanged 1 (Must be correct to score explanation)

H+ removed by A– forming HA or acid reacts with salt or more HA formed 1

(ii) [H+] = 10–3.59 = 2.57 × 10–4 or 2.6 × 10–4 1 [A–] = [pic] 1 = [pic] 1 = 0.141 (mol dm[pic]3) 1 (Allow 0.139 to 0.141 and allow 0.14) (If not used 3.59, to find [H+] can only score M2 for working) (If 3.59 used but [H+] is wrong, can score M2 for correct method and conseq M4) If wrong method and wrong expression, can only score M1)

(ii) Alternative scheme for first three marks of part (c)(ii)

pH = pKa [pic] log[pic] 1

pKa = 3.84 1

3.59 = 3.84 – log[pic] 1
[11]

10. (a) (i) pH = –log [H+] (1)

(ii) Expression for Ka: Ka = [pic] (1)

Calculation: pH = 2.56 \ [H+] = 2.75 × 10–3 (1) Ka = [pic] = [pic] = 6.32 × 10–5 (1) (mol dm–3) or [H+] = [X–] (1) 5 depending on approximate made, values of Ka = 10–5 × using [HX] = 0.12 6.30 – 6.32 using [HX] = 0.12 – 2.75... 6.45 – 6.47 using 2.8 and [HX] = 0.12 6.53 using 2.8 and [HX] = 0.12 – 2.8 6.69 upside down Ka (b) (i) Expression for Kw: Kw = [H+] [OH–] (1)

Value of Kw: (1.0 ×)10–14 (mol2 dm–6) (1) ignore units

(ii) [H+] = [pic] = 2.22 × 10–13
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