Page 52 - PRE-U STPM CHEMISTRY TERM 1

Chemistry Term 1 STPM
Quick Check 7.5 Quick Check 7.10
1 1 (a) 4.84
(b) 4.87
(a) (b) (c) (d) (e)
(c) The change is negligible because the initial solution is
pH = 2.47 pH = 0.097 pH = –0.38 pOH = 1.25 pOH = –0.45
a buffer solution.

+
2 (a) H SO  2H + SO 4 2– 2 (a) 4.69
4
2
–3
(b) 0.50 mol dm (b) (i) 4.74
(c) 0.30 (ii) 4.65
–3
3 (a) 1.0 mol dm
(b) pOH = 0 Quick Check 7.11
–3
–3
4 (a) 1.34  10 mol dm 1
(b) 2.87 Salt BaSO Al(OH) Fe S Ca (PO )

5 (a) NH + H O  NH + OH – 4 3 2 3 3 4 2
+
3 2 4 Solubility/g dm –3 2.44  10 –3 2.26  10 –33 2.04 10 –16 1.22 10 –3
–
–3
(b) [OH ] = 5.1  10 mol dm –3
pOH = 2.29
Salt Silver sulphate Barium carbonate Silver iodide
Quick Check 7.6 K 5.00  10 M 3 8.16  10 M 2 1.50  10 –16 M 2
–7
–9
1 2.93 sp
2 11.1 Quick Check 7.12
–3
–3
3 (a) pH = 3.08; α = 4.12  10 1 3.55  10 g
(b) pH = 4.08; α = 4.12  10 –2 2 1.00  10 mol dm –6
–10
2
(c) Dilution increases both the pH and the degree of 3 No
dissociation.
–6
–3
4 (a) 7.87  10 mol dm Quick Check 7.13
(b) 8.90 1 The oxalate ions react with H to form undissociated
+
5 (a) (i) 2.87 H C O . 4
2
2
2–
(ii) 0.0134 C O + 2H → H C O
+
2
4
2
2
4
(b) (i) 2.96 The sulphate ion does not react with H .
+
(ii) 0.0164
2 BaCO reacts with the dilute HCl in the stomach to form
3
–4
6 (a) 4.30  10 soluble BaCl . 2
(b) 3.70  10 –7 BaCO (s) + 2HCl(aq) → BaCl (aq) + H O(l) + CO (g)
3
–3
7 (a) 3.12  10 mol dm –3 BaSO does not react with HCl. 2 2 2
4
(b) 3.21  10 mol dm –3
–12
(c) 11.49 STPM Practice 7
Objective Questions
Quick Check 7.7 1 A 2 C 3 C 4 A 5 C
1 6 B 7 B 8 D 9 C 10 C
(a) (b) (c) (d) (e) (f) (g)
11 A 12 C 13 D 14 D 15 D
Acid Neutral Acid Base Acid Base Base 16 A 17 A 18 D 19 A 20 B
21 C 22 A 23 B 24 C 25 D
Quick Check 7.8 26 C 27 A 28 C 29 C 30 C
31 D 32 B 33 B 34 C 35 D
1 (a) NaOH + ClCH COOH → ClCH COONa + H O
2
2
2
(b) 0.063 M Structured and Essay Questions
(c) 1.56 1 (a) The solubility product of a sparingly salt is defined
2 (a) 0.143 M as the product of the molar concentration of the
(b) NaOH + HXO → NaXO + H O constituent ions, in a saturated solution, each raised
4 4 2
(c) 0.200 mol dm –3 to the power of the stoichiometric coefficient.
(d) 100.5 (b) (i) Ca(OH) + 2HCl → CaCl + 2H O
2
2
2
(e) A = 35.5; X ≡ Chlorine M × 20 1
r
0.045 × 20.2 = 2
Quick Check 7.9 M = 0.023 mol dm –3
1 (d) and (f) only – –3
[OH ] = 2  0.023 = 0.046 mol dm
2 (a) 4.91 (c) 4.22 (e) 4.86 pOH = –log(0.046) = 1.34
(b) 4.91 (d) 9.67 (f) 4.47 ∴ pH = 14 – 1.34 = 12.66
375
12 Answers.indd 375 3/26/18 4:06 PM