Answers to Practise Questions for Solubility/Solutions unit:
1.
a) mol L-1 : (4.32 x 10-2 g PbCO3)( 1 mol / 267.2 g) / 0.125 L = 1.29 x 10-3 M
b) ppm:
ppm / 1 x 106 = 4.32 x 10-2 g / 125 g
ppm = 346 ppm
c) ppb:
ppb / 1 x 109 = 4.32 x 10-2 g / 125 g
ppb = 346 000 ppb
2.
% weight / 100% = 16 g NaCl / 80 g solution = 20%
3.
5.62 ppm / 1 x 106 = g Sr2+ / g H2O
i) find g H2O in the lake: (5.61 x 105L)( 1.00 kg / 1.00 L)( 1.00 x 103 g / 1.00 kg) = 5.61 x 108 g
ii) substitute, cross multiply and solve for g Sr2+
5.62 ppm / 1 x 106 = g Sr2+ / 5.61 x 108 g
g Sr2+ = (5.62ppm)(5.61 x 108g) / 1 x 106 = 3 160 g or 3.16 kg
g Sr2+ = 3.15 kg
4.
ppm = (2.00 x 10-3 g Pb2+ / 5.00 x 103g H2O) X 1 x106
5.
a) low solubility
6.
(a) Tl2CO3 low solubility
(c) Ag2SO4 low solubility
7.
(NH4)2CO3(aq) + BaCl2(aq) 2 NH4Cl(aq) + BaCO3(s)
8.
3 Na2S(aq) + 2 Fe(NO3)3(aq) → 6 NaNO3(aq) + Fe2S3(s)
complete ionic equation:
6 Na+(aq) + 3 S2-(aq) + 2 Fe3+(aq) + 6 NO3-(aq)  Fe2S3(s) + 6 Na+(aq) + 6 NO3-(aq)
net ionic equation:
3 S2-(aq) + 2 Fe3+(aq)  Fe2S3(s)
9.
a) Pb2+(aq) + 2 OH-(aq)  Pb(OH)2(s)
b) Ca2+(aq) + CO32-(aq)  CaCO3(s)
10.
(0.2400 M)( 100 / 300)5 = 9.88 x 10-4 M
11.
3 Ni2+(aq) + 2 PO43-(aq)
12.
(d) CO32- ion cannot because it precipitates both ions.
13.
One possible answer is:
(step 1) add potassium chloride reagent, KCl(aq) to precipitate the Hg2Cl2. Centrifuge and decant, or
filter to separate ppt. from solution
(step 2) add potassium sulfate reagent, K2SO4(aq) to precipitate the SrSO4. . Centrifuge and decant, or
filter to separate ppt. from solution
(step 3) add potassium hydroxide reagent, KOH(aq) to precipitate the Zn(OH)2. . Centrifuge and
decant, or filter to separate ppt. from solution
14.
precipitate A is PbCl2
Pb2+(aq) + 2 Cl-(aq)  PbCl2(s)
precipitate B is Ca(OH)2 Ca2+(aq) + 2 OH-(aq)  Ca(OH)2(s)
precipitate C is SrSO4
Sr2+(aq) + SO42-(aq)  SrSO4(s)
b) soluble
= 0.400 ppm
c) soluble d) low solubility e) soluble
(b) K3PO4 soluble
(d) Mg(OH)2 low solubility
→
Ni3(PO4)2(s)
15. first convert mass units to grams: (30 g)( 1 g / 1 x106 g) = 3.0 x 10-5g; (6.00 kg)( 1 x 103g / 1 kg) = 6.00 x 103g
ppb = (3.0 x 10-5 g pollutant / 6.00 x 103 g air) X 1 x 109 = 5.0 ppb
16.
C1 = C2V2 / V1
= (2.00 M)(20 ml) / 600 ml = 0.067 M
17.
(a) Na2SO4(s)  2 Na+(aq) + SO42-(aq)
(b) c = n/v
i) n = (5.00 g)( 1 mol / 142.05 g) = 0.0352 mol
ii) c = 0.0352 mol / 2.00 L = 0.0176 M
(c) [Na+] = (0.0176 M)( 2 Na+ / 1 Na2SO4) = 0.0352 M
[ SO42-] = (0.0176 M)( 1 SO42- / 1 Na2SO4) = 0.0176 M
18.
C4th dilution = (12.0 M)( 20 ml / 200 m l)4 = 0.0012 M
19.
Ni2+(aq) + CO32-(aq)
20.
These are characteristic properties of the Cobalt II ion, Co2+(aq)
 NiCO3(s)
Answers to Multiple choice questions:
1.
A
Ca2+ insoluble with sulfate, soluble with chloride
2.
C
11.7 g NaCl (1 mol / 58.5 g) / 0.25 L = 0.80 M
3.
A
1 200 g crosses the sodium nitrate solubility curve at 570C
4.
B
OH- will precipitate Be2+; SO32- will precipitate Ra2+
5.
B
C1V1 = C2 V2
6.
C
Net ionic equations include reacting species only. No spectator ions are included
7.
A
Overall ionic equations show all spectator ions as well as those involved in the formation
of the precipitate.
8.
B
PbSO4 is the only combination that forms a solid
9.
C
this is the common ion effect. Any compound containing Mg2+ or SO42- ions
will cause MgSO4 to precipitate from solution
10.
D
Net ionic equations include reacting species only. No spectator ions are included.
11.
A
only Ag+ is insoluble with acetate ion. Pb2+ and Sr2+ remain
only Pb2+ is insoluble with iodide ion. Sr2+ remains.
Sr2+ is insoluble with sulfate ion.
12.
B
C1V1 = C2 V2
13.
D
Concentration after 3rd dilution = (5.0 M)(100 / 250 )3 = 0.32
(12.0 mol L-1)(V1) = (0.500 mol L-1)(5.00 L)
V1 = 0.208 L
(0.60 mol L-1)(0.025 L) = (0.15 mol L-1)(V2)
V2 = 0.100 L total volume
0.100 L – 0.025 L = added volume = 0.075 L
14.
A
(22 g Mg2+) / (115 L water ) = 22 g / 1.15 x 105 g water
1 ml = 1 g for water
5
9
22.0 g / 1.15 x 10 g = x ppb / 1 x 10
x = 1.91 x 105 ppb