Solutions (Heath Chapter 16) p.5
*T*1) Determine which of the following compounds are insoluble in water:
Refer to ‘solubility’ sheet
a) sodium hydroxide -soluble
b) ammonium acetate-soluble
c) calcium sulfate- insoluble (not soluble)
d) lead(II) chloride-insoluble
e) potassium chloride-soluble
f) calcium bromide-soluble
2) Write the net ionic equation for the following reactions:
a) lead (II) nitrate and potassium bromide
Ion box
cation
anion
outcome/products
2+
Pb
PbBr2(s) not soluble
NO3K+
Br -
KNO3(aq) soluble
Pb2+(aq) + 2Br -(aq) PbBr2(s) (milky white ppt)
b) sodium sulphate and barium chloride
Ion box
cation
anion
outcome/products
2Na+
BaSO4(s) not soluble
SO4
Ba2+
NaCl (aq) soluble
Cl Ba2+(aq) + SO42-(aq) BaSO4(s)
3) How many grams of H2SO4 would you mix with 500.0 mL of solution to
prepare a 0.500 M solution?
V = 500.0 mL = 0.5000 L
[H2SO4] =0.500 M or mol/L
X V X MM
Road Map: [ ]--- n ----- xg
xg = 0.500 mol X 0.5000 L X 98.1 g = 25.0 g
L
mol
4) Calculate the volume of solution that contains 2.00 M Cu(NO3)2 from 80.0 g of
solute.
solute: solid Cu(NO3)2
MM(Cu(NO3)2) = 63.5 + 28 + 6(16) = 187.5 g/mol
 MM  [ ]
Road Map: xg --- n ---- V
n = 80.0 g X mol
= 0.4267 mol
187.5 g
V = n = 0.4267 mol X 1 L
= 0.213 L
[]
2.00 mol
*T*5) Calculate the concentration of each ion in each of the following solutions:
a) 0.24 M Al2(SO4)3
Al2(SO4)3

2Al3+ +
3SO42- (Dissociation Eqn: breaks into ions)
1 mol
2 mol
3 mol (Mole ratio)
0.24 M
0.48 M
0.72 M
b) 0.25 M Na3PO4
Na3PO4

1 mol
0.25 M
3Na+ +
3 mol
0.75 M
1PO431 mol
0.25 M
c) 0.33 M Cr2(SO4)3
Cr2(SO4)3

2Cr3+ +
1 mol
2 mol
0.33 M
0.66 M
3SO423 mol
0.99 M
*T* d) 0.30 M NaBr mixed with an equal volume of 0.50 M CuBr2
Let's suppose that the volume is 1L, then the total volume is 2 L
NaBr 
Na+ +
Br(Dissociation Eqn)
0.30 M
0.30 M
0.30 M
(Ion Concentrations)
2+
CuBr2 
Cu
+
2Br
(Dissociation Eqn)
0.50 M
0.50 M
1.00 M
C1V1 =C2V2 
C2 = C1V1
 Dilution Eqn
V2
C2 = Final concentration
C1 = Initial concentration
V1 = Initial volume
V2 = Total volume (1 L + 1 L = 2 L)
[Na+] = 0.30 M X 1 L = 0.15 M
2L
2+
[Cu ] = 0.50 M X 1 L = 0.25 M
2L
[Br ]NaBr = 0.30 M X 1 L = 0.15 M
2L
[Br ]CuBr2 = 1.00 M X 1 L = 0.50 M
2L
[Br ]Total = [Br ]NaBr + [Br-]CuBr2 = 0.15 + 050 = 0.65 M
e) 140.0 mL of 0.36 M CaCl2 mixed with 60.0 mL of 0.22 M Ca(NO3)2
CaCl2 
Ca2+ +
2Cl1 mol
1 mol
2 mol
0.36 M
0.36 M
0.72 M
Ca(NO3)2  Ca2+ +
2NO31 mol
1 mol
2 mol
0.22 M
0.22 M
0.44 M
[Ca2+] CaCl2 = 0.36 M X 0.140 L = 0.280 M
0.180 L
[Ca2+] Ca(NO3)2 = 0.22 M X 0.060 L = 0.0733 M
0.180 L
[Ca2+]Total = 0.28 M + 0.073 M = 0.35 M
2 dp
3 dp
2 dp
[Cl ] = 0.72 M X 0.140 L = 0.56 M
0.180 L
[NO3 ] = 0.44 M X 0.060 L = 0.15 M
0.180 L
6) Calculate the volume required to make the following diluted solutions:
a) 2.5 L of 1.0 M H2SO4 from 18 M H2SO4
C1V1 =C2V2
C1= 18 M
V1= ? L
C2= 1.0 M
V2= 2.5 L
V1 = C2V2 = 1.0 mol X 2.5 L X 1 L
= 140 mL or 0.14 L
C1
L
18 mol
b) 250 mL of 0.14 M NaOH from 6.0 M NaOH
C1V1 =C2V2
C1= 6.0 M
V1= ? L
C2= 0.14 M
V2= 0.250 L
V1 = C2V2 = 6.0 mol X 0.25 L X 1 L
= 5.8 mL or 0.0058 L
C1
L
0.14 mol
Answers:
1) c and d are insoluble
2a) Pb2+ (aq) + 2Br-(aq)  PbBr2(s)
b) Ba2+(aq) + SO42-(aq) BaSO4(s)
3) 25.0 g
4) 0.213 L
3+
5a) [Al ]= 0.48 M [SO42-]= 0.72 M
b) [Na+]= 0.75 M [PO43-]= 0.25 M
c) [Cr3+]= 0.66 M [SO42]= 0.99 M
d) [Na+]= 0.15 M [Br-]= 0.65M [Cu2+]= 0.25 M
e) [Ca2+]= 0.32 M [Cl-]= 0.50 M [NO3-]= 0.13 M
6a) 1.4 X102 mL
b) 5.8 mL