Chemistry 20 – Lesson 23
Reactions in Solution
Practice problems
1.
Hydrochloric acid is added to a solution of barium hydroxide.
2 HCl (aq) + Ba(OH)2 (aq)  BaCl2 (aq) + 2 HOH (l)
(non–ionic)
2 H+ (aq) + 2 Cl– (aq) + Ba2+ (aq) + 2 OH– (aq)  Ba2+ (aq) + 2 Cl– (aq) + 2 HOH (l) (total ionic)
H+ (aq) + OH– (aq)  HOH (l)
2.
Magnesium metal is added to an aqueous solution of hydrogen bromide.
Mg (s) + 2 HBr (aq)
 MgBr2 (aq) + H2 (g)
Mg (s) + 2 H+ (aq) + 2 Br– (aq) 
Mg (s) + 2 H+ (aq) 
3.
4.
5.
(net ionic)
(non–ionic)
Mg2+ (aq) + 2 Br– (aq) + H2 (g)
(total ionic)
Mg2+ (aq) + H2 (g)
(net ionic)
Calcium metal reacts with water.
Ca (s) + 2 HOH (l)  H2 (g) + Ca(OH)2 (aq)
(non–ionic)
Ca (s) + 2 HOH (l)  H2 (g) + Ca2+ (aq) + 2 OH– (aq)
(total ionic)
Ca (s) + 2 HOH (l)  H2 (g) + Ca2+ (aq) + 2 OH– (aq)
(net ionic)
Aqueous solutions of potassium sulfate and barium chloride are mixed.
K2SO4 (aq) + BaCl2 (aq)  BaSO4 (s) + 2 KCl (aq)
(non–ionic)
2 K+(aq) + SO42(aq) + Ba2+ (aq) + 2 Cl– (aq)  BaSO4 (s) + 2 K+ (aq) + 2 Cl– (aq)
(total ionic)
SO42(aq) + Ba2+ (aq)  BaSO4 (s)
(net ionic)
An aqueous solution of washing soda, Na2CO3, is added to remove Mg2+ (aq) from water.
Mg2+(aq) + 2 Na+ (aq) + CO32(aq)  2 Na+ (aq) + MgCO3 (s)
(total ionic)
Mg2+(aq) + CO32(aq)  MgCO3 (s)
(net ionic)
Dr. Ron Licht
23 - 1
www.structuredindependentlearning.com
Assignment
/39
1.
Potassium metal reacts with water.
/3
K (s) +
2 HOH (l)  H2 (g) + 2 KOH (aq)
(non–ionic)
K (s) +
2 HOH (l)  H2 (g) + 2 K+ (aq) + 2 OH– (aq)
(total ionic)
K (s) +
2 HOH (l)  H2 (g) + 2 K+ (aq) + 2 OH– (aq)
(net ionic)
2.
A lead (II) acetate solution reacts with a sodium sulfide solution to yield a precipitate.
/3
Pb(CH3COO)2 (aq) +
Na2S (aq)  2 NaCH3COO (aq) +
PbS (s)
(non–ionic)
Pb2+(aq) + 2 CH3COO(aq) + 2 Na+(aq) + S2–(aq)  2 CH3COO(aq) + 2 Na+(aq) + PbS(s) (total)
Pb2+(aq) + S2–(aq)  PbS(s)
(net ionic)
3.
Solutions of sodium sulfate and barium bromide are added together.
/3
Na2SO4 (aq) +
BaBr2 (aq)  BaSO4 (s) +
2 NaBr (aq)
(non–ionic)
2 Na+(aq) + SO42(aq) + Ba2+ (aq) + 2 Br– (aq)  BaSO4 (s) + 2 Na+ (aq) + 2 Br– (aq) (total ionic)
SO42(aq) + Ba2+ (aq)  BaSO4 (s)
(net ionic)
4.
An aqueous solution of sodium carbonate is used to remove calcium ions from water.
/3
Ca2+(aq) + 2 Na+ (aq) + CO32(aq)  2 Na+ (aq) + CaCO3 (s)
Ca2+(aq) + CO32(aq)
 CaCO3 (s)
(total ionic)
(net ionic)
5.
A precipitate forms when potassium iodide is mixed with lead (II) nitrate.
/3
Pb(NO3)2 (aq) + 2 KI (aq)  2 KNO3 (aq) + PbI2 (s)
(non–ionic)
Pb2+(aq) + 2 NO3(aq) + 2 K+(aq) + 2 I–(aq)  2 K+(aq) + 2 NO3–(aq) + PbI2 (s)
(total ionic)
Pb2+(aq) + 2 I–(aq)  PbI2 (s)
(net ionic)
6.
A calcium nitrate solution is added to a solution of sodium carbonate.
/3
Ca(NO3)2 (aq) + Na2CO3 (aq)  2 NaNO3 (aq) +
CaCO3 (s)
(non–ionic)
Ca2+(aq) + 2 NO3(aq) + 2 Na+(aq) + CO32–(aq)  2 Na+(aq) + 2 NO3–(aq) + CaCO3 (s) (total)
Ca2+(aq) + CO32– (aq)  CaCO3 (s)
Dr. Ron Licht
(net ionic)
23 - 2
www.structuredindependentlearning.com
7.
A precipitate forms when iron (III) nitrate reacts with sodium phosphate.
/3
Fe(NO3)3 (aq) + Na3PO4 (aq)  FePO4 (s) +
3 NaNO3 (aq)
(non–ionic)
Fe3+(aq) + 3 NO3(aq) + 3 Na+(aq) + PO43(aq)  FePO4 (s) + 3 Na+(aq) + 3 NO3(aq) (total ionic)
Fe3+(aq) + PO43(aq)  FePO4 (s)
8.
/6
Pb2+(aq)
S2(aq) 
+
cS2  ?
v Pb2  0.0580L
vS2  0.100L
n Pb2  0.100 mol L (0.0580L)
n Pb2  0.00580mol
Cl2 (aq)
+
2 I(aq) 
cI  0.120 mol L
A. calculate moles
n I  0.120
mol
L
(2.50L)
n I  0.300mol
/6
B. mole ratio
n S2 n Pb2

1
1
n S2 0.00580mol

1
1
n S2  0.00580mol
2 Cl(aq)
C. calculate concentration
0.00580 mol
cS2 
0.100 L
cS2  0.0580 mol L
+
I2 (s)
m I2  ?
v I  2.50L
/6
10.
PbS (s)
cPb2  0.100 mol L
A. calculate moles
9.
(net ionic)
Fe3+(aq)
B. mole ratio
n I2 n I

1
2
n I2 0.300mol

1
2
n I2  0.150mol
3 OH–(aq)
+

cFe3  ?
cOH  0.0200 mol L
v Fe3  0.800L
vOH  0.00480L
B. mole ratio
n OH  0.0200 mol L (0.00480L)
n Fe3
n OH  0.0000960mol
Dr. Ron Licht
1
m I2  0.150mol(253.80 g mol)
m I2  38.1g
Fe(OH)3 (s)
A. calculate moles

C. calculate mass
n OH
C. calculate concentration
0.0000320 mol
c Fe3 
0.800 L
3
n Fe3 0.0000960mol c  4.00 105 mol
L

Fe3
1
3
n Fe3  0.0000320mol
23 - 3
www.structuredindependentlearning.com