UNIT 3: SOLUTIONS
Stoichiometry in Solution
Recall, stoichiometry involves calculating the amounts of reactants and products in
chemical reactions.
Thus, you can use stoichiometry to determine the amount of ions that react. You
can use the same basic steps that you learned in Unit 2 to solve stoichiometric
problems involving solutions!
General Guidelines for Stoichiometry in Solutions:
 convert mass or concentrations to moles
 write the net ionic equation or chemical equation
 use coefficients of balanced equation to determine the number of moles of
required substance
 if this is a limiting reagent problem identify the limiting reagent
 calculate the mass or concentration of required substance
example 1: Finding the minimum volume of precipitate
An aqueous solution that contains silver ions is usually treated with chloride ions
to recover silver chloride. What is the minimum volume of 0.25 mol/L magnesium
chloride, MgCl2(aq), needed to precipitate all the silver ions in 60 ml of 0.30 mol/L
silver nitrate, AgNO3 (aq)? Assume that silver chloride is completely insoluble in
water.
SOLUTION:
 write a chemical equation for the formation for the reaction
MgCl2(aq) + 2 AgNO2(aq)  2 AgCl(s) + Mg(NO3)2(aq)
 convert concentrations to moles of AgNO3
(0.018 mol)
 use coefficients of balanced equation to determine number of moles of
MgCl2 required
0.018 mol of AgNO3 x 1 mol MgCl2
= 0.0090 mol MgCl2
2 mol AgNO3
 calculate the volume of MgCl2 needed
0.0090 mol of MgCl2 x
1 L
= 0.036 L
0.25 mol
UNIT 3: SOLUTIONS
example 2: Limiting Reagent
Mercury salts have a number of important uses in industry and in chemical
analysis. Because mercury compounds are poisonous, however, the mercury ions
must be removed from the waste water. Suppose that 25.00 ml of 0.085mol/L
aqueous sodium sulphide is added to 56.5 ml of 0.10 mol/L mercury (II) nitrate.
What mass of mercury(II) sulphide, HgS(s) precipitates?
SOLUTION:
 write net ionic equation or chemical equation
Hg(NO3)2(aq) + Na2S(aq)  2NaNO3(aq) + HgS(s)
 convert mass or concentrations to moles
Amount of Hg(NO3)2 = 0.0565 L x 0.10 mol/L
= 0.00565 mol
Amount of Na2S = 0.0250 L x 0.085 mol/L
= 0.00212 mol
 use coefficients of balanced equation to determine number of moles of
required substance
since each mole of Na2S produces 1 mol of HgS, we expect either 0.00565 mol or 0.00212
mol of HgS.
 If limiting reactant problem: identify the limiting reactant
Since reactants are needed in a 1:1 ratio, the reactant present in the smallest amount is the
limiting reactant.
Thus, HgS is limiting.
 Calculate the mass or concentration of required substance
Molar mass of HgS = 200.6 + 32.1
= 232 .7 g/mol
mass of HgS = 0.00212 mol x 232.7 g
= 0.493 g
1 mol
UNIT 3: SOLUTIONS
example 3: The concentration of ions
Calculate the concentration (mol/L) of chloride ions in each solution.
a. 19.8 g of potassium chloride dissolved in 100 ml of solution
b. 26.5 g of calcium chloride dissolved in 150 ml of solution
c. a mixture of the two solutions in parts (a) and (b), assuming that the
volumes are additive.
SOLUTION:
 convert mass into moles for each solution
 write the equations for the dissociation of the substance
 use coefficients of balanced equation to determine number of moles of Clpresent
 Calculate the concentration of each solution
 Calculate the concentration of Cl- ions in the final solution
Solution
Molar mass
Amount (mol)
Dissociation
equation
Amount of Cl-
Concentration
of Cl-
KCl
CaCl2
39.10 + 35.45 = 74.55 g
40.08 + (2 x 34.45) = 110.98 g
19.8 g/ 74.55 g/mol = 0.266 mol 26.5 g / 110.98 g/mol = 0.239 mol
KCl  K+ + Cl0.266 mol x 1 mol Cl- = 0.266
mol
1 mol KCl
0.266 mol = 2.66 mol/L
0.100 L
CaCl2  Ca2+ + 2 Cl0.239 mol x 2 mol Cl- = 0.478
mol
1 mol CaCl2
0.478 mol = 3.19 mol/L
0.150 L
The concentration of Cl- of each solution is 2.66 mol/L and 3.19 mol/L respectively. The
concentration of Cl- when mixed is 2.98 mol/L
UNIT 3: SOLUTIONS
example 4: The mass Percent of Ions
The leaves of a rhubarb plant contain a relatively high concentration of oxalate
ions, C2O42-. Oxalate ions are poisonous, causing respiratory failure. To determine
the percent of oxalate ions, a student measured the mass of some leaves. Then
the student ground up the leaves and added excess calcium chloride solution to
precipitate calcium oxalate. The student tested 238.6 g of leaves. The dried mass
of calcium oxalate was 0.566 g. What was the mass percent of oxalate ions in
leaves?
SOLUTION:
 write net ionic equation for the formation of calcium oxalate
The net ionic equation is
Ca2+(aq) + C2042- (aq)  CaC2O4 (s)
 Convert mass to moles of calcium oxalate
(0.00434 mol)
 use coefficients of balanced equation to determine number of moles of
oxalate ions
0.00434 mol of CaC2O4 x 1 mol C2O42- = 0.00434 mol of CaC2O4
1 mol CaC2O4
 Calculate the mass of oxalate ions
M C2042- = 88.02 g/mol
m of C2042- = 0.00434 mol x 88.02 g = 0.382 g
1 mol C2042-
 Calculate mass percent of oxalate ions in rhubarb
mass % = mass of ions/mass of leaves x 100%
= 0.382 g/238.6 g x 100%
= 0.160%
UNIT 3: SOLUTIONS
PORTFOLIO:
4. STOICHIOMETRY IN SOLUTIONS
1. Food manufactures sometimes add calcium acetate to puddings and sweet
sauces as a thickening agent. What volume of 0.500 mol/L calcium acetate,
Ca(CH3COO)2(aq), contains 0.300 mol of acetate ions? (ans: 0.600 L)
2. Ammonium phosphate can be used as a fertilizer. If 6.00 g of ammonium
phosphate is dissolved in sufficient water to produce 300 ml of solution, what
are the concentrations (mol/L) of the ammonium ions and phosphate ions
present? (ans: phosphate: 0.134 mol/L; ammonium: 0.402 mol/L)
3. 8.76 g of sodium sulphide is added to 35.0 ml of 0.250 mol/L lead(II) nitrate
solution. Calculate the maximum mass of precipitate that can form. (ans: 2.65 g)
4. 25.0 ml of 0.400 mol/L Pb(NO3)2(aq) is mixed with 300 ml of 0.220 mol/L KI(aq).
What is the maximum mass of precipitate that can form? (ans: 4.61 g)
5. A student mixes 15.0 ml of 0.250 mol/L aqueous sodium hydroxide with 20.0 ml
of 0.400 mol/L aqueous aluminum nitrate. Calculate the maximum mass of
precipitate that forms. (ans: 0.624 g)