1. Solution Stoichiometry
(Review)
Titration: HA + BOH  H2O + AB
molar mass A
molarity HA (M)
Liters of HA
mol HA
grams A
grams A
1 mol A
mol HA
1 Liter
mol-to-mol ratio
grams B
1 mol B
mol BOH
1 Liter
molar mass B
molarity BOH (M)
Liters of BOH
grams B
mol BOH 1

2. Example #1: Liters of B  Liters of A
Hydrobromic acid is titrated to equivalence with sodium hydroxide.
What volume of 2.00 M HBr is needed to react with 10. L of 2.00 M of NaOH ?
HBr(aq) NaOH(aq)  H2O(l) + NaBr(aq)
V = ? L
V = 10. L
M = 2.00 mol/L
M = 2.00 mol/L
molarity HA
2.00
mol NaOH 1
mol HBr 1
L HBr x x
10. L NaOH x 1
L NaOH 1
mol NaOH
2.00
mol HBr
L of BOH
molarity BOH
= ____L HBr
10. L HBr

3. Example #2: Liters of B  Liters of A
Hydrobromic acid is titrated to equivalence with sodium hydroxide.
What volume of 1.50 M HBr is needed to react with 120. mL of 1.25 M of NaOH ?
HBr(aq) NaOH(aq)  H2O(l) + NaBr(aq)
V = ? L
V = .120 L
M = 1.50 mol/L
M = 1.25 mol/L
molarity HA
1.25
mol NaOH 1
mol HBr 1
L HBr x x
.120 L NaOH x 1
L NaOH 1
mol NaOH
1.50
mol HBr
L of BOH
molarity BOH
= ____L HBr
.100 L HBr or 100. mL HBr

4. Example #3: KOH(aq) + HNO3(aq)  H2O(l) + KNO3(aq)
A 20.4 mL solution of KOH was completely neutralized by 48.5 mL of M HNO3 .
What is the concentration of the KOH?
KOH(aq) HNO3(aq)  H2O(l) + KNO3(aq)
V = 20.4 mL
V = 48.5 mL
M = ? mol/L
M = mol/L
0.150
mol HNO3 1
mol KOH
mol KOH x
L HNO3 x = 1
L HNO3 1
mol HNO3
L of HA
molarity HA
molarity BOH
mol KOH
= _____ M KOH
0.357 M KOH
L KOH

5. Example #4: Liters of A  M of B
A 159 mL sample of KOH is titrated to equivalence with 100. mL of 1.50 M H2SO4 . Calculate the molarity of the of the KOH solution. H2SO4(aq) + 2 KOH(aq)  2 H2O(l) + K2SO4(aq)
V = L
V = L
M = 1.50 mol/L
M = ? mol/L
1.50
mol H2SO4 2
mol KOH
mol KOH x =
0.100 L H2SO4 x 1
L H2SO4 1
mol H2SO4
L of HA
molarity HA
molarity BOH
0.300 mol KOH
= _____ M KOH
1.89 M KOH
0.159 L KOH