1. Section 3: Titrations
Titration
Calculations
Print 1, 3-5, 7-8

2. ACID + BASE  SALT + WATER
Neutralization Reactions (REVIEW)
Neutralization –
rxn of acid and base making a salt and water
ACID + BASE  SALT + WATER =
HCl + NaOH  NaCl + HOH
H2O

3. Titration MAVA = MBVB moles A = moles B
(video clip)
The analytical technique used to calculate the moles of an unknown.
MAVA = MBVB
buret
titrant
moles A = moles B
known vol. (V)
known conc. (M)
1:1 ratio ONLY!!!
Safari Montage (2 min) Titration
analyte
known vol. (V)
unknown conc. (M)

4. equivalence point: end point: equal stoichiometric amounts
(moles) that react completely
mol H+ = mol OH–
end point:
indicator permanently changes color

5. Pink in BASE
Phenolphthalein
Clear in ACID

6. Solution Stoichiometry
(Review)
Titration: HA + BOH  H2O + AB
molar mass A
molarity HA (M)
g A
mol HA
L of HA
g A
1 mol A
mol HA
1 L
mol-to-mol ratio
g B
1 mol B
mol BOH
1 L
molar mass B
molarity BOH (M)
g B
mol BOH
L of BOH 6

7. Example #1a: L of B  L of A HBr(aq) + NaOH(aq)  H2O(l) + NaBr(aq)
Hydrobromic acid is titrated to equivalence with sodium hydroxide.
What volume of 1.50 M HBr is needed to react with 120. L 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

8. MAVA = MBVB Example #1b: L of B  L of A
HBr(aq) NaOH(aq)  H2O(l) + NaBr(aq)
V = ? L
V = 120. L
M = 1.50 mol/L
M = 1.25 mol/L
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
MAVA = MBVB
1:1 ratio ONLY!!!
(1.50 M) x VA = (1.25 M) x (120. L)
VA = (1.25 M) x (120. L) =
(1.50 M)
100 L HBr

9. MAVA = MBVB Example #2: 1:1 ratio ONLY!!!
A 20.4 mL solution of KOH was completely neutralized by 48.5 mL of M HNO3 .
What is the concentration of the KOH?
(0.150 M) x (48.5 mL) = MB x (20.4 mL)
(0.150 M) x (48.5 mL) = MB =
(20.4 mL)
0.357 M KOH

10. Example #3: L 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

11. Quick Quiz!
1) When the moles of acid equals the moles bases you have reached the…
A) titration point
B) end point
C) equivalence point
D) neutralization point 11

12. Quick Quiz.
2) If 20. mL of 0.50 M HCl is titrated to equivalence by 40. mL of NaOH, then the NaOH concentration must be ___.
A) 2.0 M
B) 1.0 M
C) 0.50 M
D) 0.25 M
moles of H+ = moles of OH–
0.020 L
0.50 M
0.040 L
??? M 12