Chemistry 12 (HL)
Unit 5 / IB Topic 18.4 (and 1.5)
Acids and Bases – Part 7
Acid-Base Titrations – Practice Problems
1.
Sketch and annotate a titration curve for these titrations:
a)
3
-3
-3
-3
-3
50.0 cm 1.0 mol dm CH3COOH is titrated with 1.0 mol dm NaOH
a rather ugly sketch!
b)
3
50.0 cm 1.0 mol dm CH3COOH is titrated with 1.0 mol dm NH3
another rather ugly sketch!
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Chemistry 12 (HL)
Unit 5 / IB Topic 18.4 (and 1.5)
2.
a) Identify the types of reactants used in this titration curve.
analyte = weak base
titrant = strong acid
b) Determine the pH at the equivalence point.
4.8 – 5.0
c) Determine the pKb of the base used.
volume to reach equivalence point = 90 mL
∴ half equivalence point occurs @ 45 mL
pH @ half equivalence point = pKa = 8.8
∴ pKb = 14.0 – pKa = 5.2
d) 25.00 mL of base was titrated in this example, with an acid with a concentration
-3
of 0.854 mol dm . Calculate the concentration of the base.
NOTE: Assume a 1:1 acid:base ratio in the neutralization equation.
from the titration curve: volume acid needed to neutralize the base = 90 mL
+
moles H added = 0.854 mol dm
-3
3
x 0.090 dm = 0.07686 mol
–
∴ moles OH = 0.077 mol = moles base
3
∴ [base] = 0.077 mol / 0.02500 dm = 3.1 mol dm
3.
–3
Given these titrations curve for three weak acids versus NaOH:
A
B
a)
C
Determine the pKa for each acid. Would you be able to identify the acid? How?
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Chemistry 12 (HL)
Unit 5 / IB Topic 18.4 (and 1.5)
pKa = pH at the half equivalence point (i.e. at 12.5 mL)
C: pKa ~ 2.6
B:
pKa ~ 4.5
A: pKa ~ 7.8
You could identify the acid from a table of pKa values.
b)
What is the effect of the “strength” of the weak acid on the pH at the equivalence point?
The higher the pKa, the weaker the acid.
∴ the order of increasing acid strength (weakest to strongest) is: A < B < C
From the titration curves, the pH at the equivalence point increases as the acid becomes
weaker.
4.
-2
Calculate the concentration of an unknown acid H2X if 30.00 mL of 7.39 x 10 mol/L KOH are
required to reach the equivalence point in a titration of 20.00 mL of H2X.
H2X + 2KOH à 2KX + 2H2O
-2
moles KOH = 7.39 x 10 mol dm
(note the complete reaction arrow!)
-3
3
x 0.03000 cm = 2.217 x 10
-4
–4
mol
-4
∴ moles H2X = 2.217 x 10 mol x ½ = 1.108 x 10 mol
-4
3
∴ [H2X] = (1.108 x 10 mol) / 0.02000 cm = 0.003695 mol dm
5.
-3
-3
3
Calculate the volume of 0.00500 mol dm NH3 required to neutralize 84.92 cm of 0.00398 mol dm
CH3COOH.
CH3COOH + NH3 à NH4CH3COO
moles CH3COOH = 0.00398 mol dm
-3
-3
(Note the complete reaction arrow, even when
weak acids and bases are used. Titration
reactions always go to completion.)
3
x 0.08492 cm = 0.0003380 mol
∴ moles NH3 = 0.0003380 mol
∴ volume NH3 = (0.0003380 mol) / 0.00500 mol dm
-3
3
= 0.06760 dm = 67.60 cm
3
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