Titration Curve Review
Titration curve - plot of pH vs volume of titrant added
For curve - important points to calculate pH and plot are:
1. initial point; 2. eq point; 3. right before eq point; 4. right after eq point; 5. well after eq pt; and
6. half-equivalence point for weak acid or base only
In class we did this…
0.10 M HCl 50 mL with 0.20 M NaOH - Strong acid and strong base
1. Initial point (0 mL of NaOH added)
2. Equivalence point
HCl -> H
+
+ Cl
-
First let's calculate what volume of NaOH was added
0.1 M -> 0.1 M using mol H
+
= mol OH
-
or MH+ Va = MOH- Vb
pH = -log (0.1) =
1
so Vb = (0.10 M)(50 mL)/(0.20 M ) = 25 mL
Since it's a strong acid-strong base titration, at eq pt
H
+
+ OH
-
-> H2O it's neutral! pH = 7
3. Before eq pt, VNaOH = 24 mL
Let's use stoichiometry -> moles
H
+
+ OH
-
-> H2O
I (initial) 5 4.8 mol H
+
= M X L* = 0.1 X 50 mL = 5 mmol
C (change) -4.8 -4.8 (Limiting Reactant) mol OH
-
= M X L = 0.2 X 24 mL = 4.8 mmol
F
(final)
0.2
0
* Could convert to L or use mL and find mmol
pH = -log (0.0.0027) = 2.6 M H
+
= mol/Vtotal = 0.2 mol/74 mL = 0.0027 M
4. After eq pt, VNaOH = 26 mL
H
+
+ OH
-
-> H2O
I
(initial)
5
5.2
C (change) -5 -5 (Limiting Reactant) mol OH
-
= M X L = 0.2 X 26 mL = 5.2 mmol
F
(final)
0
0.2
pOH = -log (0.0.0026) = 2.6; pH = 11.4 M OH
-
= mol/Vtotal = 0.2 mol/76 mL = 0.0026 M
5. Well after eq pt, I'll pick 35 mL
H
+
+ OH
-
-> H2O
I
(initial)
5
7
C (change) -5 -5 (Limiting Reactant) mol OH
-
= M X L = 0.2 X 35 mL = 7mmol
F
(final)
0
2
pOH = -log (0.0.024) = 1.6; pH = 12.4 M OH
-
= mol/Vtotal = 2 mol/85 mL = 0.024 M
Volume
NaOH (mL)
pH
0
1
24
2.6
25
7
26
11.4
35
12.4
0
2
4
6
8
10
12
14
0 5 10 15 20 25 30 35 40
pH
Volume NaOH (mL)
Strong Acid - Strong Base Titration
