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Titration Curve: Calculating pH at Different Stages of Strong Acid-Strong Base Titration, Study notes of Stoichiometry

A step-by-step calculation of ph values at various stages during the titration of a strong acid (0.10 m hcl) with a strong base (0.20 m naoh). The initial point, equivalence point, and points before and after the equivalence point. It also compares the titration curve of a strong acid-strong base system with that of a weak acid-strong base system.

What you will learn

  • What is the pH value at the initial point of a strong acid-strong base titration?
  • What is the pH value after the equivalence point in a strong acid-strong base titration?
  • How does the pH value change at the equivalence point of a strong acid-strong base titration?

Typology: Study notes

2021/2022

Uploaded on 09/27/2022

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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
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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
  2. 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 M (^) H+ V (^) a = M (^) OH- V (^) b 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-^ -> H 2 O it's neutral! pH = 7 3. Before eq pt, V (^) NaOH = 24 mL Let's use stoichiometry -> moles H+^ + OH-^ -> H 2 O 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, V (^) NaOH = 26 mL H
        • OH

          -> H 2 O 

I (initial) 5 5. C (change) -5 -5 (Limiting Reactant) mol OH - = M X L = 0.2 X 26 mL = 5.2 mmol F (final) 0 0. pOH = -log (0.0.0026) = 2.6; pH = 11.4 (^) M OH-^ = mol/Vtotal = 0.2 mol/76 mL = 0.0026 M

  1. Well after eq pt, I'll pick 35 mL H+^ + OH-^ -> H 2 O 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. 25 7 26 11. 35 12. 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

Let's compare Strong Acid/Strong Base Titration with that of Weak Acid/Strong Base (or vice versa) both with 50.0 mL 0.10 M acid with 0.20 M NaOH - Strong acid and strong base

1. Initial point (0 mL of NaOH added) 1. Initial point (0 mL of NaOH added) HCl -> H+^ + Cl-^ HA <-> H+^ + A- 0.1 M -> 0.1 M Create Ice Table, Need Ka, Solve for H+ then pH = -log (0.1) = 1 find pH 2. Equivalence point 2. Equivalence point First let's calculate what volume of NaOH was added <-- Do same to find volume of base but then… using mol H + = mol OH - or M (^) H+ V (^) a = M (^) OH- V (^) b HA^ +^ OH - -> H2O + A - so Vb = (0.10 M)(50 mL)/(0.20 M ) = 25 mL Use ICF table to determine amount of A- produced Since it's a strong acid-strong base titration, at eq pt in moles but then convert to concentration H + + OH - -> H 2 O it's neutral! pH = 7 moles HA and OH- will be 0 since eq pt Finally consider… A- + H2O <-> HA +OH- Create Ice table with A- initial conc, need to calc Kb using Kw/Ka, solve for OH-, then pOH-> pH YIKES!!! 3. Before eq pt, V (^) NaOH = 24 mL 3. Before eq pt, V (^) NaOH = 24 mL Let's use stoichiometry -> moles <-- Do same but with equation… H+^ + OH-^ -> H 2 O HA^ +^ OH - -> H2O + A - I (initial) (^5) 4.8 I C (change) -4.8 -4.8 (Limiting Reactant) C F (final) 0.2 0 F x 0 y pH = -log (0.0.0027) = 2.6 Then use Henderson-Hasselbach with y and x 4. After eq pt, V (^) NaOH = 26 mL 4. After eq pt, V (^) NaOH = 26 mL H + + OH - -> H 2 O <-- Do same but with equation… I (initial) 5 5.2 HA + OH-^ -> H2O + A- C (change) -5 -5 (Limiting Reactant) I F (final) 0 0.2 C F 0 x y pOH = -log (0.0.0026) = 2.6; pH = 11. /\ (0.2/76) Then, can try ICE with A- like you did at end of step 2 but you WILL have an initial amount of OH-. But majority of the OH- will come from the excess, unreacted OH-. So you really can do the same thing you did at the end on the left for strong acid/base

  1. Well after eq pt, I'll pick 35 mL 5. Well after eq pt, I'll pick 35 mL H + + OH - -> H 2 O <-- Do same but with equation… I (initial) 5 7 HA + OH-^ -> H2O + A- C (change) (^) -5 -5 (Limiting Reactant) I F (final) 0 2 C F 0 x y pOH = -log (0.0.024) = 1.6; pH = 12. Then just find conc of excess, unreacted OH- and do the same thing you did on the left
    1. Remember, you also have pH=pKa at half eq pt.