UNIT: 8.4 P.1 Acid-Base Titration Notes

Introduction to Titration

This video focuses on the first part of unit 8.4, covering acid-base reactions, specifically titration.
Titration will be discussed in two parts due to the extensive material.

Titration involves the gradual addition of a solution (titrant) of known concentration to another solution (analyte) of unknown concentration to determine the analyte's concentration.
The concept of titration was briefly introduced in unit four, including titration curves and the equivalence point.
In a typical titration setup, a base is added to an acid or vice versa, and the change in pH is monitored.
Titrant: The solution with a known concentration that is added during titration (e.g., base in the graph).
Analyte: The solution with an unknown concentration that is being titrated (e.g., acid in the graph).
Titration can involve an acid plus base, as seen in this video, but it can also be redox titrations.

Equivalence Point Definition: The point in a titration where the moles of titrant added are stoichiometrically equivalent to the moles of analyte in the sample.
For acid-base reactions, the equivalence point is simplified due to the one-to-one stoichiometric ratio.
Simplified Equivalence Point: Moles of acid equal moles of base.

Wommich's Problems (from Unit 8.3): ICE table problems involving reactions with water as a reactant.
- ICE tables used molarities.
- Involved adding, subtracting, and solving for x.
Neumann Problems (New Type): ICE table problems involving neutralization reactions (acid + base).
- ICE tables will use moles instead of molarities.
- Involve adding and subtracting actual numbers instead of x.
- No need for graphing calculators to solve for x.

The simplest type of titration involves a strong acid plus a strong base.
Example: 40 mL of 0.2 M perchloric acid (HClO_4) is titrated with potassium hydroxide (KOH).
Key Step: Always identify whether the acid or base is strong or weak.
Perchloric acid is a strong acid, and potassium hydroxide is a strong base.
The task is to calculate the pH after adding 10, 40, 80, and 100 mL of the base.

At the start, only the perchloric acid is present.
Strong acids dissociate completely into H^+ and ClO_4^-.
The molarity of a strong acid equals the molarity of H^+. Therefore, pH can be calculated directly from the acid concentration.
pH = -\log[H^+]
pH = -\log[0.2] = 0.699
A low pH is expected because only strong acid is present.

Update the major species list to include the strong base and its dissociation products.
The major species are water, H^+, ClO_4^-, K^+, and OH^-.
Identify the strongest acid and strongest base in the list.
- H^+ is the strongest acid.
- OH^- is the strongest base.
The reaction is a neutralization reaction: H^+ + OH^- \rightarrow H_2O
Set up an ICE table using moles or millimoles.
Calculate initial moles of H^+: 40 mL * 0.2 M = 8 millimoles
Calculate initial moles of OH^-: 10 mL * 0.2 M = 2 millimoles
Since OH^- is the limiting reactant, subtract 2 millimoles from both H^+ and OH^-.
Final moles of H^+: 8 - 2 = 6 millimoles
Final moles of OH^-: 2 - 2 = 0 millimoles
Calculate the new concentration of H^+: 6 millimoles / (40 mL + 10 mL) = 0.12 M
Calculate the pH: pH = -\log[0.12] = 0.921
The pH has increased slightly, indicating a slow change in the initial part of the titration.

Repeat the process with 40 mL of KOH.
Initial moles of H^+: 8 millimoles
Initial moles of OH^-: 40 mL * 0.2 M = 8 millimoles
H^+ + OH^- \rightarrow H_2O
Since the moles of acid and base are equal, subtract 8 millimoles from both.
Final Moles of H^+: 8 - 8 = 0 millimoles
Final Moles of OH^-: 8 - 4 = 4 millimoles
The pH has again risen, but not dramatically.
This is the halfway point or the half equivalence point.
- Half of the analyte (acid) has been neutralized at the halfway point.

Now, the problem involves 80 mL of base (KOH).
Initial moles of H^+: 8 millimoles
Initial moles of OH^-: 80 mL * 0.1 M = 8 millimoles
H^+ + OH^- \rightarrow H_2O
Since the moles of acid and base are equal subtract 8 millimoles from both
In this scenario, there is no limiting reactant, all the acid, and all the base have been completely reacted.
No leftover H^+ or OH^- is present.
The solution consists only of water which means that
The pH=7
This represents the equivalence point, where moles of acid equal moles of base.
For strong acid-strong base titrations, the equivalence point always occurs at pH 7.

The pH increases slowly at the beginning of the titration.
Approaching the equivalence point produces a dramatic and rapid increase in the pH.
Beyond the equivalence point, there is only an excess of OH^-.

Question: 30 mL of 0.03 M HNO_3 is titrated with 70 mL of 0.02 M KOH. Which is the limiting reactant, and what is the approximate pH?
Since both the acid and the base is strong we can derive the same net ionic equation from our previous example, namely:
H^+ + OH^- \rightarrow H_2O
- Initial moles of H^+: 30 mL * 0.03 M = 0.9 millimoles
Initial moles of OH^-: 70 mL * 0.02 M = 1.4 millimoles
HNO_3 is the limiting reactant, we are left with some excess hydroxide after a reaction, and answer choices (a) and (b) can be eliminated since acid can't be present in the solutions.
Find the difference between hydroxide and hydronium:
1.4-0.9=0.5
- Divide this number by the total volume to obtain the concentration of hydroxide:
[OH^-]=\frac{0.5}{30+70}= \frac{0.5}{100} = 0.005 M = 5*10^{-3} M
- Find the estimated pOH, we know that it is very close to our experient because of the logarithms:
pOH \approx 3
- To find a proper pH, subtract the pOH from 14 as follows:
pH=14-3=11
- Therefore, the answer is D, 12, since it's the closest number to the true value obtained from more precise calculations.

This video covered the basics of titration calculations with a strong acid and a strong base.
The next video will cover titrations involving weak acids or weak bases.