Acid-Base Titrations Section 17.3. Introduction Definition: – In an acid-base titration, a...
-
Upload
cornelia-watts -
Category
Documents
-
view
222 -
download
0
Transcript of Acid-Base Titrations Section 17.3. Introduction Definition: – In an acid-base titration, a...
Acid-Base Titrations
Section 17.3
Introduction
• Definition:– In an acid-base titration, a solution containing a
known concentration of a base is slowly added to an acid.
– An indicator is used to signal the equivalence point of the titration.• This is the point at which stoichiometrically equivalent
amounts of acid and base have been mixed.
– A pH meter can also be used to find the equivalence point.
Introduction
• The typical titration apparatus includes:– a buret to hold the titrant– a beaker to hold the analyte– a pH meter to measure the pH
Introduction
• In this section, we will be looking at a series of titrations in detail to understand why acids and behave the way they do.– Strong acid-strong base titration– Weak acid-strong base titration– Polyprotic acid-strong base titration
Strong Acid-Strong Base Titrations
• The titration curve of a strong acid-strong base titration has the following shape.
Strong Acid-Strong Base Titrations
• The titration curve of a strong acid-strong base titration has the following shape.
We divide the curve into four regions:1. Initial pH
2. Initial pH to eq. point
3. Equivalence point
4. After eq. point
Strong Acid-Strong Base Titrations
• The titration curve of a strong acid-strong base titration has the following shape.
We divide the curve into four regions:1. Initial pH
The pH of the solution is determined by the concentration of the strong acid.
Strong Acid-Strong Base Titrations
• The titration curve of a strong acid-strong base titration has the following shape.
We divide the curve into four regions:
As base is added, pH increases slowly and then rapidly. The pH is determined by the concentration of the acid that is not neutralized.
2. Initial pH to eq. point
Strong Acid-Strong Base Titrations
• The titration curve of a strong acid-strong base titration has the following shape.
We divide the curve into four regions:
At the equivalence point, [OH−] = [H+]. The pH = 7.00.
3. Equivalence point
Strong Acid-Strong Base Titrations
• The titration curve of a strong acid-strong base titration has the following shape.
We divide the curve into four regions:
As more base is added, pH increases rapidly and then slowly. pH is determined by the concentration of the excess base.
4. After eq. point
Strong Acid-Strong Base Titrations
• The titration curve of a strong acid-strong base titration has the following shape.
Let’s see how this works in practice.
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLb) 51.0 mL
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mL
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLThis is between the initial point and the equivalence point.
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLThis is between the initial point and the equivalence point. pH is determined by the amount of acid that has not been neutralized.
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLTherefore, we need to determine the number of mols of acid remaining, nacid, and the total volume, Vtotal, of the solution. (Remember, adding the NaOH increases the total volume.
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,i = Macid,i × Vacid,i
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,i = Macid,i × Vacid,i = (0.100 M)(0.0500 L)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,i = Macid,i × Vacid,i = 5.00 × 10−3 mol
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,i = Macid,i × Vacid,i = 5.00 × 10−3 mol
nbase,added = Mbase,added × Vbase,added
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,i = Macid,i × Vacid,i = 5.00 × 10−3 mol
nbase,added = (0.100 M)(0.0490 L)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,i = Macid,i × Vacid,i = 5.00 × 10−3 mol
nbase,added = 4.90 × 10−3 mol
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,i = Macid,i × Vacid,i = 5.00 × 10−3 mol
nbase,added = 4.90 × 10−3 mol
nacid,remaining = nacid,i − nbase,added
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,i = Macid,i × Vacid,i = 5.00 × 10−3 mol
nbase,added = 4.90 × 10−3 mol
nacid,remaining = (5.00 − 4.90) × 10−3 mol
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,i = Macid,i × Vacid,i = 5.00 × 10−3 mol
nbase,added = 4.90 × 10−3 mol
nacid,remaining = 0.10 × 10−3 mol
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,i = Macid,i × Vacid,i = 5.00 × 10−3 mol
nbase,added = 4.90 × 10−3 mol
nacid,remaining = 1.0 × 10−4 mol
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,remaining = 1.0 × 10−4 mol
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,remaining = 1.0 × 10−4 mol
Vtotal = Vacid + Vbase
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,remaining = 1.0 × 10−4 mol
Vtotal = Vacid + Vbase = 0.0500 L + 0.0490 L
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,remaining = 1.0 × 10−4 mol
Vtotal = Vacid + Vbase = 0.0990 L
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,remaining = 1.0 × 10−4 mol
Vtotal = 0.0990 L
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,remaining = 1.0 × 10−4 mol
Vtotal = 0.0990 L
[H+] = nacid,remaining/Vtotal
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,remaining = 1.0 × 10−4 mol
Vtotal = 0.0990 L
[H+] = (1.0 × 10−4 mol)/(0.0990 L)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,remaining = 1.0 × 10−4 mol
Vtotal = 0.0990 L
[H+] = 1.0 × 10−3 M
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,remaining = 1.0 × 10−4 mol
Vtotal = 0.0990 L
[H+] = 1.0 × 10−3 MpH = −log[H+]
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,remaining = 1.0 × 10−4 mol
Vtotal = 0.0990 L
[H+] = 1.0 × 10−3 MpH = −log(1.0 × 10−3)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,remaining = 1.0 × 10−4 mol
Vtotal = 0.0990 L
[H+] = 1.0 × 10−3 MpH = 3.00
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLpH = 3.00
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mL
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLb) 51.0 mL
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.
b) 51.0 mL
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mL
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLThis is beyond the equivalence point.
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLThis is beyond the equivalence point. All of the strong acid has been used up.
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLThis is beyond the equivalence point. All of the strong acid has been used up. The pH is determined by the excess base that has been added.
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLWe already determined nacid,i = 5.00 × 10−3 mol
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnacid,i = 5.00 × 10−3 mol
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnacid,i = 5.00 × 10−3 mol
nbase,added = Mbase,added × Vbase,added
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnacid,i = 5.00 × 10−3 mol
nbase,added = (0.100 M)(0.0510 L)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnacid,i = 5.00 × 10−3 mol
nbase,added = 5.10 × 10−3 mol
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnacid,i = 5.00 × 10−3 mol
nbase,added = 5.10 × 10−3 mol
nbase,remaining = nbase,added − nacid,i
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnacid,i = 5.00 × 10−3 mol
nbase,added = 5.10 × 10−3 mol
nbase,remaining = (5.10 − 5.00) × 10−3 mol
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnacid,i = 5.00 × 10−3 mol
nbase,added = 5.10 × 10−3 mol
nbase,remaining = 0.10 × 10−3 mol
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnbase,remaining = 1.0 × 10−4 mol
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnbase,remaining = 1.0 × 10−4 mol
Vtotal = Vacid + Vbase
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnbase,remaining = 1.0 × 10−4 mol
Vtotal = Vacid + Vbase = 0.0500 L + 0.0510 L
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnbase,remaining = 1.0 × 10−4 mol
Vtotal = Vacid + Vbase = 0.1010 L
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnbase,remaining = 1.0 × 10−4 mol
Vtotal = 0.1010 L
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnbase,remaining = 1.0 × 10−4 mol
Vtotal = 0.1010 L
[OH−] = nbase,remaining/Vtotal
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnbase,remaining = 1.0 × 10−4 mol
Vtotal = 0.1010 L
[OH−] = (1.0 × 10−4 mol)/(0.1010 L)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnbase,remaining = 1.0 × 10−4 mol
Vtotal = 0.1010 L
[OH−] = 9.9 × 10−4 M
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnbase,remaining = 1.0 × 10−4 mol
Vtotal = 0.1010 L
[OH−] = 9.9 × 10−4 MpOH = −log[OH−]
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnbase,remaining = 1.0 × 10−4 mol
Vtotal = 0.1010 L
[OH−] = 9.9 × 10−4 MpOH = −log(9.9 × 10−4)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnbase,remaining = 1.0 × 10−4 mol
Vtotal = 0.1010 L
[OH−] = 9.9 × 10−4 MpOH = 3.00
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnbase,remaining = 1.0 × 10−4 mol
Vtotal = 0.1010 L
[OH−] = 9.9 × 10−4 MpOH = 3.00 pH = 14.00 − 3.00⇒
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnbase,remaining = 1.0 × 10−4 mol
Vtotal = 0.1010 L
[OH−] = 9.9 × 10−4 MpOH = 3.00 pH = 14.00 − 3.00 = 11.00⇒
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnbase,remaining = 1.0 × 10−4 mol
Vtotal = 0.1010 L
[OH−] = 9.9 × 10−4 MpH = 11.00
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLpH = 11.00
Weak Acid-Strong Base Titrations
• The curve for a weak acid-strong base titration is similar to that for a strong acid-weak base.
Weak Acid-Strong Base Titrations
• The curve for a weak acid-strong base titration is similar to that for a strong acid-weak base.
1. The initial pH is determined by the Ka of the acid.
Weak Acid-Strong Base Titrations
• The curve for a weak acid-strong base titration is similar to that for a strong acid-weak base.
2. To determine the pH from the initial point to the eq. point, we first neutralize the weak acid and then use the Henderson-Hasselbach equation.
Weak Acid-Strong Base Titrations
• The curve for a weak acid-strong base titration is similar to that for a strong acid-weak base.
3. At the eq. point, we have no HX, only X−. We need to use the Kb value to find pH.
Weak Acid-Strong Base Titrations
• The curve for a weak acid-strong base titration is similar to that for a strong acid-weak base.
4. Beyond the eq. point, we use the excess base to calculate pH.
Weak Acid-Strong Base Titrations
• The curve for a weak acid-strong base titration is similar to that for a strong acid-weak base.
Let’s see how this works in practice.
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
First, calculate the concentrations of materials before neutralization reaction.
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
First, calculate the concentrations of materials before neutralization reaction.
CH3COOH + OH− CH➙ 3COO− + H2O
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
First, calculate the concentrations of materials before neutralization reaction.
CH3COOH + OH− CH➙ 3COO− + H2O
ninitial,acid = Macid × Vacid
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
First, calculate the concentrations of materials before neutralization reaction.
CH3COOH + OH− CH➙ 3COO− + H2O
ninitial,acid = Macid × Vacid = (0.100 M)(0.0500 L)
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
First, calculate the concentrations of materials before neutralization reaction.
CH3COOH + OH− CH➙ 3COO− + H2O
ninitial,acid = Macid × Vacid = 5.00 × 10−3 mol
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
First, calculate the concentrations of materials before neutralization reaction.
CH3COOH + OH− CH➙ 3COO− + H2O
ninitial,acid = 5.00 × 10−3 mol
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
First, calculate the concentrations of materials before neutralization reaction.
CH3COOH + OH− CH➙ 3COO− + H2O
ninitial,acid = 5.00 × 10−3 mol
nbase = Mbase × Vbase
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
First, calculate the concentrations of materials before neutralization reaction.
CH3COOH + OH− CH➙ 3COO− + H2O
ninitial,acid = 5.00 × 10−3 mol
nbase = Mbase × Vbase = (0.100 M)(0.0450 L)
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
First, calculate the concentrations of materials before neutralization reaction.
CH3COOH + OH− CH➙ 3COO− + H2O
ninitial,acid = 5.00 × 10−3 mol
nbase = Mbase × Vbase = 4.50 × 10−3 mol
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
First, calculate the concentrations of materials before neutralization reaction.
CH3COOH + OH− CH➙ 3COO− + H2O
ninitial,acid = 5.00 × 10−3 mol
nbase = 4.50 × 10−3 mol
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
Next, use the values to determine the concentrations after the neutralization.
CH3COOH + OH− CH➙ 3COO− + H2O
ninitial,acid = 5.00 × 10−3 mol
nbase = 4.50 × 10−3 mol
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
Next, use the values to determine the concentrations after the neutralization.
CH3COOH + OH− CH➙ 3COO− + H2Oninitial 5.0 × 10−3 mol 4.5 × 10−3 mol 0.0 mol
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
Next, use the values to determine the concentrations after the neutralization.
CH3COOH + OH− CH➙ 3COO− + H2Oninitial 5.0 × 10−3 mol 4.5 × 10−3 mol 0.0 mol
nchange −4.5 × 10−3 mol −4.5 × 10−3 mol +4.5 × 10−3 mol
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
Next, use the values to determine the concentrations after the neutralization.
CH3COOH + OH− CH➙ 3COO− + H2Oninitial 5.0 × 10−3 mol 4.5 × 10−3 mol 0.0 mol
nchange −4.5 × 10−3 mol −4.5 × 10−3 mol +4.5 × 10−3 mol
nfinal 5.0 × 10−4 mol 0.0 mol 4.5 × 10−3 mol
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
Next, use the values to determine the concentrations after the neutralization.
CH3COOH + OH− CH➙ 3COO− + H2O
nacid,final = 5.0 × 10−4 mol
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
Next, use the values to determine the concentrations after the neutralization.
CH3COOH + OH− CH➙ 3COO− + H2O
nacid,final = 5.0 × 10−4 mol
[acid] = nacid,final/Vtotal
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
Next, use the values to determine the concentrations after the neutralization.
CH3COOH + OH− CH➙ 3COO− + H2O
nacid,final = 5.0 × 10−4 mol
[acid] = (5.0 × 10−4 mol)/(0.0950 L)
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
Next, use the values to determine the concentrations after the neutralization.
CH3COOH + OH− CH➙ 3COO− + H2O
nacid,final = 5.0 × 10−4 mol
[acid] = 5.3 × 10−3 M
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
Next, use the values to determine the concentrations after the neutralization.
CH3COOH + OH− CH➙ 3COO− + H2O
[acid] = 5.3 × 10−3 M[base] = nbase,final/Vtotal
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
Next, use the values to determine the concentrations after the neutralization.
CH3COOH + OH− CH➙ 3COO− + H2O
[acid] = 5.3 × 10−3 M[base] = (4.50 × 10−3 mol)/(0.0950 L)
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
Next, use the values to determine the concentrations after the neutralization.
CH3COOH + OH− CH➙ 3COO− + H2O
[acid] = 5.3 × 10−3 M[base] = 4.74 × 10−2 M
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
Next, use the values to determine the concentrations after the neutralization.
CH3COOH + OH− CH➙ 3COO− + H2O
[acid] = 5.3 × 10−3 M [base] = 4.74 × 10−2 M
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
Next, use the values to determine the concentrations after the neutralization.
CH3COOH + OH− CH➙ 3COO− + H2O
[acid] = 5.3 × 10−3 M [base] = 4.74 × 10−2 MpKa = −logKa
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
Next, use the values to determine the concentrations after the neutralization.
CH3COOH + OH− CH➙ 3COO− + H2O
[acid] = 5.3 × 10−3 M [base] = 4.74 × 10−2 MpKa = −log(1.8 × 10−5)
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
Next, use the values to determine the concentrations after the neutralization.
CH3COOH + OH− CH➙ 3COO− + H2O
[acid] = 5.3 × 10−3 M [base] = 4.74 × 10−2 MpKa = 4.74
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
Next, use the values to determine the concentrations after the neutralization.
CH3COOH + OH− CH➙ 3COO− + H2O
pH = pKa + log([base]/[acid)
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
Next, use the values to determine the concentrations after the neutralization.
CH3COOH + OH− CH➙ 3COO− + H2O
pH = 4.74 + log([4.74 × 10−2 M]/[5.3 × 10−3 M])
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
Next, use the values to determine the concentrations after the neutralization.
CH3COOH + OH− CH➙ 3COO− + H2O
pH = 4.74 + log(9.00)
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
Next, use the values to determine the concentrations after the neutralization.
CH3COOH + OH− CH➙ 3COO− + H2O
pH = 4.74 + 0.954
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
Next, use the values to determine the concentrations after the neutralization.
CH3COOH + OH− CH➙ 3COO− + H2O
pH = 5.69
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
At the equivalence point, all of the weak acid is converted to its conjugate weak base.
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
At the equivalence point, all of the weak acid is converted to its conjugate weak base. This means that we will be using the base hydrolysis expression to find [OH−], pOH, and pH.
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
The number of mols of acetate in solution at the equivalence point is equal to the number of mols of acetic acid at the start of the titration.
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
nbase,eq. pt. = ninitial,acid
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
nbase,eq. pt. = ninitial,acid = (0.100 M)(0.0500 L)
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
nbase,eq. pt. = ninitial,acid = 5.00 × 10−3 mol
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
nbase,eq. pt. = 5.00 × 10−3 mol
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
nbase,eq. pt. = 5.00 × 10−3 mol
The concentration of the acetate is given by:
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
nbase,eq. pt. = 5.00 × 10−3 mol
The concentration of the acetate is given by:[base] = (nbase,eq. pt.)/(Vtotal)
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
nbase,eq. pt. = 5.00 × 10−3 mol
The concentration of the acetate is given by:[base] = (5.00 × 10−3 mol)/(0.100 L)
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
nbase,eq. pt. = 5.00 × 10−3 mol
The concentration of the acetate is given by:[base] = 5.00 × 10−2 M
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
[base] = 5.00 × 10−2 M
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
[base] = 5.00 × 10−2 MNext, we do an i-c-e table on the base hydrolysis.
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
[base] = 5.00 × 10−2 MCH3COO− + H2O CH➙ 3COOH + OH−
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
[base] = 5.00 × 10−2 MCH3COO− + H2O CH➙ 3COOH + OH−
i 5.00 × 10−2 0.0 0.0
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
[base] = 5.00 × 10−2 MCH3COO− + H2O CH➙ 3COOH + OH−
i 5.00 × 10−2 0.0 0.0c −x +x +x
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
[base] = 5.00 × 10−2 MCH3COO− + H2O CH➙ 3COOH + OH−
i 5.00 × 10−2 0.0 0.0c −x +x +xe 5.00 × 10−2 − x x x
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
[base] = 5.00 × 10−2 M, [acid] = [OH−] = xCH3COO− + H2O CH➙ 3COOH + OH−
i 5.00 × 10−2 0.0 0.0c −x +x +xe 5.00 × 10−2 − x x x
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
[base] = 5.00 × 10−2 M, [acid] = [OH−] = xKb = Kw/Ka = (1.0 × 10−14)/(1.8 × 10−5)
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
[base] = 5.00 × 10−2 M, [acid] = [OH−] = xKb = Kw/Ka = 5.6 × 10 −10
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
[base] = 5.00 × 10−2 M, [acid] = [OH−] = xKb = 5.6 × 10 −10
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
[base] = 5.00 × 10−2 M, [acid] = [OH−] = xKb = 5.6 × 10 −10 = ([acid][OH−])/[base]
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
[base] = 5.00 × 10−2 M, [acid] = [OH−] = xKb = 5.6 × 10 −10 = (x2)/(5.00 × 10−2)
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
[base] = 5.00 × 10−2 M, [acid] = [OH−] = xKb = 5.6 × 10 −10 = (x2)/(5.00 × 10−2)
x2 = (5.6 × 10 −10)(5.00 × 10−2) = 2.8 × 10−11
x = (2.8 × 10−11)½
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
[base] = 5.00 × 10−2 M, [acid] = [OH−] = xKb = 5.6 × 10 −10 = (x2)/(5.00 × 10−2)
x2 = (5.6 × 10 −10)(5.00 × 10−2) = 2.8 × 10−11
x = (2.8 × 10−11)½ = 5.3 × 10−6
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
[base] = 5.00 × 10−2 M, [acid] = [OH−] = xKb = 5.6 × 10 −10 = (x2)/(5.00 × 10−2)
x2 = (5.6 × 10 −10)(5.00 × 10−2) = 2.8 × 10−11
x = (2.8 × 10−11)½ = 5.3 × 10−6 = [OH−]
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
[OH−] = 5.3 × 10−6 M
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
[OH−] = 5.3 × 10−6 MpOH = −log[OH−]
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
[OH−] = 5.3 × 10−6 MpOH = −log[OH−] = −log(5.3 × 10−6)
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
[OH−] = 5.3 × 10−6 MpOH = −log[OH−] = −log(5.3 × 10−6) = 5.28
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
[OH−] = 5.3 × 10−6 MpOH = 5.28
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
[OH−] = 5.3 × 10−6 MpOH = 5.28pH = 14.00 – pOH
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
[OH−] = 5.3 × 10−6 MpOH = 5.28pH = 14.00 – pOH = 14.00 − 5.28
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
[OH−] = 5.3 × 10−6 MpOH = 5.28pH = 14.00 – pOH = 14.00 − 5.28 = 8.72
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
pH = 8.72
• The pH of the titration at the equivalence point depends on the type of acids and bases being titrated.– Strong acid-Strong base: eq. pt. = 7.0– Weak acid-Strong base: eq. pt. > 7.0– Strong acid-Weak base: eq. pt. < 7.0
Titrations of Polyprotic Acids
• When weak acid contain more than one ionizable H atom, as in phosporous acid, H3PO3, reaction with OH− occurs in a series of steps.– H3PO3(aq) + H2O(l) H➙ 2PO3
−(aq) + H3O+(aq)
– H2PO3−(aq) + H2O(l) HPO➙ 3
2−(aq) + H3O+(aq)
– HPO32−(aq) + H2O(l) PO➙ 3
3−(aq) + H3O+(aq)
Titrations of Polyprotic Acids
• When the neutralization steps are sufficiently separated, the substance exhibits multiple equivalence points.