Mr Cole Chemistry

Buffer Questions

pH of Buffers

Acids and Bases Worksheet

1. Standard Buffer Calculations (Acid + Salt)

a. Calculate the pH of a buffer that was formed from 50.0 cm³ of 0.500 mol dm⁻³ ethanoic acid (Ka of 1.74×10⁻⁵) and 30.0 cm³ of 0.400 mol dm⁻³ sodium ethanoate.
Answer

Step 1: Calculate Moles
Moles Acid (HA) = 0.500 × 0.050 = 0.025 mol
Moles Salt (A⁻) = 0.400 × 0.030 = 0.012 mol

Step 2: Calculate [H⁺]
[H⁺] = Kₐ × (mol HA / mol A⁻)
[H⁺] = 1.74×10⁻⁵ × (0.025 / 0.012) = 3.625×10⁻⁵

Step 3: Calculate pH
pH = -log(3.625×10⁻⁵) = 4.44

b. Calculate the pH of a buffer that was formed from the addition of 1.00 g of sodium propanoate to 200 cm³ of 0.200 mol dm⁻³ propanoic acid. The Ka of propanoic acid is 1.35×10⁻⁵.
Answer

Step 1: Calculate Moles
Moles Acid (HA) = 0.200 × 0.200 = 0.040 mol
Moles Salt (A⁻) = Mass / Mr = 1.00 / 96.1 = 0.0104 mol
(Mr C₂H₅COONa = 96.1)

Step 2: Calculate [H⁺]
[H⁺] = 1.35×10⁻⁵ × (0.040 / 0.0104) = 5.19×10⁻⁵

Step 3: Calculate pH
pH = -log(5.19×10⁻⁵) = 4.28

c. Calculate the pH of a buffer that was formed from the addition of 100 cm³ of 0.500 cm³ sodium propanoate to 500 cm³ of 1.00 mol dm⁻³ propanoic acid. The Ka of propanoic acid is 1.35×10⁻⁵.
Answer

(Note: The question states “0.500 cm³” for sodium propanoate, but the calculation assumes “0.500 mol dm⁻³”.)

Step 1: Calculate Moles
Moles Acid (HA) = 1.00 × 0.500 = 0.500 mol
Moles Salt (A⁻) = 0.500 × 0.100 = 0.050 mol

Step 2: Calculate [H⁺]
[H⁺] = 1.35×10⁻⁵ × (0.500 / 0.050) = 1.35×10⁻⁴

Step 3: Calculate pH
pH = -log(1.35×10⁻⁴) = 3.87

d. Calculate the pH of a buffer that was formed from the addition of 80.0 cm³ of 0.100 mol dm⁻³ methanoic acid and 30.0 cm³ of 0.0500 mol dm⁻³ potassium methanoate. The Ka of methanoic acid is 1.78×10⁻⁴.
Answer

Step 1: Calculate Moles
Moles Acid (HA) = 0.100 × 0.080 = 0.008 mol
Moles Salt (A⁻) = 0.0500 × 0.030 = 0.0015 mol

Step 2: Calculate [H⁺]
[H⁺] = 1.78×10⁻⁴ × (0.008 / 0.0015) = 9.49×10⁻⁴

Step 3: Calculate pH
pH = -log(9.49×10⁻⁴) = 3.02

2. Partial Neutralisation (Acid + Strong Base)

a. Calculate the pH of a buffer made from the addition of 100 cm³ of 0.500 mol dm⁻³ ethanoic acid and 80.0 cm³ of 0.550 mol dm⁻³ sodium hydroxide. The Ka of ethanoic acid is 1.74×10⁻⁵.
Answer

Step 1: Initial Moles
Acid (HA) = 0.500 × 0.100 = 0.050 mol
Base (OH⁻) = 0.550 × 0.080 = 0.044 mol

Step 2: ICE Table (Moles)

HA+OH⁻→A⁻+H₂O
Initial0.0500.0440–
Change-0.044-0.044+0.044–
Final0.00600.044–

Step 3: Calculate pH
[H⁺] = 1.74×10⁻⁵ × (0.006 / 0.044) = 2.37×10⁻⁶
pH = -log(2.37×10⁻⁶) = 5.62

b. Calculate the pH of a buffer made from 500 cm³ of 0.400 mol dm⁻³ methanoic acid and 100 cm³ of 0.500 mol dm⁻³ sodium hydroxide. The Ka of methanoic acid is 1.78×10⁻⁴.
Answer

Step 1: Initial Moles
Acid (HA) = 0.400 × 0.500 = 0.200 mol
Base (OH⁻) = 0.500 × 0.100 = 0.050 mol

Step 2: ICE Table (Moles)

HA+OH⁻→A⁻+H₂O
Initial0.2000.0500–
Change-0.050-0.050+0.050–
Final0.15000.050–

Step 3: Calculate pH
[H⁺] = 1.78×10⁻⁴ × (0.150 / 0.050) = 5.34×10⁻⁴
pH = -log(5.34×10⁻⁴) = 3.27

c. Calculate the pH of a buffer made from 600 cm³ of 1.00 mol dm⁻³ ethanoic acid and 1.80 g of sodium hydroxide. The Ka of ethanoic acid is 1.74×10⁻⁵.
Answer

Step 1: Initial Moles
Acid (HA) = 1.00 × 0.600 = 0.600 mol
Base (OH⁻) = 1.80 / 40.0 = 0.045 mol

Step 2: ICE Table (Moles)

HA+OH⁻→A⁻+H₂O
Initial0.6000.0450–
Change-0.045-0.045+0.045–
Final0.55500.045–

Step 3: Calculate pH
[H⁺] = 1.74×10⁻⁵ × (0.555 / 0.045) = 2.15×10⁻⁴
pH = -log(2.15×10⁻⁴) = 3.67

d. Calculate the pH of a buffer made from 800 cm³ of 0.500 mol dm⁻³ propanoic acid and 20.0 g of potassium hydroxide. The Ka of propanoic acid is 1.35×10⁻⁵.
Answer

Step 1: Initial Moles
Acid (HA) = 0.500 × 0.800 = 0.400 mol
Base (OH⁻) = 20.0 / 56.1 = 0.3565 mol
(Mr KOH = 56.1)

Step 2: ICE Table (Moles)

HA+OH⁻→A⁻+H₂O
Initial0.4000.35650–
Change-0.3565-0.3565+0.3565–
Final0.043500.3565–

Step 3: Calculate pH
[H⁺] = 1.35×10⁻⁵ × (0.0435 / 0.3565) = 1.65×10⁻⁶
pH = -log(1.65×10⁻⁶) = 5.78

3. Calculating Quantities from pH

a. Calculate the mass of sodium ethanoate that was added to 500 cm³ of 0.400 mol dm⁻³ ethanoic acid to produce a buffer with a pH of 4.50. Ethanoic acid has a Ka of 1.74×10⁻⁵.
Answer

Step 1: Find [H⁺] and Ratios
[H⁺] = 10⁻⁴·⁵ = 3.16×10⁻⁵
Moles Acid (HA) = 0.400 × 0.500 = 0.200 mol

Step 2: Solve for Moles Salt (A⁻)
[H⁺] = Kₐ × (mol HA / mol A⁻)
mol A⁻ = Kₐ × mol HA / [H⁺]
mol A⁻ = (1.74×10⁻⁵ × 0.200) / 3.16×10⁻⁵ = 0.110 mol

Step 3: Calculate Mass
Mass = 0.110 × 82.0 = 9.02 g
(Mr CH₃COONa = 82.0)

b. Calculate the mass sodium methanoate added to 200 cm³ of 0.200 mol dm⁻³ methanoic acid to produce a buffer with a pH of 3.80. The Ka of methanoic acid is 1.78×10⁻⁴.
Answer

Step 1: Find [H⁺] and Ratios
[H⁺] = 10⁻³·⁸ = 1.585×10⁻⁴
Moles Acid (HA) = 0.200 × 0.200 = 0.040 mol

Step 2: Solve for Moles Salt (A⁻)
mol A⁻ = (1.78×10⁻⁴ × 0.040) / 1.585×10⁻⁴ = 0.0449 mol

Step 3: Calculate Mass
Mass = 0.0449 × 68.0 = 3.05 g
(Mr HCOONa = 68.0)

c. A buffer was made from 500 cm³ ethanoic acid and 2.70g sodium ethanoate. The resulting buffer had a pH of 4.10 and the Ka of ethanoic acid is 1.74×10⁻⁵. Calculate the concentration of the ethanoic acid used.
Answer

Step 1: Moles of Salt
Mol A⁻ = 2.70 / 82.0 = 0.0329 mol

Step 2: Solve for Moles Acid (HA)
[H⁺] = 10⁻⁴·¹ = 7.94×10⁻⁵
mol HA = [H⁺] × mol A⁻ / Kₐ
mol HA = (7.94×10⁻⁵ × 0.0329) / 1.74×10⁻⁵ = 0.150 mol

Step 3: Calculate Concentration
[HA] = 0.150 / 0.500 = 0.300 mol dm⁻³

d. A buffer was made from 200 cm³ propanoic acid and 8.80 g of potassium propanoate. The resulting buffer had a pH of 4.57 and the Ka of propanoic acid is 1.35×10⁻⁵. Calculate the concentration of the propanoic acid used.
Answer

Step 1: Moles of Salt
Mol A⁻ = 8.80 / 112.2 = 0.0784 mol

Step 2: Solve for Moles Acid (HA)
[H⁺] = 10⁻⁴·⁵⁷ = 2.69×10⁻⁵
mol HA = (2.69×10⁻⁵ × 0.0784) / 1.35×10⁻⁵ = 0.156 mol

Step 3: Calculate Concentration
[HA] = 0.156 / 0.200 = 0.781 mol dm⁻³

4. Mixed Buffer Problems

a. A buffer is made from 250 cm³ of 0.600 mol dm⁻³ ethanoic acid and 100 cm³ of 0.600 mol dm⁻³ sodium hydroxide. The Ka of ethanoic acid is 1.74×10⁻⁵. Calculate the pH of the buffer solution formed.
Answer

Step 1: Initial Moles (Partial Neutralisation)
Acid (HA) = 0.600 × 0.250 = 0.150 mol
Base (OH⁻) = 0.600 × 0.100 = 0.060 mol

Step 2: ICE Table (Moles)

HA+OH⁻→A⁻+H₂O
Initial0.1500.0600–
Change-0.060-0.060+0.060–
Final0.09000.060–

Step 3: Calculate pH
[H⁺] = 1.74×10⁻⁵ × (0.090 / 0.060) = 2.61×10⁻⁵
pH = -log(2.61×10⁻⁵) = 4.58

b. A buffer is made from 250 cm³ of 0.0500 mol dm⁻³ ethanoic acid and 1.00 g of sodium ethanoate. The Ka of ethanoic acid is 1.74×10⁻⁵. Calculate the pH of the buffer solution formed.
Answer

Step 1: Calculate Moles
Acid (HA) = 0.050 × 0.250 = 0.0125 mol
Salt (A⁻) = 1.00 / 82.0 = 0.0122 mol

Step 2: Calculate [H⁺]
[H⁺] = 1.74×10⁻⁵ × (0.0125 / 0.0122) = 1.78×10⁻⁵

Step 3: Calculate pH
pH = -log(1.78×10⁻⁵) = 4.75

c. Calculate the pH of a buffer that was formed from the addition of 150 cm³ of 1.00 mol dm⁻³ methanoic acid and 50.0 cm³ of 0.500 mol dm⁻³ potassium methanoate. The Ka of methanoic acid is 1.78×10⁻⁴.
Answer

Step 1: Calculate Moles
Acid (HA) = 1.00 × 0.150 = 0.150 mol
Salt (A⁻) = 0.500 × 0.050 = 0.025 mol

Step 2: Calculate [H⁺]
[H⁺] = 1.78×10⁻⁴ × (0.150 / 0.025) = 1.07×10⁻³

Step 3: Calculate pH
pH = -log(1.07×10⁻³) = 2.97

d. A buffer made from 500 cm³ of 0.100 mol dm⁻³ ethanoic acid and an unknown mass of sodium ethanoate had a pH of 4.85. The Ka of ethanoic acid is 1.74×10⁻⁵. Calculate the mass of sodium ethanoate added to form the buffer.
Answer

Step 1: Find [H⁺] and Ratios
[H⁺] = 10⁻⁴·⁸⁵ = 1.41×10⁻⁵
Moles Acid (HA) = 0.100 × 0.500 = 0.050 mol

Step 2: Solve for Moles Salt (A⁻)
mol A⁻ = (1.74×10⁻⁵ × 0.050) / 1.41×10⁻⁵ = 0.0617 mol

Step 3: Calculate Mass
Mass = 0.0617 × 82.0 = 5.06 g

e. Calculate the pH of a buffer made from 500 cm³ of 1.50 mol dm⁻³ ethanoic acid and 500cm³ of 0.600 mol dm⁻³ potassium hydroxide.
Answer

Step 1: Initial Moles (Partial Neutralisation)
Acid (HA) = 1.50 × 0.500 = 0.750 mol
Base (OH⁻) = 0.600 × 0.500 = 0.300 mol

Step 2: ICE Table (Moles)

HA+OH⁻→A⁻+H₂O
Initial0.7500.3000–
Change-0.300-0.300+0.300–
Final0.45000.300–

Step 3: Calculate pH
[H⁺] = 1.74×10⁻⁵ × (0.450 / 0.300) = 2.61×10⁻⁵
pH = -log(2.61×10⁻⁵) = 4.58
(Assuming Ka for ethanoic acid is 1.74×10⁻⁵)

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