Adding to Buffers

Acids and Bases Worksheet

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1. A buffer is created from 500cm³ of 0.200 mol dm⁻³ ethanoic acid (Kₐ of 1.74 × 10⁻⁵) and 5.00g of sodium ethanoate.

a) Calculate the pH of the buffer.

Step 1: Calculate moles of Acid (HA)
Moles = conc × vol = 0.200 × (500/1000) = 0.100 mol

Step 2: Calculate moles of Salt (A^–)
Mr(CH₃COONa) = 12 + 3 + 12 + 16 + 16 + 23 = 82.0
Moles = mass / Mr = 5.00 / 82.0 = 0.0610 mol

Step 3: Calculate [H⁺]
[H⁺] = K_a × ([Acid] / [Salt])
[H⁺] = 1.74 × 10^-5 × (0.100 / 0.0610) = 2.85 × 10^-5 mol dm^-3

Step 4: Calculate pH
pH = -log(2.85 × 10^-5) = 4.54 (to 3 s.f.)

b) 10cm³ of 0.5 mol dm⁻³ hydrochloric acid is added to the buffer. Calculate the new pH.

Step 1: Calculate moles of H⁺ added
Moles H⁺ = 0.5 × (10/1000) = 0.005 mol

Step 2: Adjust moles of HA and A^–
Adding Acid reacts with the salt (A^–) to form more acid (HA).
New moles A^– = 0.0610 – 0.005 = 0.0560 mol
New moles HA = 0.100 + 0.005 = 0.105 mol

Step 3: Calculate new [H⁺]
[H⁺] = 1.74 × 10^-5 × (0.105 / 0.0560) = 3.26 × 10^-5 mol dm^-3

Step 4: Calculate pH
pH = -log(3.26 × 10^-5) = 4.49

c) Explain why the buffer resists changing pH on the addition of hydrochloric acid.

There is an equilibrium set up:

CH₃COOH ⇌ CH₃COO^– + H⁺

When HCl is added, [H⁺] increases. As there is a large reservoir of ethanoate ions (CH₃COO^–), the equilibrium shifts to the left to remove the added H⁺ ions. This prevents a significant change in pH.

2. A buffer is made from 200cm³ of 0.5 mol dm⁻³ methanoic acid (Kₐ of 1.78 × 10⁻⁴) and 200cm³ of 0.3 mol dm⁻³ sodium hydroxide.

a) Calculate the pH of the buffer.

Note: This is a partial neutralisation question. We must construct an ICE table to find the moles of Acid remaining and Salt formed.

Step 1: Calculate Initial Moles
Moles HCOOH (HA) = 0.5 × 0.200 = 0.100 mol
Moles NaOH (OH^–) = 0.3 × 0.200 = 0.060 mol

Equation	HCOOH	+ NaOH	→ HCOONa	+ H₂O
Initial Moles	0.100	0.060	0	–
Change	-0.060	-0.060	+0.060	–
Equilibrium	0.040	0	0.060	–

Step 2: Calculate [H⁺]
[H⁺] = 1.78 × 10^-4 × (0.040 / 0.060) = 1.19 × 10^-4 mol dm^-3

Step 3: Calculate pH
pH = -log(1.19 × 10^-4) = 3.92

b) 20cm³ of 0.2 mol dm⁻³ sodium hydroxide is added to the buffer. Calculate the new pH.

Step 1: Calculate moles OH^– added
Moles = 0.2 × (20/1000) = 0.004 mol

Step 2: Adjust moles
Adding base reacts with the acid (HA) to form more salt (A^–).
Old Equilibrium moles: HA = 0.040, A^– = 0.060
New moles HA = 0.040 – 0.004 = 0.036 mol
New moles A^– = 0.060 + 0.004 = 0.064 mol

Step 3: Calculate new [H⁺]
[H⁺] = 1.78 × 10^-4 × (0.036 / 0.064) = 1.00 × 10^-4 mol dm^-3

Step 4: Calculate pH
pH = -log(1.00 × 10^-4) = 4.00

c) Explain why the buffer resists changing pH on the addition of sodium hydroxide.

There is an equilibrium set up:

HCOOH ⇌ HCOO^– + H⁺

On addition of NaOH, the OH^– ions react with H⁺ to form water. This decreases [H⁺]. The equilibrium shifts to the right to replace the H⁺ ions, minimising the change in pH.

3. A buffer is made from 250cm³ of 0.2 mol dm⁻³ ethanoic acid (Kₐ: 1.74 × 10⁻⁵) and 250cm³ of 0.2 mol dm⁻³ sodium ethanoate.

a) Calculate the pH of the buffer.

Observation:
The concentration and volume of the acid and salt are identical. Therefore, [Acid] = [Salt].

Calculation:
[H⁺] = K_a × ([HA] / [A^–])
[H⁺] = 1.74 × 10^-5 × 1
pH = -log(1.74 × 10^-5) = 4.76

b) Distilled water is added to the buffer so that the buffer solution now has a volume of 600cm³. Calculate the new pH.

Step 1: Calculate Moles
Moles Acid = 0.2 × (250/1000) = 0.050 mol
Moles Salt = 0.2 × (250/1000) = 0.050 mol

Step 2: Calculate New Concentrations
New Volume = 600cm³ = 0.600 dm³
New [HA] = 0.050 / 0.600 = 0.0833 mol dm⁻³
New [A⁻] = 0.050 / 0.600 = 0.0833 mol dm⁻³

Step 3: Calculate pH
[H⁺] = 1.74 × 10⁻⁵ × (0.0833 / 0.0833)
[H⁺] = 1.74 × 10⁻⁵ mol dm⁻³
pH = -log(1.74 × 10⁻⁵) = 4.76

c) Explain why the pH didn’t change when diluted.

The ratio of [Acid] to [Salt] remains constant because both components are diluted by the same factor. As K_a is a constant (at constant temperature), [H⁺] remains unchanged.

4. A buffer made from 200cm³ 0.75 mol dm⁻³ propanoic acid (Kₐ of 1.35 × 10⁻⁵) and 100cm³ of 0.5 mol dm⁻³ sodium hydroxide.

a) Calculate the pH of the buffer.

Step 1: Initial Moles (Partial Neutralisation)
Moles Acid (HA) = 0.75 × 0.200 = 0.150 mol
Moles Base (OH^–) = 0.5 × 0.100 = 0.050 mol

Equation	HA	+ NaOH	→ NaA	+ H₂O
Initial	0.150	0.050	0	–
Change	-0.050	-0.050	+0.050	–
Equilibrium	0.100	0	0.050	–

Step 2: Calculate pH
[H⁺] = 1.35 × 10^-5 × (0.100 / 0.050) = 2.70 × 10^-5
pH = -log(2.70 × 10^-5) = 4.57

b) Calculate the pH if 20cm³ of 0.4 mol dm⁻³ H₂SO₄ is added.

Step 1: Calculate H⁺ added (Diprotic Acid!)
Moles H₂SO₄ = 0.4 × 0.020 = 0.008 mol
Moles H⁺ = 0.008 × 2 = 0.016 mol

Step 2: Adjust moles
Added acid reacts with Salt (A^–) to form Acid (HA).
New moles A^– = 0.050 – 0.016 = 0.034 mol
New moles HA = 0.100 + 0.016 = 0.116 mol

Step 3: Calculate pH
[H⁺] = 1.35 × 10^-5 × (0.116 / 0.034) = 4.60 × 10^-5
pH = -log(4.60 × 10^-5) = 4.34

5. A buffer is made from 800cm³ of 1.00 mol dm⁻³ ethanoic acid (Kₐ of 1.74 × 10⁻⁵) and 22.0g of sodium hydroxide.

a) Calculate the pH of the buffer.

Step 1: Initial Moles
Moles Acid (HA) = 1.00 × 0.800 = 0.800 mol
Moles NaOH = 22.0 / 40.0 = 0.550 mol

Equation	HA	+ NaOH	→ NaA
Initial	0.800	0.550	0
Change	-0.550	-0.550	+0.550
Equilibrium	0.250	0	0.550

Step 2: Calculate pH
[H⁺] = 1.74 × 10^-5 × (0.250 / 0.550) = 7.91 × 10^-6
pH = -log(7.91 × 10^-6) = 5.10

b) 1.50g of barium hydroxide is added to the buffer. Calculate the new pH.

Step 1: Calculate OH^– added
Mr Ba(OH)₂ = 137.3 + 2(16+1) = 171.3
Moles Ba(OH)₂ = 1.50 / 171.3 = 0.008757 mol
Moles OH^– = 0.008757 × 2 = 0.0175 mol (Ba(OH)₂ releases 2 OH^–)

Step 2: Adjust moles
Added base reacts with Acid (HA) to form Salt (A^–).
New moles HA = 0.250 – 0.0175 = 0.2325 mol
New moles A^– = 0.550 + 0.0175 = 0.5675 mol

Step 3: Calculate pH
[H⁺] = 1.74 × 10^-5 × (0.2325 / 0.5675) = 7.13 × 10^-6
pH = -log(7.13 × 10^-6) = 5.15

Mr Cole Chemistry

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Adding to Buffers