Mixture of pH Questions
Acids and Bases Worksheet
1. In the following reactions, state which substance is the acid; which is the base; which is the conjugate acid; and which is the conjugate base.
a) H₂SO₄ + Ca(OH)₂ → CaSO₄ + 2H₂O
H₂SO₄ is the acid.
Ca(OH)₂ is the base.
H₂O is the conjugate acid and conjugate base.
b) H₂SO₄ + HNO₃ → HSO₄⁻ + H₂NO₃⁺
H₂SO₄ is the acid.
HNO₃ is the base.
HSO₄⁻ is the conjugate base.
H₂NO₃⁺ is the conjugate acid.
c) H₂NCH₂COOH + H₂NCHClCOOH → ⁺H₃NCH₂COOH + H₂NCHClCOO⁻
H₂NCHClCOOH is the acid.
H₂NCH₂COOH is the base.
⁺H₃NCH₂COOH is the conjugate acid.
H₂NCHClCOO⁻ is the conjugate base.
2. 200cm³ of 0.5 mol dm⁻³ sodium hydroxide is added to 150cm³ of 0.4 mol dm⁻³ propanoic acid. The Kₐ of propanoic acid is 1.35×10⁻⁵.
a) Calculate the pH of the propanoic acid before the reaction occurs.
Step 1: Use weak acid approximation
[H⁺]² = Kₐ × [HA]
[H⁺]² = Kₐ × [HA]
Step 2: Calculate [H⁺]
[H⁺]² = 1.35×10⁻⁵ × 0.4 = 5.4×10⁻⁶
[H⁺] = 2.32×10⁻³ mol dm⁻³
[H⁺]² = 1.35×10⁻⁵ × 0.4 = 5.4×10⁻⁶
[H⁺] = 2.32×10⁻³ mol dm⁻³
Step 3: Calculate pH
pH = -log(2.32×10⁻³) = 2.63
pH = -log(2.32×10⁻³) = 2.63
b) Calculate the pH of the sodium hydroxide before the reaction occurs.
Step 1: Calculate [H⁺] using K_w
[H⁺] = K_w / [OH⁻]
[H⁺] = 1×10⁻¹⁴ / 0.5 = 2×10⁻¹⁴ mol dm⁻³
[H⁺] = K_w / [OH⁻]
[H⁺] = 1×10⁻¹⁴ / 0.5 = 2×10⁻¹⁴ mol dm⁻³
Step 2: Calculate pH
pH = -log(2×10⁻¹⁴) = 13.70
pH = -log(2×10⁻¹⁴) = 13.70
c) Calculate the pH after 50cm³ of sodium hydroxide is added.
Step 1: Calculate Initial Moles
Acid (HA) = 0.4 × (150/1000) = 0.060 mol
Base (OH⁻) = 0.5 × (50/1000) = 0.025 mol
Acid (HA) = 0.4 × (150/1000) = 0.060 mol
Base (OH⁻) = 0.5 × (50/1000) = 0.025 mol
| Equation | HA | + OH⁻ | → A⁻ | + H₂O |
|---|---|---|---|---|
| Initial | 0.060 | 0.025 | 0 | – |
| Change | -0.025 | -0.025 | +0.025 | – |
| Equilibrium | 0.035 | 0 | 0.025 | – |
Step 2: Calculate [H⁺]
[H⁺] = Kₐ × (Moles Acid / Moles Salt)
[H⁺] = 1.35×10⁻⁵ × (0.035 / 0.025) = 1.89×10⁻⁵ mol dm⁻³
[H⁺] = Kₐ × (Moles Acid / Moles Salt)
[H⁺] = 1.35×10⁻⁵ × (0.035 / 0.025) = 1.89×10⁻⁵ mol dm⁻³
Step 3: Calculate pH
pH = -log(1.89×10⁻⁵) = 4.72
pH = -log(1.89×10⁻⁵) = 4.72
d) Calculate the volume of sodium hydroxide needed to reach the equivalence point.
Moles of acid = 0.060 mol
For equivalence, moles OH⁻ = moles Acid.
Volume = Moles / Conc = 0.060 / 0.5 = 0.12 dm³ (120 cm³)
e) Calculate the pH at the half neutralisation point.
At half neutralisation, pH = pKₐ.
pH = -log(1.35×10⁻⁵) = 4.87
f) Suggest a suitable pH indicator.
Phenolphthalein (as the equivalence point will be > pH 7 for a weak acid / strong base titration).
3. This question is about the dissociation of water.
a) Write a K_c expression for the dissociation of water.
K_c = [H⁺][OH⁻] / [H₂O]
b) Write the K_w expression and explain why it is different to the K_c expression.
K_w = [H⁺][OH⁻]
Water is left out from the K_w expression because the concentration of water is effectively constant as it is so large compared to the ions.
c) The K_w of water at 298k is 1.00×10⁻¹⁴ and at 323k is 5.48×10⁻¹⁴.
Explain the difference:
The dissociation reaction H₂O ⇌ H⁺ + OH⁻ is endothermic. As temperature increases, the equilibrium shifts to the right, increasing [H⁺] and [OH⁻], thus increasing K_w.
The dissociation reaction H₂O ⇌ H⁺ + OH⁻ is endothermic. As temperature increases, the equilibrium shifts to the right, increasing [H⁺] and [OH⁻], thus increasing K_w.
Calculate pH at 298K:
[H⁺] = √(1.00×10⁻¹⁴) = 1.00×10⁻⁷
pH = -log(1.00×10⁻⁷) = 7.00
[H⁺] = √(1.00×10⁻¹⁴) = 1.00×10⁻⁷
pH = -log(1.00×10⁻⁷) = 7.00
Calculate pH at 323K:
[H⁺] = √(5.48×10⁻¹⁴) = 2.34×10⁻⁷
pH = -log(2.34×10⁻⁷) = 6.63
[H⁺] = √(5.48×10⁻¹⁴) = 2.34×10⁻⁷
pH = -log(2.34×10⁻⁷) = 6.63
Neutrality:
They are both neutral because in pure water, [H⁺] always equals [OH⁻]. Neutral does not strictly mean pH 7; it means equal concentrations of hydrogen and hydroxide ions.
They are both neutral because in pure water, [H⁺] always equals [OH⁻]. Neutral does not strictly mean pH 7; it means equal concentrations of hydrogen and hydroxide ions.
4. A scientist makes a 500cm³ solution of 1 mol dm⁻³ ethanoic acid (Kₐ of 1.74×10⁻⁵).
a) Calculate the pH of the solution.
Step 1: Calculate [H⁺]
[H⁺] = √(Kₐ × [HA])
[H⁺] = √(1.74×10⁻⁵ × 1) = 4.17×10⁻³ mol dm⁻³
[H⁺] = √(Kₐ × [HA])
[H⁺] = √(1.74×10⁻⁵ × 1) = 4.17×10⁻³ mol dm⁻³
Step 2: Calculate pH
pH = -log(4.17×10⁻³) = 2.38
pH = -log(4.17×10⁻³) = 2.38
b) The scientist wants to make the solution have a pH of 3. Calculate the mass of sodium ethanoate that would need to be added to achieve this.
Step 1: Calculate Target [H⁺]
[H⁺] = 10⁻³·⁰⁰ = 0.001 mol dm⁻³
[H⁺] = 10⁻³·⁰⁰ = 0.001 mol dm⁻³
Step 2: Calculate [A⁻] needed
Kₐ = ([H⁺][A⁻]) / [HA]
[A⁻] = (Kₐ × [HA]) / [H⁺]
[A⁻] = (1.74×10⁻⁵ × 1) / 0.001 = 0.0174 mol dm⁻³
Kₐ = ([H⁺][A⁻]) / [HA]
[A⁻] = (Kₐ × [HA]) / [H⁺]
[A⁻] = (1.74×10⁻⁵ × 1) / 0.001 = 0.0174 mol dm⁻³
Step 3: Calculate Moles and Mass
Moles A⁻ = 0.0174 × (500/1000) = 0.0087 mol
Mr(CH₃COONa) = 82.0
Mass = 0.0087 × 82.0 = 0.71 g
Moles A⁻ = 0.0174 × (500/1000) = 0.0087 mol
Mr(CH₃COONa) = 82.0
Mass = 0.0087 × 82.0 = 0.71 g
5. A buffer is made from an unknown acid that has a concentration of 0.3 mol dm⁻³ and sodium hydroxide. A 50cm³ sample of the acid is tested, and it is found to have a pH of 2.55. 100cm³ of the acid is then mixed with 40cm³ of 0.5 mol dm⁻³ sodium hydroxide.
a) Calculate the pH of the buffer formed.
Step 1: Determine Kₐ of the acid
We must find the accurate equilibrium concentration of HA.
[H⁺] = 10⁻².⁵⁵ = 0.00282 mol dm⁻³
[HA]eq = 0.014859 / 0.050 = 0.297 mol dm⁻³
Kₐ = [H⁺]² / [HA]eq = (0.00282)² / 0.297 = 2.67×10⁻⁵
We must find the accurate equilibrium concentration of HA.
[H⁺] = 10⁻².⁵⁵ = 0.00282 mol dm⁻³
Dissociation ICE Table (in 50cm³ sample):
| Equation | HA | ⇌ H⁺ | + A⁻ |
|---|---|---|---|
| Initial Moles | 0.015 (0.3 × 0.050) |
0 | 0 |
| Change | -0.000141 ([H⁺] × 0.050) |
+0.000141 | +0.000141 |
| Equilibrium | 0.014859 | 0.000141 | 0.000141 |
Kₐ = [H⁺]² / [HA]eq = (0.00282)² / 0.297 = 2.67×10⁻⁵
Step 2: Buffer Moles Calculation
Moles Acid (HA) = 0.3 × 0.100 = 0.030 mol
Moles Base (OH⁻) = 0.5 × 0.040 = 0.020 mol
Moles Acid (HA) = 0.3 × 0.100 = 0.030 mol
Moles Base (OH⁻) = 0.5 × 0.040 = 0.020 mol
| Equation | HA | + OH⁻ | → A⁻ |
|---|---|---|---|
| Initial | 0.030 | 0.020 | 0 |
| Change | -0.020 | -0.020 | +0.020 |
| Equilibrium | 0.010 | 0 | 0.020 |
Step 3: Calculate pH
[H⁺] = Kₐ × (Moles Acid / Moles Salt)
[H⁺] = 2.67×10⁻⁵ × (0.010 / 0.020) = 1.34×10⁻⁵
pH = -log(1.34×10⁻⁵) = 4.87
[H⁺] = Kₐ × (Moles Acid / Moles Salt)
[H⁺] = 2.67×10⁻⁵ × (0.010 / 0.020) = 1.34×10⁻⁵
pH = -log(1.34×10⁻⁵) = 4.87
b) 25 cm³ of 0.1 mol dm⁻³ hydrochloric acid is added to the buffer. Calculate the new pH.
Step 1: Moles of H⁺ added
Moles = 0.1 × 0.025 = 0.0025 mol
Moles = 0.1 × 0.025 = 0.0025 mol
Step 2: Adjust Buffer Moles
Acid reacts with Salt (A⁻) to form Acid (HA).
New Moles A⁻ = 0.020 – 0.0025 = 0.0175 mol
New Moles HA = 0.010 + 0.0025 = 0.0125 mol
Acid reacts with Salt (A⁻) to form Acid (HA).
New Moles A⁻ = 0.020 – 0.0025 = 0.0175 mol
New Moles HA = 0.010 + 0.0025 = 0.0125 mol
Step 3: Calculate pH
[H⁺] = 2.67×10⁻⁵ × (0.0125 / 0.0175) = 1.91×10⁻⁵
pH = -log(1.91×10⁻⁵) = 4.72
[H⁺] = 2.67×10⁻⁵ × (0.0125 / 0.0175) = 1.91×10⁻⁵
pH = -log(1.91×10⁻⁵) = 4.72
c) The original buffer solution is made again, and 50cm³ of 0.05 mol dm⁻³ of barium hydroxide is added to the buffer. Calculate the new pH.
Step 1: Moles of OH⁻ added
Moles Ba(OH)₂ = 0.05 × 0.050 = 0.0025 mol
Moles OH⁻ = 0.0025 × 2 = 0.005 mol
Moles Ba(OH)₂ = 0.05 × 0.050 = 0.0025 mol
Moles OH⁻ = 0.0025 × 2 = 0.005 mol
Step 2: Adjust Buffer Moles (Original Buffer)
Original: HA = 0.010, A⁻ = 0.020
Base reacts with Acid (HA) to form Salt (A⁻).
New Moles HA = 0.010 – 0.005 = 0.005 mol
New Moles A⁻ = 0.020 + 0.005 = 0.025 mol
Original: HA = 0.010, A⁻ = 0.020
Base reacts with Acid (HA) to form Salt (A⁻).
New Moles HA = 0.010 – 0.005 = 0.005 mol
New Moles A⁻ = 0.020 + 0.005 = 0.025 mol
Step 3: Calculate pH
[H⁺] = 2.67×10⁻⁵ × (0.005 / 0.025) = 5.34×10⁻⁶
pH = -log(5.34×10⁻⁶) = 5.27
[H⁺] = 2.67×10⁻⁵ × (0.005 / 0.025) = 5.34×10⁻⁶
pH = -log(5.34×10⁻⁶) = 5.27