Mr Cole Chemistry

pH of Strong Mixtures

Acids and Bases Worksheet

1. Calculating pH of Mixtures

Calculate the pH of the following mixtures.
1. 100cm³ of 0.6 mol dm⁻³ HCl added to 80cm³ of 0.5 mol dm⁻³ NaOH.

Moles of H⁺ (HCl): 0.1 × 0.6 = 0.06
Moles of OH⁻ (NaOH): 0.08 × 0.5 = 0.04
Excess: 0.06 – 0.04 = 0.02 moles of H⁺
New Volume: 100 + 80 = 180 cm³ = 0.18 dm³
[H⁺]: 0.02 / 0.18 = 0.111 mol dm⁻³
pH: -log(0.111) = 0.954

2. 250cm³ of 1.0 mol dm⁻³ NaOH added to 200cm³ of 0.5 mol dm⁻³ H₂SO₄.

Moles of H₂SO₄: 0.2 × 0.5 = 0.1
Moles of H⁺: 0.1 × 2 = 0.2
Moles of OH⁻ (NaOH): 0.25 × 1 = 0.25
Excess: 0.25 – 0.20 = 0.05 moles of OH⁻
New Volume: 200 + 250 = 450 cm³ = 0.45 dm³
[OH⁻]: 0.05 / 0.45 = 0.111 mol dm⁻³
[H⁺]: 1×10⁻¹⁴ / 0.111 = 9×10⁻¹⁴ mol dm⁻³
pH: -log(9×10⁻¹⁴) = 13.05

3. 100cm³ of 2.0 mol dm⁻³ H₂SO₄ added to 100cm³ of 2.5 mol dm⁻³ LiOH.

Moles of H₂SO₄: 0.1 × 2 = 0.2
Moles of H⁺: 0.2 × 2 = 0.4
Moles of OH⁻ (LiOH): 0.1 × 2.5 = 0.25
Excess: 0.40 – 0.25 = 0.15 moles of H⁺
New Volume: 200 cm³ = 0.2 dm³
[H⁺]: 0.15 / 0.2 = 0.75 mol dm⁻³
pH: -log(0.75) = 0.12

4. 6g of solid NaOH added to 100cm³ of 0.5 mol dm⁻³ HNO₃.

Moles of H⁺ (HNO₃): 0.1 × 0.5 = 0.05
Moles of OH⁻ (NaOH): 6 / 40 = 0.15
Excess: 0.15 – 0.05 = 0.1 moles of OH⁻
Volume: 100 cm³ = 0.1 dm³ (assume constant)
[OH⁻]: 0.1 / 0.1 = 1 mol dm⁻³
[H⁺]: 1×10⁻¹⁴ / 1 = 1×10⁻¹⁴ mol dm⁻³
pH: -log(1×10⁻¹⁴) = 14.00

5. Calculate the change in pH when 50cm³ of 2 mol dm⁻³ H₂SO₄ is added to 100cm³ of 1 mol dm⁻³ KOH.

Initial pH (KOH): [OH⁻]=1, pOH=0, pH=14.00

Final pH:
Moles H₂SO₄: 0.05 × 2 = 0.1 → Moles H⁺: 0.2
Moles OH⁻ (KOH): 0.1 × 1 = 0.1
Excess: 0.2 – 0.1 = 0.1 moles of H⁺
New Volume: 150 cm³ = 0.15 dm³
[H⁺]: 0.1 / 0.15 = 0.667 mol dm⁻³
Final pH: -log(0.667) = 0.18

Change in pH: 0.18 – 14.00 = -13.82

2. Reverse Calculations & Solid Bases

6. 150cm³ of NaOH was added to 100cm³ of 1.5 mol dm⁻³ HBr. Final pH was 13.68. Calculate [NaOH].
Answer

Step 1: Analyze Final Mixture
pH 13.68 → pOH = 0.32
[OH⁻] = 10⁻⁰·³² = 0.479 mol dm⁻³
Total Volume = 0.25 dm³
Excess OH⁻ moles = 0.479 × 0.25 = 0.120 moles

Step 2: Account for Neutralization
Initial HBr moles (H⁺) = 1.5 × 0.1 = 0.150 moles
Total OH⁻ added = Neutralized H⁺ + Excess OH⁻
Total OH⁻ = 0.150 + 0.120 = 0.270 moles

Step 3: Calculate Concentration
[NaOH] = 0.270 / 0.150 = 1.80 mol dm⁻³

7. 100cm³ of LiOH added to 100cm³ of HCl raised pH from -0.08 to 12.06. Calculate [LiOH].
Answer

Step 1: Initial Acid Moles
Initial pH -0.08 → [H⁺] = 10⁰·⁰⁸ = 1.20 mol dm⁻³
Initial H⁺ moles = 1.2 × 0.1 = 0.120 moles

Step 2: Final Mixture Excess
Final pH 12.06 → pOH = 1.94
[OH⁻] = 10⁻¹.⁹⁴ = 0.0115 mol dm⁻³
Total Volume = 0.2 dm³
Excess OH⁻ moles = 0.0115 × 0.2 = 0.0023 moles

Step 3: Calculate Concentration
Total OH⁻ added = 0.120 (neutralized) + 0.0023 (excess) = 0.1223 moles
[LiOH] = 0.1223 / 0.1 = 1.22 mol dm⁻³

8. Solid NaOH added to 100cm³ of 1.0 mol dm⁻³ H₂SO₄. Final pH 13.34. Calculate mass of NaOH.
Answer

Step 1: Initial Acid Moles
Moles H₂SO₄ = 0.1 × 1.0 = 0.1
Moles H⁺ to neutralize = 0.1 × 2 = 0.2 moles

Step 2: Final Excess Base
Final pH 13.34 → pOH = 0.66
[OH⁻] = 10⁻⁰·⁶⁶ = 0.219 mol dm⁻³
Excess OH⁻ moles = 0.219 × 0.1 = 0.0219 moles

Step 3: Calculate Mass
Total OH⁻ moles = 0.2 (neutralized) + 0.0219 (excess) = 0.2219 moles
Mass NaOH = 0.2219 × 40 = 8.88 g

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