pH calculations for acids and alkalis: strong bases, dilution and mixtures

Overview

Calculating the pH of strong bases such as sodium and barium hydroxide using Kw, the effect of diluting acids, and the pH after mixing measured volumes of acid and alkali.

What you need to know

Answer each one before you reveal it.

Questions from this subtopic

Each one with the model answer underneath.

1. 1

   A weak acid solution is diluted tenfold. By how much does its pH rise, and why is the answer not one unit as it is for a strong acid?

   Model answer

   For a weak acid [H⁺] = √(Ka × c), so dividing c by ten divides [H⁺] by √10 rather than by 10. The pH therefore rises by log₁₀√10 = 0.50 rather than 1.00, because dilution also shifts the dissociation equilibrium to the right and partly offsets the loss of acid.

2. 2

   Calculate the pH of 0.0500 mol dm⁻³ sodium hydroxide at 298 K.

   Model answer

   Sodium hydroxide is a strong base, so [OH⁻] = 0.0500 mol dm⁻³. Then [H⁺] = Kw/[OH⁻] = 1.00 × 10⁻¹⁴ / 0.0500 = 2.00 × 10⁻¹³ mol dm⁻³, giving pH = −log₁₀(2.00 × 10⁻¹³) = 12.70.

3. 3

   Calculate the pH of 2.50 × 10⁻³ mol dm⁻³ potassium hydroxide at 298 K.

   Model answer

   Potassium hydroxide is strong, so [OH⁻] = 2.50 × 10⁻³ mol dm⁻³. [H⁺] = 1.00 × 10⁻¹⁴ / (2.50 × 10⁻³) = 4.00 × 10⁻¹² mol dm⁻³, so pH = −log₁₀(4.00 × 10⁻¹²) = 11.40.

4. 4

   Barium hydroxide is a strong base. Calculate the pH of a 0.0200 mol dm⁻³ solution at 298 K.

   Model answer

   Each formula unit provides two hydroxide ions, so [OH⁻] = 2 × 0.0200 = 0.0400 mol dm⁻³. Then [H⁺] = 1.00 × 10⁻¹⁴ / 0.0400 = 2.50 × 10⁻¹³ mol dm⁻³ and pH = −log₁₀(2.50 × 10⁻¹³) = 12.60.

5. 5

   4.00 g of sodium hydroxide is dissolved in water and made up to 500 cm³. Calculate the pH of the solution at 298 K. (Molar mass of NaOH = 40.0 g mol⁻¹.)

   Model answer

   Moles of NaOH = 4.00 / 40.0 = 0.100 mol, in 0.500 dm³, so the concentration is 0.200 mol dm⁻³. Sodium hydroxide is strong, so [OH⁻] = 0.200 mol dm⁻³ and [H⁺] = 1.00 × 10⁻¹⁴ / 0.200 = 5.00 × 10⁻¹⁴ mol dm⁻³. Hence pH = −log₁₀(5.00 × 10⁻¹⁴) = 13.30.

6. 6

   A solution of sodium hydroxide has pH 12.50 at 298 K. Calculate its concentration.

   Model answer

   [H⁺] = 10⁻¹²·⁵⁰ = 3.16 × 10⁻¹³ mol dm⁻³, so [OH⁻] = Kw/[H⁺] = 1.00 × 10⁻¹⁴ / (3.16 × 10⁻¹³) = 0.0316 mol dm⁻³. Since the base is strong and provides one hydroxide per formula unit, the concentration of sodium hydroxide is 0.0316 mol dm⁻³.

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