OCR A-Level Chemistry AS Breadth in chemistry (01), November 2021: Question 6

Calculating Hydroxide Ion Concentration in Mixed Solutions

OCR AS Level Chemistry • Multiple Choice Question

What this question tests

This question tests your ability to calculate the concentration of hydroxide ions (OH⁻) when two different alkaline solutions are mixed together. It assesses your competency in using moles equations (moles = concentration × volume), recognizing stoichiometry (calculating the exact mole ratio of OH⁻ ions produced per mole of base, especially for dibasic bases like Ca(OH)₂), and correctly combining volumes to find a final concentration.

Question 6 Breakdown

Full Worked Solution & Mark Scheme

✅ Correct Answer: D (0.175)

The correct option is D. Top-level responses quickly identify that calcium hydroxide is dibasic, correctly calculate the total moles of hydroxide ions, divide by the total combined volume in dm³, and arrive cleanly at 0.175 mol dm⁻³ .

💡 Key Knowledge

Monobasic vs Dibasic: KOH releases 1 mole of OH⁻ per mole of compound. Ca(OH)₂ releases 2 moles of OH⁻ per mole of compound: Ca(OH)₂ -> Ca²⁺ + 2OH⁻ .
Concentration formula: Concentration = Moles / Volume (in dm³)
Total Volume: When mixing solutions, always add the individual volumes together to get the new total volume of the mixture.

🧠 Exam Technique

Don't panic when two solutions are mixed—break the problem down sequentially: find moles in solution 1, find moles in solution 2, sum them up, then divide by total volume.
Always double-check volume units. Converting cm³ to dm³ by dividing by 1000 is where many multiple-choice marks are dropped under time pressure.

❌ Common Errors & Misconceptions

Forgetting the stoichiometry of Ca(OH)₂: Treating calcium hydroxide as a monobasic alkali (assuming 1:1 ratio) leads straight to distracter answer C (0.150).
Averaging concentrations: Many students incorrectly add the two concentrations together or take a mean, completely ignoring the different volumes mixed.

📐 Step-by-Step Calculation Guide

Calculate moles of OH⁻ from KOH:
Volume = 250.0 cm³ = 0.250 dm³
Moles of KOH = 0.100 mol dm⁻³ × 0.250 dm³ = 0.0250 mol
Since KOH -> K⁺ + OH⁻, moles of OH⁻ = 0.0250 mol
Calculate moles of OH⁻ from Ca(OH)₂:
Volume = 750.0 cm³ = 0.750 dm³
Moles of Ca(OH)₂ = 0.100 mol dm⁻³ × 0.750 dm³ = 0.0750 mol
Since Ca(OH)₂ releases two OH⁻ ions, multiply by 2:
Moles of OH⁻ = 0.0750 × 2 = 0.150 mol
Find total moles of OH⁻:
Total moles = 0.0250 + 0.150 = 0.175 mol
Calculate total volume of the mixture:
Total volume = 250.0 cm³ + 750.0 cm³ = 1000 cm³ = 1.00 dm³
Calculate final OH⁻ concentration:
Concentration = Total moles / Total volume in dm³ = 0.175 mol / 1.00 dm³ = 0.175 mol dm⁻³ (Option D)

Mark Scheme Allocation: 1 mark awarded for selecting option D (AO2.4: Applying quantitative chemical knowledge and problem-solving skills to unfamiliar contexts).