WJEC A-Level Chemistry 1410QS
Formulae and Equations
Basic Ideas About Atoms
Chemical Calculations
Bonding
Solid Structures
The Periodic Table
Simple Equilibria and Acid-Base Reactions
Thermochemistry
Rates of Reaction
The Wider Impact of Chemistry
Organic Compounds
Hydrocarbons
Halogenoalkanes
Alcohols and Carboxylic Acids
Instrumental Analysis
Redox and Standard Electrode Potential
Redox Reactions
Chemistry of the P-Block
Chemistry of the D-Block Transition Metals
Chemical Kinetics
Enthalpy Changes for Solids and Solutions
Entropy and Feasibility of Reactions
Equilibrium Constants
Acid-Base Equilibria
Stereoisomerism
Aromaticity
Alcohols and Phenols
Aldehydes and Ketones
Carboxylic Acids and Their Derivatives
Amines
Amino Acids, Peptides and Proteins
Organic Synthesis and Analysis
Practical Skills and Techniques
Planning, Analysis and Evaluation
Core Experiments and Their Chemistry
Mathematical Skills in Chemistry
Acid-Base Equilibria
pH calculations for acids and alkalis: how to use pH
- 3 things to know
- A2 Unit 3
Overview
Fourteen recall questions on pH calculations for acids and alkalis: how to use pH.
What you need to know
Answer each one before you reveal it.
Hydrocyanic acid has Ka = 4.9 × 10⁻¹⁰ mol dm⁻³. Calculate the pH of a 0.100 mol dm⁻³ solution.
Calculate the pH of 0.100 mol dm⁻³ chloroethanoic acid, Ka = 1.38 × 10⁻³ mol dm⁻³, and comment on the reliability of the usual approximation.
Questions from this subtopic
Each one with the model answer underneath.
- 1
Calculate the pH of 0.0250 mol dm⁻³ nitric acid.
Model answer
Nitric acid is a strong monobasic acid, so it is fully dissociated and [H⁺] = 0.0250 mol dm⁻³. pH = −log₁₀(0.0250) = 1.60.
- 2
Calculate the pH of 2.50 × 10⁻³ mol dm⁻³ hydrochloric acid.
Model answer
Hydrochloric acid is strong and monobasic, so [H⁺] = 2.50 × 10⁻³ mol dm⁻³. pH = −log₁₀(2.50 × 10⁻³) = 2.60.
- 3
A solution of sulfuric acid is 0.0500 mol dm⁻³. Treating both ionisations as complete, calculate its pH.
Model answer
Sulfuric acid is dibasic, so each mole releases two moles of H⁺ and [H⁺] = 2 × 0.0500 = 0.100 mol dm⁻³. pH = −log₁₀(0.100) = 1.00, which is 0.30 units lower than the pH 1.30 a monobasic acid of the same concentration would give, because doubling the hydrogen ion concentration lowers the pH by log₁₀ 2.
- 4
25.0 cm³ of 0.100 mol dm⁻³ hydrochloric acid is diluted with water to 250 cm³. What is the pH of the diluted solution?
Model answer
Moles of HCl = 0.100 × 0.0250 = 2.50 × 10⁻³ mol, and this is now in 0.250 dm³, so the concentration is 2.50 × 10⁻³ / 0.250 = 0.0100 mol dm⁻³. The acid is strong, so [H⁺] = 0.0100 mol dm⁻³ and pH = 2.00. A tenfold dilution of a strong acid raises the pH by exactly one unit.
- 5
A strong monobasic acid has pH 1.40. Calculate its concentration.
Model answer
[H⁺] = 10⁻ᵖᴴ = 10⁻¹·⁴⁰ = 0.0398 mol dm⁻³. Because the acid is strong and monobasic, the acid concentration is the same, 0.0398 mol dm⁻³.
- 6
Calculate the pH of 0.100 mol dm⁻³ ethanoic acid, for which Ka = 1.74 × 10⁻⁵ mol dm⁻³.
Model answer
For a weak acid, Ka = [H⁺][A⁻]/[HA]. Since [H⁺] = [A⁻] and very little acid dissociates, [HA] is taken as 0.100 mol dm⁻³, giving [H⁺] = √(Ka × c) = √(1.74 × 10⁻⁵ × 0.100) = √(1.74 × 10⁻⁶) = 1.32 × 10⁻³ mol dm⁻³. pH = −log₁₀(1.32 × 10⁻³) = 2.88.
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