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
Titrations: curve shapes, salt pH and choosing an indicator
- 1 thing to know
- A2 Unit 3
Overview
Recall questions on curve shapes, salt pH and choosing an indicator.
What you need to know
Answer each one before you reveal it.
Questions from this subtopic
Each one with the model answer underneath.
- 1
In a titration, what is the difference between the equivalence point and the end point?
Model answer
The equivalence point is the moment when the acid and base have reacted in exactly the stoichiometric ratio, which is a property of the chemistry. The end point is the moment the indicator changes colour, which is a property of the indicator; a well-chosen indicator makes the two coincide closely enough for accurate work.
- 2
Why does the pH change so abruptly near the equivalence point of a strong acid and strong base titration?
Model answer
Very close to equivalence almost all the acid has been used up, so each further drop of alkali makes a very large fractional change in the tiny remaining [H⁺]. Because pH is a logarithmic scale, that fractional change appears as a jump of several pH units for a fraction of a cubic centimetre of titrant.
- 3
What salt is formed when hydrochloric acid neutralises sodium hydroxide, and what is the pH of its solution?
Model answer
Sodium chloride is formed. Neither Na⁺ nor Cl⁻ reacts appreciably with water, because Cl⁻ is the conjugate base of a strong acid and so is an extremely poor proton acceptor. The solution is therefore neutral, with pH 7 at 298 K.
- 4
Why is a solution of ammonium chloride acidic?
Model answer
It is the salt of a strong acid and a weak base, so the ammonium ion is the conjugate acid of a weak base and donates protons to water: NH₄⁺(aq) + H₂O(l) ⇌ NH₃(aq) + H₃O⁺(aq). The chloride ion does nothing, so the extra hydrogen ions make the solution acidic, typically around pH 5.
- 5
Why is a solution of sodium ethanoate alkaline?
Model answer
It is the salt of a weak acid and a strong base, so the ethanoate ion is the conjugate base of a weak acid and takes protons from water: CH₃COO⁻(aq) + H₂O(l) ⇌ CH₃COOH(aq) + OH⁻(aq). The sodium ion has no effect, so the hydroxide ions released make the solution alkaline.
- 6
Calculate the pH of 0.100 mol dm⁻³ sodium ethanoate at 298 K, given Ka for ethanoic acid = 1.74 × 10⁻⁵ mol dm⁻³ and Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶.
Model answer
For CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻ the constant is Kw/Ka = 1.00 × 10⁻¹⁴ / (1.74 × 10⁻⁵) = 5.75 × 10⁻¹⁰ mol dm⁻³. Then [OH⁻] = √(5.75 × 10⁻¹⁰ × 0.100) = 7.58 × 10⁻⁶ mol dm⁻³, so [H⁺] = 1.00 × 10⁻¹⁴ / (7.58 × 10⁻⁶) = 1.32 × 10⁻⁹ mol dm⁻³ and pH = 8.88.
Other ways to work this
The same content, four other ways.
What is inside Levo
Six ways to work the same material.
Every question in Acid-Base Equilibria, on a schedule.
Attempt a question from memory, write the correct version, then attempt it again. Levo records the jump between your first attempt and your last, and brings each question back on the day you are due to forget it.
Or £46 a year, about £3.83 a month. Cancel anytime.