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
Chemical Calculations
Concentration, solubility and titration calculations: titration calculations and estimating a titre
- 3 things to know
- AS Unit 1
Overview
Recall questions on titration calculations and estimating a titre.
What you need to know
Answer each one before you reveal it.
Calculate the volume of 0.0500 mol dm⁻³ sulfuric acid needed to neutralise 25.0 cm³ of 0.100 mol dm⁻³ sodium hydroxide.
25.0 cm³ of 0.0500 mol dm⁻³ sodium carbonate solution is titrated with 0.100 mol dm⁻³ hydrochloric acid. What titre is expected?
Questions from this subtopic
Each one with the model answer underneath.
- 1
Why is the concentration of a solution of sodium hydroxide usually found by titration rather than by weighing out the solid?
Model answer
Sodium hydroxide absorbs water vapour and carbon dioxide from the air, so a weighed sample is not pure and the concentration calculated from its mass would be too high. Titrating it against a solution of accurately known concentration measures the amount actually present.
- 2
In a titration, 25.0 cm³ of sodium hydroxide solution required 23.60 cm³ of 0.100 mol dm⁻³ hydrochloric acid for neutralisation. Calculate the concentration of the alkali.
Model answer
n(HCl) = 0.100 mol dm⁻³ × 0.02360 dm³ = 2.36 × 10⁻³ mol. The equation NaOH + HCl → NaCl + H₂O gives a 1 : 1 ratio, so n(NaOH) = 2.36 × 10⁻³ mol. Then c = 2.36 × 10⁻³ ÷ 0.0250 = 0.0944 mol dm⁻³, quoted to three significant figures to match the least precise data.
- 3
Express a titration result of 0.0944 mol dm⁻³ sodium hydroxide as a concentration in g dm⁻³.
Model answer
Multiply by the molar mass: 0.0944 mol dm⁻³ × 40.0 g mol⁻¹ = 3.78 g dm⁻³, to three significant figures.
- 4
A student records titres of 24.10, 23.55, 23.60 and 23.65 cm³. Which values should be used for the mean titre, and what is it?
Model answer
The first is the rough titration and is discarded. The remaining three lie within 0.10 cm³ of one another and so are concordant. The mean is (23.55 + 23.60 + 23.65) ÷ 3 = 70.80 ÷ 3 = 23.60 cm³.
- 5
What does concordant mean when applied to titration results, and why does it matter?
Model answer
It means the titres agree closely, conventionally within 0.10 cm³ of each other. Averaging only concordant results excludes the rough titration and any run in which the end-point was overshot, so the mean reflects the reproducible value rather than being dragged off by an outlier.
- 6
Before starting, a student needs to know roughly what titre to expect when 25.0 cm³ of 0.100 mol dm⁻³ sodium hydroxide is titrated with 0.100 mol dm⁻³ hydrochloric acid. What estimate should they make, and why is it useful?
Model answer
The reaction is 1 : 1 and the concentrations are equal, so about 25 cm³ of acid will be needed. Knowing this confirms that a 50 cm³ burette is the right size and that the titre will fall in a well-measured part of its range, and it warns the student if the first result is wildly different.
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