Problem Solving with Acids, Bases and Salts

Introduction

Longer exam questions rarely test one idea on its own. A single question might ask you to choose a method for making a salt, write its equation, calculate a concentration from titration data and predict the pH of the final solution. Success depends less on knowing more facts and more on having a strategy: read carefully, identify each stage, and link them. This page models that strategy using problems that draw on the whole unit.

Core explanation

A general strategy.

1. Read the whole question first. Underline the substances, quantities and what is being asked for. 2. Identify the chemistry. Is it neutralisation, reaction with a metal or carbonate, precipitation, or a pH calculation? 3. Write a balanced equation. Almost every quantitative step depends on it. 4. Convert to moles. Use n = c × V (V in dm³) for solutions and n = m ÷ Mᵣ for solids. 5. Use the mole ratio from the equation. 6. Convert back to the quantity asked for, with units. 7. Check that the answer is sensible.

Choosing a salt preparation method. This decision tree covers most questions:

- Is the salt insoluble ? Use precipitation : mix two solutions that each supply one of its ions, filter, wash and dry. - Is the salt soluble and is the base insoluble (metal, oxide, carbonate)? Add excess solid to the acid, filter off the excess, then crystallise the filtrate. - Is the salt soluble and the base soluble (alkali or ammonia)? Use titration to find the exact volumes, repeat without indicator, then crystallise.

Solubility rules are essential: all sodium, potassium, ammonium and nitrate salts are soluble; most chlorides are soluble except silver and lead(II) chloride; most sulfates are soluble except barium, lead(II) and (sparingly) calcium sulfate; most carbonates are insoluble except those of sodium, potassium and ammonium.

Linking pH to reactions. When an acid is gradually neutralised, pH rises. If a strong acid is in excess, the solution is acidic; at exact neutralisation of a strong acid by a strong alkali the pH is 7; if the alkali is in excess, the solution is alkaline. Knowing which reactant is in excess lets you predict the pH qualitatively without any logarithms.

Titration calculations. Use the mean of concordant titres, convert cm³ to dm³ by dividing by 1000, and pay attention to mole ratios other than 1:1, such as sulfuric acid reacting with sodium hydroxide (1:2).

Formulae

n = c × V (n in mol, c in mol/dm³, V in dm³); n = m ÷ Mᵣ; mass concentration (g/dm³) = c (mol/dm³) × Mᵣ

Step-by-step reasoning

A full multi-step problem: 25.0 cm³ of sodium hydroxide solution is neutralised by a mean titre of 20.0 cm³ of 0.100 mol/dm³ sulfuric acid. Find the concentration of the sodium hydroxide, then name a method to make pure sodium sulfate crystals.

1. Equation: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. 2. Moles of H₂SO₄ = 0.100 × 0.0200 = 0.00200 mol. 3. Mole ratio 1:2, so moles of NaOH = 0.00400 mol. 4. Concentration of NaOH = 0.00400 ÷ 0.0250 = 0.160 mol/dm³. 5. Both reactants are soluble, so repeat the titration with the same volumes but no indicator, then evaporate and crystallise the solution.

Visual explanation

Picture a flow chart with three boxes in a row: "Equation" → "Moles" → "Answer". Below it, a decision tree starts with "Is the salt soluble?" and branches to "Precipitation" on the no side and, on the yes side, to "Is the base soluble?", which branches to "Titration" or "Excess solid base, filter". The two diagrams together cover most multi-step questions.

Real-world analogy

Solving a multi-step question is like following a route with several changes of train. Each step (equation, moles, ratio, answer) is a separate journey, and getting off at the wrong station at any stage means you arrive in the wrong place. Planning the whole route first prevents that.

Real-world example

Quality-control chemists at a food factory check the acidity of vinegar by titrating a sample with standard sodium hydroxide. They then convert moles of ethanoic acid into grams per 100 cm³ to check the label is accurate. The job requires exactly the chain of reasoning used in exam questions.

Why?

Why must you always write the balanced equation before calculating? The coefficients give the mole ratio. Assuming a 1:1 ratio when the real ratio is 1:2 doubles or halves the answer, which is one of the most common errors in titration calculations.

Common misconception

"Any soluble salt can be made by adding excess base and filtering." This only works if the excess base is insoluble and can be filtered off. An excess of a soluble alkali would stay dissolved and contaminate the salt, so titration is needed instead.

Worked example

Question: A student wants copper(II) sulfate crystals. 0.050 mol of sulfuric acid is available. What mass of copper(II) oxide is needed to react exactly, and why should the student add a little more than this? (Mᵣ CuO = 79.5)

Reasoning: CuO + H₂SO₄ → CuSO₄ + H₂O, ratio 1:1, so 0.050 mol of CuO is needed. Mass = 0.050 × 79.5 = 3.98 g. Adding an excess of the insoluble oxide ensures all the acid reacts; the unreacted oxide is filtered off.

Answer: About 4.0 g; excess ensures all the acid is used up, and the leftover solid is removed by filtration.

Quick check

1. Which method would you use to make barium sulfate? Answer: Precipitation, because barium sulfate is insoluble.

Exam focus

In long questions, set out each stage clearly with labels such as "moles of acid" so that method marks can be awarded even if a later number is wrong. Give answers to a sensible number of significant figures (usually three) with units. Justify the choice of salt preparation method using solubility.

Advanced insight

Back titration is used when a base is insoluble or reacts slowly, such as calcium carbonate in an indigestion tablet. The solid is dissolved in a known excess of acid, then the leftover acid is titrated with alkali. Subtracting gives the acid used by the solid. It is the same mole reasoning applied twice.

Summary

Multi-step problems are solved by reading carefully, identifying the chemistry, writing a balanced equation, converting to moles, applying the mole ratio and converting back. Choose salt preparation using solubility: precipitation for insoluble salts, excess insoluble base for soluble salts from insoluble bases, and titration when both reactants are soluble. Predict pH by deciding which reactant is in excess.

Practice questions

1. 25.0 cm³ of hydrochloric acid is neutralised by 18.75 cm³ of 0.200 mol/dm³ sodium hydroxide. Calculate the concentration of the acid. Answer: Moles NaOH = 0.200 × 0.01875 = 0.00375 mol; ratio 1:1; concentration HCl = 0.00375 ÷ 0.0250 = 0.150 mol/dm³. 2. Suggest how to prepare potassium nitrate crystals and justify your method. Answer: Titrate potassium hydroxide with nitric acid, repeat without indicator, then crystallise; both reactants are soluble, so an excess could not be filtered off. 3. 0.010 mol of hydrochloric acid is mixed with 0.012 mol of sodium hydroxide. Is the final solution acidic, neutral or alkaline? Explain. Answer: Alkaline, because the reaction is 1:1 and 0.002 mol of sodium hydroxide is left in excess. 4. Which two solutions could be mixed to make silver chloride, and what would you do next? Answer: Silver nitrate and sodium chloride solutions; filter off the precipitate, wash it with distilled water and leave it to dry.