Question: Write the ionic equation, with state symbols, for the neutralisation of dilute sulfuric acid by potassium hydroxide solution.Step 1: write the balanced full equation.H2SO4(aq) + 2KOH(aq) → K2SO4(aq) + 2H2O(l) Step 2: split the aqueous ionic substances into ions. The dilute acid is treated as H+(aq) and SO42-(aq) ions. Water is a covalent liquid, so it is not split.2H+(aq) + SO42-(aq) + 2K+(aq) + 2OH-(aq) → 2K+(aq) + SO42-(aq) + 2H2O(l) Step 3: cancel the spectator ions, K+(aq) and SO42-(aq).2H+(aq) + 2OH-(aq) → 2H2O(l) Step 4: divide by any common factor so the equation uses the simplest whole numbers. Here every coefficient is divided by 2.H+(aq) + OH-(aq) → H2O(l) Step 5: check. Charge on the left is (+1) + (-1) = 0, and water is neutral.This is the ionic equation for the reaction of any strong acid with any alkali, for example hydrochloric acid with sodium hydroxide or nitric acid with potassium hydroxide. The anion from the acid and the metal ion from the alkali are always spectator ions.
Ionic Equations - Question 2
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Specification: Edexcel Chemistry (8CH0), AS-level, Topic 5: Formulae, Equations and Amounts of Substance. Topic: Amount of Substance and Calculations.
Key terms in this lesson
- spectator ion
- An ion present in a reaction mixture that appears unchanged on both sides of the equation and takes no part in it.
- neutralisation
- A reaction between an acid and a base that produces a salt and water.
More in Amount of Substance and Calculations
- Chemical Formulae
- Balanced Equations
- Ionic Equations
- Ionic Equations - Question1
- Ionic Equations - Question3
- Ionic Equations - Question 4
- Calculations of Reacting Masses
- Working out Equations from Reacting Masses
All 39 lessons in Amount of Substance and Calculations · All Edexcel AS-level Chemistry topics