Chemistry - Acids, Bases, and Salts

Properties of Acids and Bases

Acids

An acid is a substance that produces hydrogen ions ($\mathrm{H^+}$) when dissolved in water.

General properties:

• Turn blue litmus red
• Have a sour taste
• React with metals to produce hydrogen gas
• React with bases in neutralisation reactions
• Have pH values below 7

Bases

A base is a substance that neutralises an acid. A soluble base is called an alkali and produces hydroxide ions ($\mathrm{OH^-}$) in solution.

General properties of alkalis:

• Turn red litmus blue
• Have a soapy feel
• React with acids in neutralisation reactions
• Have pH values above 7

Worked Example 1

Write balanced equations for the reaction of hydrochloric acid with (a) zinc and (b) calcium carbonate.

(a) $\mathrm{Zn} + 2\mathrm{HCl} \to \mathrm{ZnCl_2} + \mathrm{H_2}$

(b) $\mathrm{CaCO_3} + 2\mathrm{HCl} \to \mathrm{CaCl_2} + \mathrm{H_2O} + \mathrm{CO_2}$

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Strong and Weak Acids

Strong Acids

A strong acid is completely dissociated in aqueous solution. The equilibrium lies entirely to the right:

$\mathrm{HCl} \to \mathrm{H^+} + \mathrm{Cl^-}$

$\mathrm{H_2SO_4} \to 2\mathrm{H^+} + \mathrm{SO_4^{2-}}$

$\mathrm{HNO_3} \to \mathrm{H^+} + \mathrm{NO_3^-}$

For a strong acid of concentration $c$, $[\mathrm{H^+}] = c$ (for monoprotic) or $[\mathrm{H^+}] = 2c$ (for diprotic like $\mathrm{H_2SO_4}$).

Weak Acids

A weak acid is only partially dissociated in aqueous solution. A dynamic equilibrium is established:

$\mathrm{CH_3COOH} \rightleftharpoons \mathrm{H^+} + \mathrm{CH_3COO^-}$

For a weak acid of concentration $c$, $[\mathrm{H^+}] \ll c$.

Key point: A weak acid is not the same as a dilute acid. Concentration refers to how much acid is dissolved; strength refers to the degree of dissociation.

Comparison	Strong Acid	Weak Acid
Dissociation	Complete	Partial
Equilibrium	Lies far to the right	Lies to the left
pH (same conc)	Lower (more acidic)	Higher (less acidic)
Conductivity	Higher (more ions)	Lower (fewer ions)
Reaction rate	Faster	Slower

Worked Example 2

Calculate the pH of a $0.050 \mathrm{ mol/dm^3}$ solution of $\mathrm{HCl}$.

$\mathrm{HCl}$ is a strong acid, so $[\mathrm{H^+}] = 0.050 \mathrm{ mol/dm^3}$.

$\mathrm{pH} = -\log_{10}[\mathrm{H^+}] = -\log_{10}(0.050) = 1.30$

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The pH Scale

The pH scale measures the acidity or alkalinity of a solution:

$\mathrm{pH} = -\log_{10}[\mathrm{H^+}]$

where $[\mathrm{H^+}]$ is in $\mathrm{mol/dm^3}$.

$[\mathrm{H^+}] = 10^{-\mathrm{pH}}$

At $25^\circ\mathrm{C}$: $\mathrm{pH} + \mathrm{pOH} = 14$

pH Range	Classification
$0 - 6$	Acidic
7	Neutral
$8 - 14$	Alkaline

Worked Example 3

A solution has $\mathrm{pH} = 3.40$. Calculate $[\mathrm{H^+}]$ and $[\mathrm{OH^-}]$.

$[\mathrm{H^+}] = 10^{-3.40} = 3.98 \times 10^{-4} \mathrm{ mol/dm^3}$

$[\mathrm{OH^-}] = \frac{K_w}{[\mathrm{H^+}]} = \frac{1.0 \times 10^{-14}}{3.98 \times 10^{-4}} = 2.51 \times 10^{-11} \mathrm{ mol/dm^3}$

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Indicators

Indicators are substances that change colour depending on the pH of the solution.

Indicator	Colour in Acid	Colour in Alkali	pH Range
Litmus	Red	Blue	$4.5 - 8.3$
Methyl orange	Red	Yellow	$3.1 - 4.4$
Phenolphthalein	Colourless	Pink	$8.3 - 10.0$
Universal	Varies	Varies	$1 - 14$

For titrations:

• Strong acid vs. strong base: use methyl orange or phenolphthalein
• Strong acid vs. weak base: use methyl orange
• Weak acid vs. strong base: use phenolphthalein

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Neutralisation

Neutralisation is the reaction between an acid and a base to produce a salt and water:

$\mathrm{Acid} + \mathrm{Base} \to \mathrm{Salt} + \mathrm{Water}$

Ionic Equation for Neutralisation

$\mathrm{H^+}(aq) + \mathrm{OH^-}(aq) \to \mathrm{H_2O}(l)$

This is the net ionic equation for all acid-base neutralisations in aqueous solution.

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Common Acids and Their Properties

Acid	Formula	Type	Key Property
Hydrochloric acid	$\mathrm{HCl}$	Monoprotic, strong	Used in stomach, reacts with metals, carbonates
Sulphuric acid	$\mathrm{H_2SO_4}$	Diprotic, strong	Dehydrating agent, diprotic (2 $\mathrm{H^+}$ per molecule)
Nitric acid	$\mathrm{HNO_3}$	Monoprotic, strong	Oxidising agent, reacts with metals to produce NO/NO$_2$
Ethanoic acid	$\mathrm{CH_3COOH}$	Monoprotic, weak	Found in vinegar, partial dissociation
Carbonic acid	$\mathrm{H_2CO_3}$	Diprotic, weak	Formed when $\mathrm{CO_2}$ dissolves in water

Worked Example 4

Write the equation for the reaction of dilute nitric acid with copper.

Unlike $\mathrm{HCl}$, nitric acid is an oxidising agent and does not produce hydrogen with copper:

$3\mathrm{Cu} + 8\mathrm{HNO_3}(\mathrm{dilute}) \to 3\mathrm{Cu(NO_3)_2} + 2\mathrm{NO} + 4\mathrm{H_2O}$

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Common Bases

Base	Formula	Type	Use
Sodium hydroxide	$\mathrm{NaOH}$	Strong alkali	Soap manufacture, neutralisation
Potassium hydroxide	$\mathrm{KOH}$	Strong alkali	Making soft soaps
Calcium hydroxide	$\mathrm{Ca(OH)_2}$	Weak alkali	Limewater, agriculture
Ammonia solution	$\mathrm{NH_3(aq)}$	Weak alkali	Cleaning, fertiliser production

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Salt Preparation Methods

The method used to prepare a salt depends on the solubility of the salt and the reactivity of the metal or base.

Solubility Rules

Ion	Soluble?
$\mathrm{Na^+}$, $\mathrm{K^+}$, $\mathrm{NH_4^+}$	Always soluble
$\mathrm{NO_3^-}$	Always soluble
$\mathrm{Cl^-}$	Soluble except $\mathrm{AgCl}$, $\mathrm{PbCl_2}$
$\mathrm{SO_4^{2-}}$	Soluble except $\mathrm{BaSO_4}$, $\mathrm{PbSO_4}$, $\mathrm{CaSO_4}$
$\mathrm{CO_3^{2-}}$	Insoluble except $\mathrm{Na^+}$, $\mathrm{K^+}$, $\mathrm{NH_4^+}$
$\mathrm{OH^-}$	Insoluble except $\mathrm{Na^+}$, $\mathrm{K^+}$, $\mathrm{Ca^{2+}}$ (slightly)

Method 1: Acid + Metal

For reactive metals (above hydrogen in the reactivity series) with soluble salts:

$\mathrm{Zn} + \mathrm{H_2SO_4} \to \mathrm{ZnSO_4} + \mathrm{H_2}$

Excess metal is added, then filtered to remove the excess, and the filtrate is crystallised.

Method 2: Acid + Insoluble Base (Base Addition)

For metals that are not reactive enough to displace hydrogen (e.g., copper), or for metal oxides and hydroxides:

$\mathrm{CuO} + \mathrm{H_2SO_4} \to \mathrm{CuSO_4} + \mathrm{H_2O}$

Excess base is added, then filtered, and the filtrate is crystallised.

Method 3: Acid + Alkali (Titration)

For soluble salts of reactive metals, especially when the base is an alkali:

$\mathrm{NaOH} + \mathrm{HCl} \to \mathrm{NaCl} + \mathrm{H_2O}$

The exact volume of alkali needed to neutralise the acid is found by titration. The solution is then evaporated to crystallise the salt.

Method 4: Precipitation

For insoluble salts, mix two soluble solutions:

$\mathrm{AgNO_3}(aq) + \mathrm{NaCl}(aq) \to \mathrm{AgCl}(s) + \mathrm{NaNO_3}(aq)$

The precipitate is collected by filtration, washed, and dried.

Worked Example 5

Describe how to prepare pure, dry crystals of copper(II) sulphate from copper(II) oxide and dilute sulphuric acid.

1. Add excess $\mathrm{CuO}$ to dilute $\mathrm{H_2SO_4}$ in a beaker.
2. Warm gently and stir until no more $\mathrm{CuO}$ dissolves (excess ensures all acid is consumed).
3. Filter to remove the excess $\mathrm{CuO}$.
4. Heat the filtrate to concentrate it by evaporation until crystals begin to form.
5. Allow the solution to cool slowly for crystallisation.
6. Filter to collect the crystals and pat dry between filter papers.

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Ionic Equations

Ionic equations show only the species that actually participate in a reaction (spectator ions are cancelled).

Steps to Write Ionic Equations

1. Write the full balanced equation with state symbols.
2. Split all soluble ionic compounds into their ions.
3. Cancel spectator ions (ions that appear on both sides unchanged).
4. Write the net ionic equation.

Worked Example 6

Write the ionic equation for the reaction between lead(II) nitrate and potassium iodide.

Full equation:

$\mathrm{Pb(NO_3)_2}(aq) + 2\mathrm{KI}(aq) \to \mathrm{PbI_2}(s) + 2\mathrm{KNO_3}(aq)$

Ionic equation:

$\mathrm{Pb^{2+}}(aq) + 2\mathrm{NO_3^-}(aq) + 2\mathrm{K^+}(aq) + 2\mathrm{I^-}(aq) \to \mathrm{PbI_2}(s) + 2\mathrm{K^+}(aq) + 2\mathrm{NO_3^-}(aq)$

Net ionic equation (cancel $\mathrm{K^+}$ and $\mathrm{NO_3^-}$):

$\mathrm{Pb^{2+}}(aq) + 2\mathrm{I^-}(aq) \to \mathrm{PbI_2}(s)$

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Carbonates and Their Reactions

Carbonates react with acids to produce a salt, water, and carbon dioxide gas:

$\mathrm{Carbonate} + \mathrm{Acid} \to \mathrm{Salt} + \mathrm{H_2O} + \mathrm{CO_2}$

Test for Carbon Dioxide

Bubble the gas through limewater ($\mathrm{Ca(OH)_2}$ solution). If $\mathrm{CO_2}$ is present, the limewater turns milky:

$\mathrm{CO_2} + \mathrm{Ca(OH)_2} \to \mathrm{CaCO_3}(s) + \mathrm{H_2O}$

Worked Example 7

$5.00 \mathrm{ g}$ of calcium carbonate is reacted with excess hydrochloric acid. Calculate the volume of $\mathrm{CO_2}$ produced at RTP.

$\mathrm{CaCO_3} + 2\mathrm{HCl} \to \mathrm{CaCl_2} + \mathrm{H_2O} + \mathrm{CO_2}$

$n(\mathrm{CaCO_3}) = \frac{5.00}{100} = 0.0500 \mathrm{ mol}$

$n(\mathrm{CO_2}) = 0.0500 \mathrm{ mol}$

$V(\mathrm{CO_2}) = 0.0500 \times 24.0 = 1.20 \mathrm{ dm^3}$

Thermal Decomposition of Carbonates

Group 1 carbonates (except lithium carbonate) do not decompose on heating. Other metal carbonates decompose:

$\mathrm{CaCO_3} \xrightarrow{\Delta} \mathrm{CaO} + \mathrm{CO_2}$

$\mathrm{CuCO_3} \xrightarrow{\Delta} \mathrm{CuO} + \mathrm{CO_2}$

$\mathrm{ZnCO_3} \xrightarrow{\Delta} \mathrm{ZnO} + \mathrm{CO_2}$

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Common Pitfalls

• Confusing strong acids with concentrated acids. Strength refers to the degree of dissociation; concentration refers to the amount dissolved.
• Writing $\mathrm{H^+}$ instead of $\mathrm{H_3O^+}$ (hydronium ion). While $\mathrm{H^+}$ is acceptable in most exam answers, remember that protons in water exist as $\mathrm{H_3O^+}$.
• Forgetting state symbols in ionic equations. State symbols are essential for correctly identifying spectator ions.
• Using the wrong indicator for a titration. Remember: strong acid + weak base uses methyl orange; weak acid + strong base uses phenolphthalein.
• Assuming all metal oxides are basic. Some metal oxides (e.g., $\mathrm{Al_2O_3}$, $\mathrm{ZnO}$) are amphoteric and react with both acids and bases.
• Forgetting that $\mathrm{NH_3}$ is a weak base, not a strong base.

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Summary Table

Concept	Key Point
pH	$\mathrm{pH} = -\log_{10}[\mathrm{H^+}]$
Strong acid	Completely dissociated in water
Weak acid	Partially dissociated; equilibrium lies to the left
Neutralisation	$\mathrm{H^+} + \mathrm{OH^-} \to \mathrm{H_2O}$
Soluble salts	$\mathrm{Na^+}$, $\mathrm{K^+}$, $\mathrm{NH_4^+}$, $\mathrm{NO_3^-}$ always soluble
Carbonate + acid	Produces salt + water + $\mathrm{CO_2}$
Limewater test	$\mathrm{CO_2}$ turns limewater milky

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Additional Worked Examples

Worked Example: Weak Acid pH

Calculate the pH of a $0.10 \mathrm{ mol/dm^3}$ solution of ethanoic acid ($K_a = 1.8 \times 10^{-5}$).

Solution

$\mathrm{CH_3COOH} \rightleftharpoons \mathrm{H^+} + \mathrm{CH_3COO^-}$

$K_a = \frac{[\mathrm{H^+}]^2}{[\mathrm{CH_3COOH}]} = \frac{x^2}{0.10}$

$x = \sqrt{1.8 \times 10^{-6}} = 1.34 \times 10^{-3} \mathrm{ mol/dm^3}$

$\mathrm{pH} = -\log_{10}(1.34 \times 10^{-3}) = 2.87$

Worked Example: Titration with Diprotic Acid

$25.0 \mathrm{ cm^3}$ of $\mathrm{NaOH}$ solution is titrated with $0.100 \mathrm{ mol/dm^3}$ $\mathrm{H_2SO_4}$. The average titre is $18.75 \mathrm{ cm^3}$. Find the concentration of $\mathrm{NaOH}$.

Solution

$\mathrm{H_2SO_4} + 2\mathrm{NaOH} \to \mathrm{Na_2SO_4} + 2\mathrm{H_2O}$

$n(\mathrm{H_2SO_4}) = 0.100 \times \frac{18.75}{1000} = 1.875 \times 10^{-3} \mathrm{ mol}$

$n(\mathrm{NaOH}) = 2 \times 1.875 \times 10^{-3} = 3.750 \times 10^{-3} \mathrm{ mol}$

$[\mathrm{NaOH}] = \frac{3.750 \times 10^{-3}}{25.0/1000} = 0.150 \mathrm{ mol/dm^3}$

Worked Example: Limiting Reactant with Carbonate

$3.00 \mathrm{ g}$ of $\mathrm{MgCO_3}$ is added to $50.0 \mathrm{ cm^3}$ of $2.00 \mathrm{ mol/dm^3}$ $\mathrm{HCl}$. Calculate the volume of $\mathrm{CO_2}$ produced at RTP.

Solution

$\mathrm{MgCO_3} + 2\mathrm{HCl} \to \mathrm{MgCl_2} + \mathrm{H_2O} + \mathrm{CO_2}$

$n(\mathrm{MgCO_3}) = \frac{3.00}{84.3} = 0.0356 \mathrm{ mol}$

$n(\mathrm{HCl}) = 2.00 \times 0.0500 = 0.100 \mathrm{ mol}$

$\mathrm{HCl}$ needed for $0.0356 \mathrm{ mol}$ $\mathrm{MgCO_3}$: $2 \times 0.0356 = 0.0712 \mathrm{ mol}$. Since $0.100 \mathrm{ mol}$ is available, $\mathrm{MgCO_3}$ is the limiting reactant.

$n(\mathrm{CO_2}) = 0.0356 \mathrm{ mol}$

$V(\mathrm{CO_2}) = 0.0356 \times 24.0 = 0.854 \mathrm{ dm^3}$

Worked Example: Choosing a Salt Preparation Method

A student wants to prepare pure, dry crystals of lead(II) iodide. Describe the method and write the relevant equation.

Solution

$\mathrm{PbI_2}$ is insoluble, so the precipitation method is used:

1. Prepare aqueous solutions of lead(II) nitrate and potassium iodide.
2. Mix the two solutions: $\mathrm{Pb(NO_3)_2}(aq) + 2\mathrm{KI}(aq) \to \mathrm{PbI_2}(s) + 2\mathrm{KNO_3}(aq)$
3. A bright yellow precipitate of $\mathrm{PbI_2}$ forms.
4. Filter the precipitate.
5. Wash with distilled water to remove soluble impurities.
6. Dry between filter papers.

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Exam-Style Practice Questions

Question 1: Calculate the pH of $0.0020 \mathrm{ mol/dm^3}$ $\mathrm{HNO_3}$.

$\mathrm{HNO_3}$ is a strong monoprotic acid, so $[\mathrm{H^+}] = 0.0020 \mathrm{ mol/dm^3}$.

$\mathrm{pH} = -\log_{10}(0.0020) = 2.70$

Question 2: Write the ionic equation for the reaction between silver nitrate solution and sodium chloride solution.

$\mathrm{Ag^+}(aq) + \mathrm{Cl^-}(aq) \to \mathrm{AgCl}(s)$

($\mathrm{Na^+}$ and $\mathrm{NO_3^-}$ are spectator ions.)

Question 3: Describe how to prepare a pure sample of lead(II) iodide.

Mix aqueous solutions of lead(II) nitrate and potassium iodide:

$\mathrm{Pb(NO_3)_2}(aq) + 2\mathrm{KI}(aq) \to \mathrm{PbI_2}(s) + 2\mathrm{KNO_3}(aq)$

A bright yellow precipitate of $\mathrm{PbI_2}$ forms. Filter, wash with distilled water, and dry.

Question 4: $2.50 \mathrm{ g}$ of $\mathrm{CaCO_3}$ is added to $100 \mathrm{ cm^3}$ of $1.00 \mathrm{ mol/dm^3}$ $\mathrm{HCl}$. Calculate the volume of $\mathrm{CO_2}$ produced at RTP.

$n(\mathrm{CaCO_3}) = 2.50/100 = 0.0250 \mathrm{ mol}$

$n(\mathrm{HCl}) = 1.00 \times 0.100 = 0.100 \mathrm{ mol}$

$\mathrm{CaCO_3}$ is the limiting reactant.

$n(\mathrm{CO_2}) = 0.0250 \mathrm{ mol}$

$V(\mathrm{CO_2}) = 0.0250 \times 24.0 = 0.600 \mathrm{ dm^3}$

Question 5: Explain why a $0.10 \mathrm{ mol/dm^3}$ solution of ethanoic acid has a higher pH than a $0.10 \mathrm{ mol/dm^3}$ solution of $\mathrm{HCl}$.

$\mathrm{HCl}$ is a strong acid and dissociates completely: $[\mathrm{H^+}] = 0.10 \mathrm{ mol/dm^3}$, giving $\mathrm{pH} = 1.0$. Ethanoic acid is a weak acid and only partially dissociates: $[\mathrm{H^+}] \lt 0.10 \mathrm{ mol/dm^3}$, giving a pH greater than 1.0.

Question 6: A student wants to prepare sodium chloride. Which method is most appropriate, and why?

Titration method: react $\mathrm{NaOH}$ with $\mathrm{HCl}$ using a suitable indicator (methyl orange or phenolphthalein). Both reactants are soluble, and the salt $\mathrm{NaCl}$ is soluble, so evaporation of the neutralised solution yields pure crystals.

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Problem Set

Problem 1: Calculate the pH of $0.0050 \mathrm{ mol/dm^3}$ $\mathrm{HNO_3}$.

If you get this wrong, revise: Strong and Weak Acids

Solution

$\mathrm{HNO_3}$ is a strong monoprotic acid: $[\mathrm{H^+}] = 0.0050 \mathrm{ mol/dm^3}$

$\mathrm{pH} = -\log_{10}(0.0050) = 2.30$

Problem 2: A solution has $\mathrm{pH} = 4.55$. Find $[\mathrm{H^+}]$ and $[\mathrm{OH^-}]$.

If you get this wrong, revise: The pH Scale

Solution

$[\mathrm{H^+}] = 10^{-4.55} = 2.82 \times 10^{-5} \mathrm{ mol/dm^3}$

$[\mathrm{OH^-}] = \frac{10^{-14}}{2.82 \times 10^{-5}} = 3.55 \times 10^{-10} \mathrm{ mol/dm^3}$

Problem 3: Which indicator should be used for titrating ethanoic acid with sodium hydroxide? Explain your choice.

If you get this wrong, revise: Indicators

Solution

Phenolphthalein (pH range 8.3--10.0). Ethanoic acid is a weak acid and $\mathrm{NaOH}$ is a strong base, so the equivalence point has $\mathrm{pH} \gt 7$. Phenolphthalein changes colour within this alkaline range. Methyl orange (3.1--4.4) would change colour too early, before the equivalence point.

Problem 4: Write the ionic equation for the reaction between barium chloride and sodium sulphate.

If you get this wrong, revise: Ionic Equations

Solution

Full: $\mathrm{BaCl_2}(aq) + \mathrm{Na_2SO_4}(aq) \to \mathrm{BaSO_4}(s) + 2\mathrm{NaCl}(aq)$

Ionic: $\mathrm{Ba^{2+}}(aq) + \mathrm{SO_4^{2-}}(aq) \to \mathrm{BaSO_4}(s)$

$\mathrm{Na^+}$ and $\mathrm{Cl^-}$ are spectator ions.

Problem 5: Describe how to prepare pure, dry lead(II) sulphate.

If you get this wrong, revise: Salt Preparation Methods

Solution

$\mathrm{PbSO_4}$ is insoluble, so use precipitation:

1. Mix aqueous lead(II) nitrate with aqueous sodium sulphate
2. $\mathrm{Pb(NO_3)_2}(aq) + \mathrm{Na_2SO_4}(aq) \to \mathrm{PbSO_4}(s) + 2\mathrm{NaNO_3}(aq)$
3. White precipitate forms; filter, wash with distilled water, and dry

Problem 6: $10.0 \mathrm{ g}$ of $\mathrm{CaCO_3}$ is heated strongly. Calculate the mass of $\mathrm{CaO}$ produced and the volume of $\mathrm{CO_2}$ at RTP.

If you get this wrong, revise: Thermal Decomposition of Carbonates

Solution

$\mathrm{CaCO_3} \xrightarrow{\Delta} \mathrm{CaO} + \mathrm{CO_2}$

$n(\mathrm{CaCO_3}) = \frac{10.0}{100} = 0.100 \mathrm{ mol}$

$m(\mathrm{CaO}) = 0.100 \times 56.1 = 5.61 \mathrm{ g}$

$V(\mathrm{CO_2}) = 0.100 \times 24.0 = 2.40 \mathrm{ dm^3}$

Problem 7: Explain why zinc oxide is described as amphoteric, giving equations for its reaction with both an acid and a base.

If you get this wrong, revise: Common Pitfalls

Solution

$\mathrm{ZnO}$ reacts with both acids and bases:

With acid: $\mathrm{ZnO} + 2\mathrm{HCl} \to \mathrm{ZnCl_2} + \mathrm{H_2O}$

With base: $\mathrm{ZnO} + 2\mathrm{NaOH} \to \mathrm{Na_2ZnO_2} + \mathrm{H_2O}$

Problem 8: $25.0 \mathrm{ cm^3}$ of $0.200 \mathrm{ mol/dm^3}$ $\mathrm{HCl}$ is mixed with $25.0 \mathrm{ cm^3}$ of $0.200 \mathrm{ mol/dm^3}$ $\mathrm{NaOH}$. What is the pH of the resulting solution?

If you get this wrong, revise: Neutralisation

Solution

$n(\mathrm{HCl}) = 0.200 \times 0.0250 = 0.00500 \mathrm{ mol}$

$n(\mathrm{NaOH}) = 0.200 \times 0.0250 = 0.00500 \mathrm{ mol}$

Equal moles of strong monoprotic acid and strong monoprotic base exactly neutralise each other. The product $\mathrm{NaCl}$ is a neutral salt, so $\mathrm{pH} = 7$.

Problem 9: State the solubility rules for chlorides, sulphates, and carbonates, including exceptions.

If you get this wrong, revise: Solubility Rules

Solution

Chlorides: most soluble. Exceptions: $\mathrm{AgCl}$, $\mathrm{PbCl_2}$, $\mathrm{Hg_2Cl_2}$ (insoluble).

Sulphates: most soluble. Exceptions: $\mathrm{BaSO_4}$, $\mathrm{PbSO_4}$ (insoluble); $\mathrm{CaSO_4}$ (slightly soluble).

Carbonates: most insoluble. Exceptions: Group 1 and $\mathrm{NH_4^+}$ salts are soluble.

Problem 10: A student tests an unknown solution with litmus (turns red), then adds magnesium ribbon and observes bubbling. The gas produced turns limewater milky. Identify the unknown solution and write two equations for the reactions observed.

If you get this wrong, revise: Properties of Acids and Carbonates

Solution

The solution is an acid (red litmus). The gas from the metal reaction turns limewater milky, confirming $\mathrm{CO_2}$, so the acid contains carbonate or the metal is reacting to produce $\mathrm{H_2}$ while a separate carbonate reaction produces $\mathrm{CO_2}$. However, since magnesium reacts with acid to produce $\mathrm{H_2}$ (not $\mathrm{CO_2}$), the limewater must have turned milky due to $\mathrm{CO_2}$ already dissolved in the acid. The unknown solution is likely carbonic acid ($\mathrm{H_2CO_3}$) or a solution of $\mathrm{CO_2}$ in water.

Reaction with Mg: $\mathrm{Mg} + 2\mathrm{H^+} \to \mathrm{Mg^{2+}} + \mathrm{H_2}$

Limewater: $\mathrm{CO_2} + \mathrm{Ca(OH)_2} \to \mathrm{CaCO_3}(s) + \mathrm{H_2O}$