Bases and Alkalis for the ESAT

Defining Bases and Alkalis

In chemical terms, a base is the functional opposite of an acid. A base is defined as a substance that has the ability to accept $H^+$ ions (protons), or one that can form hydroxide ions, $OH^-(aq)$, when in aqueous solution. While the terms are often used interchangeably in introductory contexts, it is important to note that an alkali is specifically a base that is soluble in water.

Strong and Weak Bases

The terms strong and weak refer to the extent to which a base dissociates or reacts with water to produce ions.

1. Strong Bases: A strong base or alkali undergoes full dissociation in water, meaning it breaks down completely into its constituent ions. Common laboratory examples include sodium hydroxide ($NaOH$) and potassium hydroxide ($KOH$). The reaction is represented with a single forward arrow to show it goes to completion:

$NaOH(aq) \rightarrow Na^+(aq) + OH^-(aq)$

1. Weak Bases: A weak base only partially forms ions in solution. Ammonia ($NH_3$) is a classic example. Because the reaction is incomplete, it exists in an equilibrium state represented by the reversible reaction arrow:

$NH_3(aq) + H_2O(l) \rightleftharpoons NH_4^+(aq) + OH^-(aq)$

Concentration vs Strength

It is vital to distinguish between chemical strength (dissociation) and concentration (the amount of solute in a volume). These terms refer to the ratio of moles of the base to the volume of the solvent, typically measured in $mol\ dm^{-3}$.

Unlike pH values, there is no fixed cut-off between dilute and concentrated solutions; they are relative terms. For example, a $2.0\ mol\ dm^{-3}$ solution is generally considered dilute, whereas a $6.0\ mol\ dm^{-3}$ solution is considered concentrated.

Crucially, these properties are independent: it is entirely possible to have a concentrated solution of a weak base (many moles of ammonia in a small volume) or a dilute solution of a strong base (very few moles of sodium hydroxide in a large volume).

Formation of Alkaline Solutions

Alkaline solutions contain aqueous hydroxide ions, $OH^-(aq)$. These ions can be produced when certain metal hydroxides or metal oxides interact with water.

Metal Hydroxides

Some metal hydroxides are soluble and dissolve to form alkaline solutions. For instance, sodium hydroxide dissolves completely:

$NaOH(s) + H_2O(l) \rightarrow Na^+(aq) + OH^-(aq)$

Calcium hydroxide is less soluble and exists in equilibrium:

$Ca(OH)_2(s) + H_2O(l) \rightleftharpoons Ca^{2+}(aq) + 2OH^-(aq)$

However, not all metal hydroxides are soluble. Copper(II) hydroxide, $Cu(OH)_2$, is insoluble. While it is chemically a base, it cannot form an alkaline solution because it does not dissolve in water.

Metal Oxides

Certain metal oxides react chemically with water to produce hydroxide ions. Examples include sodium oxide and barium oxide:

$Na_2O(s) + H_2O(l) \rightarrow 2Na^+(aq) + 2OH^-(aq)$

$BaO(s) + H_2O(l) \rightarrow Ba^{2+}(aq) + 2OH^-(aq)$

Just as with hydroxides, some oxides are basic but do not react with water. Copper(II) oxide ($CuO$), for example, is insoluble and does not form an alkaline solution.

Neutralisation Reactions

When an acid reacts with a base, the process is known as neutralisation. This reaction is typically exothermic, meaning it releases heat energy to the surroundings.

In a simple neutralisation between an acidic solution (containing $H^+(aq)$) and an alkaline solution (containing $OH^-(aq)$), the underlying process is the formation of a bond to create water. This is represented by the ionic equation:

$H^+(aq) + OH^-(aq) \rightarrow H_2O(l)$

Because bond formation releases energy and there is no bond breaking required in this specific ionic process, neutralisation is inherently exothermic.

Worked Examples

Example 1: Ionic Equations

Write the full and simplest ionic equations for the reaction between aqueous potassium hydroxide and sulfuric acid.

Step 1: Write the balanced molecular equation. $2KOH(aq) + H_2SO_4(aq) \rightarrow K_2SO_4(aq) + 2H_2O(l)$

Step 2: Identify the simplest ionic equation. Since this is a reaction between a strong aqueous acid and a strong aqueous alkali, the net change is the formation of water: $H^+(aq) + OH^-(aq) \rightarrow H_2O(l)$

Example 2: Stoichiometry with Bases

Calculate the mass of zinc sulfate formed when $20\ cm^3$ of $2.0\ mol\ dm^{-3}$ sulfuric acid reacts with excess zinc oxide.

Step 1: Write the balanced equation. $ZnO(s) + H_2SO_4(aq) \rightarrow ZnSO_4(aq) + H_2O(l)$

Step 2: Calculate moles of $H_2SO_4$. $Moles = concentration \times volume = 2.0 \times (20 / 1000) = 0.04\ mol$

Step 3: Determine moles of $ZnSO_4$. The ratio is $1:1$, so $0.04\ mol$ of $ZnSO_4$ is produced.

Step 4: Calculate mass. The $M_r$ of $ZnSO_4$ is $161.5$. $Mass = 0.04 \times 161.5 = 6.46\ g$

Example 3: Titration Calculation

A student titrates $21.5\ cm^3$ of sulfuric acid against $25.0\ cm^3$ of $2.0\ mol\ dm^{-3}$ sodium hydroxide. Calculate the concentration of the acid.

Step 1: Write the balanced equation. $H_2SO_4(aq) + 2NaOH(aq) \rightarrow Na_2SO_4(aq) + 2H_2O(l)$ Sulfuric acid is diprotic, meaning the mole ratio of $H_2SO_4$ to $NaOH$ is $1:2$.

Step 2: Calculate moles of $NaOH$. $Moles = 2.0 \times (25.0 / 1000) = 0.050\ mol$

Step 3: Calculate moles of $H_2SO_4$. $Moles = 0.050 / 2 = 0.025\ mol$

Step 4: Calculate concentration of $H_2SO_4$. $Concentration = 0.025 / (21.5 / 1000) = 1.16\ mol\ dm^{-3}$