Acids and bases form conjugate pairs

When Brønsted-Lowry acids and bases react together, they form conjugate acid-base pairs on opposite sides of the reaction equation:

A conjugate acid-base pair consists of two species that are interconverted by the transfer of a proton (H⁺).

• In the forward reaction, HA acts as an acid, donating a proton to form its conjugate base (A^-).
• In the reverse reaction, A^- acts as a base, accepting a proton from BH⁺ to reform the acid (HA).
• Similarly, B and BH⁺ form another conjugate pair, with B being the base (proton acceptor) and BH⁺ its conjugate acid.

For example, when ethanoic acid (CH₃COOH) reacts with water:

In this reaction:

• CH₃COOH and CH₃COO^- are a conjugate pair. CH₃COOH is the acid (proton donor) and CH₃COO^- is its conjugate base.
• H₂O and H₃O⁺ form the other conjugate pair, with H₂O acting as the base (proton acceptor) and H₃O⁺ as its conjugate acid.

Key points:

• The conjugate base always has one less H⁺ than its conjugate acid.
• The conjugate acid always has one more H⁺ than its conjugate base.

Water can act as an acid or a base

Water is an amphiprotic substance, meaning it can behave as both an acid and a base depending on the reaction.

1. Water as a base - When reacting with acids, water accepts a proton to form a hydronium ion (H₃O⁺), which is water's conjugate acid. For example:

2. Water as an acid - When interacting with bases, water donates a proton to form a hydroxide ion (OH^-), which is water's conjugate base. For example:

Acids react with metals and bases

1. Acids react with reactive metals to produce a salt and hydrogen gas:

• Metal + acid ➔ salt + hydrogen
• Example: Mg_(s) + 2HCl_(aq) ➔ MgCl_2(aq) + H_2(g)

2. Acids react with metal carbonates to produce a salt, water, and carbon dioxide gas:

• Metal carbonate + acid ➔ salt + water + carbon dioxide
• Example: CaCO_3(s) + 2HCl_(aq) ➔ CaCl_2(aq) + H₂O_(l) + CO_2(g)

3. Acids react with alkalis (soluble bases) to yield a salt and water:

• Acid + alkali ➔ salt + water
• Example: HCl_(aq) + NaOH_(aq) ➔ NaCl_(aq) + H₂O_(l)

4. Acids react with insoluble bases (usually metal oxides) to form a salt and water:

• Acid + metal oxide ➔ salt + water
• Example: 2HCl_(aq) + CuO_(s) ➔ CuCl_2(aq) + H₂O_(l)

The ionic product of water (Kw)

A small fraction of water molecules spontaneously transfer protons between themselves, a process called autoionisation:

2H₂O_(l) ⇌ H₃O⁺_(aq) + OH^-_(aq)

Or, more simply:

H₂O_(l) ⇌ H⁺_(aq) + OH^-_(aq)

The equilibrium constant (K_c) for this reaction is:

${\text{K}}_{\text{c}}=\frac{[{\text{H}}^{+}][{\text{OH}}^{-}]}{[{\text{H}}_2\text{O}]}$

However, because water is present in vast excess compared to H⁺ and OH^-, its concentration [H₂O] is essentially constant.

Multiplying both sides of the K_c expression by [H₂O] gives the ionic product of water (K_w):

K_w = [H⁺][OH^-]

${\text{At 298 K, K}}_{\text{w}}=1.00\times {10}^{-14}{\text{ mol}}^2{\text{ dm}}^{-6}$

The value of K_w varies with temperature.

H+ and OH- concentrations in pure water

In pure water, the concentrations of H⁺ and OH^- are equal:

${\text{K}}_{\text{w}}={[{\text{H}}^{+}]}^2={[{\text{OH}}^{-}]}^2=1.00\times {10}^{-14}{\text{ mol}}^2{\text{ dm}}^{-6}$

Therefore, in pure water at 298 K:

$[{\text{H}}^{+}]=[{\text{OH}}^{-}]=\sqrt{(1.00\times {10}^{-14})}=1.00\times {10}^{-7}{\text{ mol dm}}^{-3}$