A water-treatment engineer neutralises a spillage of dilute hydrochloric acid by adding sodium hydroxide solution until the mixture is exactly neutral. (a) Define the term proton acceptor, and explain why sodium hydroxide is classified as a base according to this definition. [2]…

Answer: ((a)) A proton acceptor is a substance that accepts (gains) a hydrogen ion, H+, from another substance. Sodium hydroxide provides hydroxide ions, OH-, which accept H+ ions from the acid to form water, so sodium hydroxide is classified as a base. ((b)) NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) ((c)) H^+(aq) + OH^-(aq) → H2O(l); sodium ions, Na+(aq), and chloride ions, Cl-(aq), are unchanged in identity and state on both sides of the full equation (spectator ions), so they are cancelled out and do not appear in the ionic equation.. Part (a): Proton acceptor A proton acceptor is a substance that accepts (gains) a hydrogen ion, H^+, from another substance. Sodium hydroxide dissolves in water to give hydroxide ions, OH^-. When it reacts with an acid, these OH^- ions acc…

IGCSE Chemistry 0620: Proton Acceptors and the Ionic Equation for Neutralisation

Part (a): Proton acceptor

A proton acceptor is a substance that accepts (gains) a hydrogen ion, $\text{H}^+$ , from another substance.

Sodium hydroxide dissolves in water to give hydroxide ions, $\text{OH}^-$ . When it reacts with an acid, these $\text{OH}^-$ ions accept $\text{H}^+$ ions donated by the acid, combining to form water. Because sodium hydroxide accepts $\text{H}^+$ ions in this way, it is classified as a base.

Part (b): Full symbol equation

Sodium hydroxide and hydrochloric acid are both soluble, and their reaction is a straightforward acid + alkali neutralisation, forming a salt (sodium chloride) and water:

$\text{NaOH(aq)} + \text{HCl(aq)} \rightarrow \text{NaCl(aq)} + \text{H}_2\text{O(l)}$

Counting atoms: 1 Na, 1 O, 2 H (1 from NaOH, 1 from HCl), 1 Cl on the left; 1 Na, 1 Cl (from NaCl), 1 O, 2 H (from $\text{H}_2\text{O}$ ) on the right, already balanced with no extra coefficients needed.

Part (c): Ionic equation and spectator ions

In solution, both $\text{NaOH}$ and $\text{HCl}$ exist fully as separate ions: $\text{Na}^+\text{(aq)}$ , $\text{OH}^-\text{(aq)}$ , $\text{H}^+\text{(aq)}$ and $\text{Cl}^-\text{(aq)}$ . Writing the full equation in terms of these ions:

$\text{Na}^+\text{(aq)} + \text{OH}^-\text{(aq)} + \text{H}^+\text{(aq)} + \text{Cl}^-\text{(aq)} \rightarrow \text{Na}^+\text{(aq)} + \text{Cl}^-\text{(aq)} + \text{H}_2\text{O(l)}$

$\text{Na}^+\text{(aq)}$ and $\text{Cl}^-\text{(aq)}$ appear, unchanged, on both sides of this equation, they do not react and remain dissolved ions throughout. Ions like these, that take no real part in the reaction, are called spectator ions, and they can be cancelled from both sides. What remains is the ionic equation, showing only the species that actually react:

$\text{H}^+\text{(aq)} + \text{OH}^-\text{(aq)} \rightarrow \text{H}_2\text{O(l)}$

This is the true reaction taking place: the proton donor ( $\text{H}^+$ ) combining with the proton acceptor ( $\text{OH}^-$ ) to form water. The same reaction occurs whichever strong acid and strong alkali are used, since $\text{Na}^+$ and $\text{Cl}^-$ never actually change.

Final answers

(a) A proton acceptor gains an $\text{H}^+$ ion; sodium hydroxide’s $\text{OH}^-$ ions accept $\text{H}^+$ from the acid, so it is a base.
(b) $\text{NaOH(aq)} + \text{HCl(aq)} \rightarrow \text{NaCl(aq)} + \text{H}_2\text{O(l)}$
(c) $\text{H}^+\text{(aq)} + \text{OH}^-\text{(aq)} \rightarrow \text{H}_2\text{O(l)}$ ; $\text{Na}^+\text{(aq)}$ and $\text{Cl}^-\text{(aq)}$ are spectator ions, unchanged on both sides, so they are cancelled out.