6.7 - Neutralisation

Meaning and Ionic Equation

• Neutralisation is a reaction between an acid and an alkali or base to produce salt and water only.

$\mathrm{HN}{\mathrm{O}}_3(\mathrm{aq})+\mathrm{KOH}(\mathrm{aq})\to \mathrm{KN}{\mathrm{O}}_3(\mathrm{aq})+{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$

• ${\mathrm{K}}^{+}$ and $\mathrm{N}{\mathrm{O}}_3^{-}$ are spectator ions. The net ionic equation is:

${\mathrm{H}}^{+}(\mathrm{aq})+\mathrm{O}{\mathrm{H}}^{-}(\mathrm{aq})\to {\mathrm{H}}_2\mathrm{O}(\mathrm{l})$

Applications

• Medicine: milk of magnesia, $\mathrm{Mg}(\mathrm{OH}{)}_2$, neutralises excess stomach acid.
• Hair care: weak alkali in shampoo neutralises acid on hair.
• Dental care: a base in toothpaste neutralises lactic acid formed by mouth bacteria.
• Agriculture: slaked lime, $\mathrm{Ca}(\mathrm{OH}{)}_2$, treats acidic soil.
• Neutralisation is used to manufacture fertilisers such as ammonium nitrate and potassium sulphate; urea is produced from ammonia and carbon dioxide.

Acid–Base Titration

• Titration quantitatively determines the volume of acid required to neutralise a measured volume of alkali, or vice versa.
• The end point is when the acid–base indicator changes colour.

Determine unknown $\mathrm{KOH}$ concentration:

1. Rinse a pipette with $\mathrm{KOH}$, then transfer exactly $25.0\mathrm{c}{\mathrm{m}}^3$ to a conical flask.
2. Add a few drops of phenolphthalein and place the flask on a white tile.
3. Rinse and fill a burette with $1.0\mathrm{mol}\mathrm{d}{\mathrm{m}}^{-3}$ $\mathrm{HN}{\mathrm{O}}_3$; record the initial reading at eye level.
4. Add acid while swirling until the solution changes from pink to colourless; record a rough titre.
5. Repeat, adding acid rapidly to within $5\mathrm{c}{\mathrm{m}}^3$ of the rough end point, then dropwise. Obtain concordant titres and calculate the average, excluding the rough result.

• Distilled water may wash flask walls because it changes concentration but not reacting moles.

Indicator	Colours
Phenolphthalein	Acidic: colourless; neutral: colourless; alkaline: pink
Methyl orange	Acidic: red; neutral: orange; alkaline: yellow

Neutralisation Calculations

For $a$ mol acid $A$ reacting with $b$ mol alkali $B$:

$\frac{M_a{V}_a}{M_b{V}_b}=\frac{a}{b}$

• Always balance the equation first and use its mole ratio.

• If a reactant is given by mass, convert it to moles before applying the stoichiometric ratio.

• $M_a$ and $M_b$ are acid and alkali molarities; $V_a$ and $V_b$ are their volumes in the same unit.

• Example: $20\mathrm{c}{\mathrm{m}}^3$ of $0.25\mathrm{mol}\mathrm{d}{\mathrm{m}}^{-3}$ $\mathrm{NaOH}$ needs $V_a=0.25\times 20/0.2=25\mathrm{c}{\mathrm{m}}^3$ of $0.2\mathrm{mol}\mathrm{d}{\mathrm{m}}^{-3}$ $\mathrm{HCl}$, using the $1:1$ ratio.

• Mass example: $4.05\mathrm{g}$ $\mathrm{ZnO}$ is $4.05/81=0.05\mathrm{mol}$. The equation $2HN{\mathrm{O}}_3+\mathrm{ZnO}\to \mathrm{Zn}(\mathrm{N}{\mathrm{O}}_3{)}_2+{\mathrm{H}}_2\mathrm{O}$ requires $0.10\mathrm{mol}$ acid. In $50\mathrm{c}{\mathrm{m}}^3$, its molarity is $0.10\times 1000/50=2.0\mathrm{mol}\mathrm{d}{\mathrm{m}}^{-3}$.