Groups 1 and 2, ion tests and titration

What you’ll learn

• Explain reactivity, solubility and thermal-stability trends.
• Link diagnostic observations to ionic equations.
• Use acid–base stoichiometry in a titration calculation.

Understand the idea

Reactivity generally increases down Groups 1 and 2 as increased shielding and larger outer-shell distance make electron removal easier. Metals form ionic chlorides; reaction with water can release hydrogen and form hydroxides, although magnesium is slow with cold water and behaves differently with steam. With oxygen, lithium mainly forms oxide, sodium readily forms peroxide and potassium readily forms superoxide under suitable oxygen conditions, so avoid treating all Group 1 products as identical.

Group 2 hydroxides become more soluble down the group, while sulfates become less soluble. Larger cations polarise nitrate and carbonate anions less strongly, so their salts are more thermally stable. Group 2 carbonates give oxide and carbon dioxide; their nitrates give oxide, nitrogen dioxide and oxygen. Most Group 1 nitrates instead give nitrite and oxygen, with lithium an important exception. Lithium carbonate also decomposes more readily than the other Group 1 carbonates.

Flame colours arise from electron transitions after excitation: lithium is crimson, sodium yellow, potassium lilac, calcium brick-red, strontium crimson and barium apple-green; magnesium has no characteristic visible flame-test colour. Carbonate or hydrogencarbonate with acid produces carbon dioxide, which clouds limewater. Acidified barium ions precipitate sulfate as white barium sulfate. Ammonium ions react with hydroxide to release ammonia, identified using appropriate supervised gas tests.

In titration, use an accurately known solution to determine an unknown concentration from the balanced equation. A standard solution requires quantitative transfer and a known final volume. Use a pipetted aliquot, consistent endpoints and concordant titres; methyl orange and phenolphthalein change colour over different pH ranges. A titre is the difference of two burette readings, so include both reading uncertainties when estimating its maximum absolute uncertainty.

Key formulae

$$\mathrm{Ca(s)+2H_2O(l)\rightarrow Ca(OH)_2(aq)+H_2(g)}$$

$$\mathrm{MCO_3(s)\rightarrow MO(s)+CO_2(g)}\quad(M\text{ is a Group 2 metal})$$

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

$$\frac{n(\mathrm{acid})}{\nu_{\mathrm{acid}}}=\frac{n(\mathrm{base})}{\nu_{\mathrm{base}}}$$

Worked example

A $25.0\ \mathrm{cm^3}$ sodium carbonate aliquot requires $20.0\ \mathrm{cm^3}$ of $0.100\ \mathrm{mol\,dm^{-3}}$ hydrochloric acid for complete neutralisation. Find the carbonate concentration.

1. The equation needs two moles of hydrogen ions per mole of $\mathrm{CO_3^{2-}}$.
2. Acid amount is $0.100(0.0200)=0.00200$ mol, so carbonate amount is $0.00100$ mol.
3. Divide by the aliquot volume: $0.00100/0.0250=0.0400\ \mathrm{mol\,dm^{-3}}$.

Answer: $0.0400\ \mathrm{mol\,dm^{-3}}$.