Chemistry · Lesson

Alkali and alkaline-earth element chemistry

Connect common alkali and alkaline-earth ion patterns with composition and solution inventories. Assess solubility and reaction comparisons with explicit phase, solvation and equilibrium assumptions.

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By the end, you can

Relate selected outer-shell patterns to common +1 and +2 ion assignments.
Balance a stipulated dissolved-salt inventory and its charges.
Distinguish equal masses from equal amounts.
Calculate an invented hydroxide solubility with and without a common-ion background.

Shared ion patterns leave important differences

Alkali atoms have a selected ns¹ outer configuration and commonly form +1 cations. Alkaline-earth atoms have ns² and commonly form +2 cations. These patterns organize many salt formulae, while smaller first members and changes down a group can differ in bonding, hydration, coordination and other behavior.

An aqueous reducing-power comparison includes solvation and other thermodynamic terms; observed reaction vigor can additionally depend on surfaces, phase and kinetics. Lithium illustrates why a single gas-phase ionization trend does not settle every aqueous comparison. Lattice and hydration effects both matter to a salt’s dissolution, and temperature and phase must accompany solubility claims.

A fully dissociated inventory is an imposed assumption

Stipulate that 0.100 mol MgCl₂ is fully dissociated in final solution volume 0.500 L, with no other solute. The ion amounts are 0.100 mol Mg²⁺ and 0.200 mol Cl⁻; amount concentrations are 0.200 and 0.400 mol L⁻¹. Their charge-weighted amount sum is 2× 0.100−0.200=0.

This is an inventory assumption, not evidence that actual ion pairing or activity corrections vanish at these concentrations. An ionic concentration is not automatically its dimensionless activity. Hydration structures and concentration-dependent nonideality need additional data; the simple count is still useful when its scope is stated.

Equal element masses do not give equal amounts

For a selected mass of 24.0 g each, use supplied approximate molar masses Mg=24.305 and Ca=40.078 g mol⁻¹. Their amounts are approximately 0.987451 and 0.598832 mol, giving Mg/Ca amount ratio approximately 1.64896. Values displayed to three significant figures are 0.987 mol, 0.599 mol and ratio 1.65.

This amount calculation does not compare reaction speed, reduction potential or useful product yield. An actual material’s composition and any impurities would affect the amount of the named element present. Keep the original mass and its molar-mass convention visible.

An invented solubility model needs its own activity assumptions

Use an unnamed solid M(OH)₂ in a stipulated ideal solution at 298.15 K with dimensionless K_sp=4.00× 10⁻¹² and concentration standard c°=1 mol L⁻¹. Ignore water autoprotolysis and other equilibria in the selected regime, and retain the solid phase. If dissolution amount concentration is s and no OH⁻ background is present, K_sp=(s/c°)(2s/c°)², giving s=1.00× 10⁻⁴ mol L⁻¹.

With initial background OH⁻ amount concentration b=0.0100 mol L⁻¹, retain the added hydroxide from dissolution: K_sp=(s/c°)((b+2s)/c°)². The positive root is approximately 3.999936× 10⁻⁸ mol L⁻¹. Neglecting 2s relative to b gives 4.00× 10⁻⁸ mol L⁻¹, about 0.001600% above the exact root.

These are invented equilibrium inputs, not the solubility of a named real alkaline-earth hydroxide. The comparison requires fixed temperature, ideal activities, a clamped initial background and no competing complexes or phases. A real salt trend cannot be inferred from charge alone or transferred from this one toy.

An unnamed hydroxide’s common-ion effect under stated ideal assumptions

Initial OH⁻ background / mol L⁻¹	Equation for positive s	s / mol L⁻¹	Approximation scope
0	K_sp=4(s/c°)³	1.00× 10⁻⁴	No other OH⁻ source in this toy
0.0100	K_sp=(s/c°)((0.0100+2s)/c°)²	3.999936× 10⁻⁸	Positive exact root
0.0100	Neglect 2s relative to background	4.00× 10⁻⁸	Approximately 0.001600% above exact root

Original equilibrium arithmetic for retained M(OH)₂(s) at fixed 298.15 K, dimensionless K_sp=4.00× 10⁻¹², c°=1 mol L⁻¹ and ideal activities. Water autoprotolysis and competing equilibria are neglected in the selected regime. The OH⁻ amount from dissolution is retained in the exact background case. No real salt constant is imported.

A net phase change has a stoichiometric mass balance

For the formal net CaCO₃ → CaO + CO₂, use supplied rounded atomic molar masses Ca=40.078, C=12.011 and O=15.999 g mol⁻¹. Molar masses are 100.086, 56.077 and 44.009 g mol⁻¹ respectively. A stipulated complete conversion of 5.00 g pure carbonate gives approximately 2.80144 g CaO and 2.19856 g CO₂.

Their sum is 5.00 g. This is theoretical complete-conversion arithmetic, rather than operating temperatures, proof of actual conversion or a measured gas collection. Phase equilibrium, kinetics, impurities and material recovery determine an actual outcome.

Worked examples

Keep formula amounts and ionic charges together

Use the imposed complete dissociation of 0.100 mol MgCl₂ in 0.500 L.

Mg²⁺ amount 0.100 mol gives 0.200 M.
Cl⁻ amount 0.200 mol gives 0.400 M.
Charge-weighted amount sum: 2× 0.100−0.200=0.

Neutral inventory with the stipulated ion concentrations; actual activities are not determined.

Divide each equal mass by its own molar mass

Use 24.0 g Mg and 24.0 g Ca with the supplied molar masses.

n_Mg = 24.0 / 24.305 ≈ 0.987451 mol.
n_Ca = 24.0 / 40.078 ≈ 0.598832 mol.
n_Mg/n_Ca ≈ 1.64896.

Approximately 0.987 mol Mg and 0.599 mol Ca to three significant figures; not a reactivity comparison.

Check the background-dominated approximation

Use K_sp=4.00× 10⁻¹², c°=1 M and OH⁻ background 0.0100 M in the invented model.

Approximate s = K_sp(c°)³ / b² = 4.00× 10⁻⁸ M.
Its added OH⁻, 2s≈ 8.00× 10⁻⁸ M, is small compared with b.
Retaining that term gives s≈ 3.999936× 10⁻⁸ M.

A bounded common-ion approximation in this specified toy, rather than a real named hydroxide measurement.

Check both products of the theoretical carbonate net

Use 5.00 g pure CaCO₃ and the supplied rounded masses.

Carbonate amount = 5.00 / 100.086 ≈ 0.0499570 mol.
CaO mass ≈ 0.0499570 × 56.077 = 2.80144 g.
CO₂ mass ≈ 0.0499570 × 44.009 = 2.19856 g.

Theoretical complete-conversion products sum to 5.00 g; actual conversion and recovery were not measured.

Try it yourself

Does the fully dissociated inventory prove that concentration equals activity for actual MgCl₂ at these amounts?
Show the answer and reasoning
No.
Dissociation and thermodynamic nonideality are separate questions.
Stipulate complete dissociation of 0.0600 mol MgCl₂ in final volume 0.300 L. Find the Mg²⁺ and Cl⁻ concentrations.
Show the answer and reasoning
Mg²⁺: 0.200 M; Cl⁻: 0.400 M.
Ion amounts 0.0600 and 0.120 mol divided by 0.300 L give these concentrations. Their charge-weighted amount sum is zero; actual activities are unspecified.
Use Ca molar mass 40.078 g mol⁻¹. How much Ca is in 12.0 g of the stipulated pure element?
Show the answer and reasoning
Approximately 0.299 mol.
12.0/40.078≈ 0.299416 mol before rounding to the mass precision. This does not predict reaction rate.
Does the invented M(OH)₂ K_sp identify a measured real salt’s solubility?
Show the answer and reasoning
No.
The unnamed equilibrium inputs and ideal assumptions are explicitly stipulated.

What to carry forward

Use common ion patterns with their limits.
Balance dissolved amounts and charges separately from activities.
Convert equal masses through their own molar masses.
Retain the common-ion and phase assumptions in a solubility model.
Separate theoretical conversion masses from actual outcomes.

References

Teaching resources and reference material supporting this lesson’s explanations and examples.