The following table lists the solubilities of the hydroxides and carbonates of some of the Group 2 elements, M, at 25 °C - CIE - A-Level Chemistry - Question 2 - 2019 - Paper 1

The general decrease in the solubility of Group 2 carbonates down the group can be attributed to the increasing lattice enthalpy compared to the decrease in hydration enthalpy. As the size of the M²⁺ ions increases, the lattice energy needed to break the solid structure becomes larger than the hydration energy gained, resulting in lower solubility.

For Magnesium (Mg), the solubility is low, thus a saturated solution of Mg(OH)₂ would not effectively test for CO₂, as it has a very low potential to form a precipitate. For Strontium (Sr), the higher solubility means that a saturated solution can be used to test for CO₂, as it would likely form a precipitate of CaCO₃. For Barium (Ba), similar to Sr, the saturated solution of Ba(OH)₂ would also effectively test for CO₂.

The solubility product, Ksp, can be calculated using the formula:

$K_{sp} = [Mg^{2+}][OH^{-}]^2$

Given the solubility of Mg(OH)₂ is 2.0 × 10⁻⁴ mol dm⁻³, we find: $[Mg^{2+}] = 2.0 imes 10^{-4}$ $[OH^{-}] = 2 imes 2.0 imes 10^{-4} = 4.0 imes 10^{-4}$

Thus, $K_{sp} = (2.0 imes 10^{-4})(4.0 imes 10^{-4})^2 = 3.2 imes 10^{-12}$

Observation: A white precipitate will appear.

Explanation: The addition of barium hydroxide will increase the concentration of Ba²⁺ ions in the solution, which will react with CO₃²⁻ ions from the barium carbonate to form a precipitate of barium carbonate.

To calculate ΔHf⁰(OH⁻(g)), we construct a Born-Haber cycle and use the relevant enthalpic values:

$ext{ΔH}f^0(OH⁻(g)) = ext{lattice energy} + 2 imes ext{enthalpy change} + ext{atomization of Mg}$ Taking the values from the data:

• Lattice energy of Mg(OH)₂(s): -2993 kJ mol⁻¹
• Atomization of Mg(s): +148 kJ mol⁻¹
• Formation of Mg(OH)₂(s): -925 kJ mol⁻¹, We calculate: $ΔHf^0(OH⁻(g)) = -2993 + 148 - 925 = -333 ext{ kJ mol}⁻¹$