Describe and explain how the solubility of the hydroxides changes down Group 2.
Calcium reacts vigorously with hydrochloric acid according to the equation $\text{Ca}(s) + 2\text{HCl}(aq) \rightarrow \text{CaCl}_2(aq) + \text{H}_2(g)$. Compare the enthalpy change for this reaction with that for the reaction in which $\text{HNO}_3(aq)$ is used instead of $\text{HCl}$, while every other condition remains the same. Explain your answer.
Calcium reacts vigorously with $\text{HCl(aq)}$, producing $\text{H}_2\text{(g)}$.\n\n$\text{Ca(s)} + 2\text{HCl(aq)} \\rightarrow \\text{CaCl}_2\text{(aq)} + \\text{H}_2\text{(g)}$
The ionic equation for this reaction is given below.\n\n$\text{Ca(s)} + 2\text{H}^+\text{(aq)} \rightarrow \text{Ca}^{2+}\text{(aq)} + \text{H}_2\text{(g)} \qquad \Delta H = x\ \text{kJ mol}^{-1}$\n\nDraw a complete, labelled Hess’ Law cycle that connects each side of the equation to the appropriate gas phase ions.\n\nUsing your cycle, the data below, and values from the Data Booklet, calculate $x$.\n\nStandard enthalpy of atomisation of $\text{Ca(s)}$, $\Delta H^{\circ}_{at}(\text{Ca}) = +178\ \text{kJ mol}^{-1}$\n\nStandard enthalpy of hydration of $\text{Ca}^{2+}\text{(g)}$, $\Delta H^{\circ}_{hyd}(\text{Ca}^{2+}) = -1576\ \text{kJ mol}^{-1}$\n\nStandard enthalpy of hydration of $\text{H}^+\text{(g)}$, $\Delta H^{\circ}_{hyd}(\text{H}^+) = -1090\ \text{kJ mol}^{-1}$
The standard enthalpy change for the reaction of $\text{Ca(s)}$ with $\text{CH}_3\text{CO}_2\text{H(aq)}$ is $2\ \text{kJ mol}^{-1}$ less negative than $x$. Suggest an explanation for this.