Question

What is the effect on the amount of CaHPO₄ that dissolves and the concentrations of Ca₂+ and HPO₄2− when each of the following are added to a mixture of solid CaHPO₄ and water at equilibrium?

a) Increase in CaHPO₄, [Ca₂+] increases, [HPO₄2−] decreases
b) Decrease in CaHPO₄, [Ca₂+] decreases, [HPO₄2−] increases
c) No effect on CaHPO₄, [Ca₂+] and [HPO₄2−] remain unchanged
d) No effect on CaHPO₄, but [Ca₂+] and [HPO₄2−] both increase

Answer 1

The addition of a compound like CaCl2 decreases CaHPO4 solubility due to the common ion effect, increasing Ca2+ concentration and decreasing HPO42- concentration. Conversely, adding HCl increases the solubility due to the formation of H2PO4-.

Step-by-step explanation:

Understanding the effect of various additions on the solubility of CaHPO4 and the concentrations of Ca2+ and HPO42- involves applying the principle of Le Châtelier and the concept of the common ion effect. Adding a compound like CaCl2 to a solution at equilibrium with solid CaHPO4 will introduce additional Ca2+ ions, hence driving the dissolution equilibrium backward and leading to a decrease in solubility of CaHPO4 (precipitation).

In this case, the concentration of Ca2+ ions increases and that of HPO42- ions decreases. Conversely, adding an acid like HCl can increase the solubility by protonating HPO42- to form H2PO4-, thus removing HPO42- ions from the solution and shifting the equilibrium towards the dissolution of more CaHPO4.