15.1 Precipitation and dissolution  (Page 2/17)

Writing equations and solubility products

Write the ionic equation for the dissolution and the solubility product expression for each of the following slightly soluble ionic compounds:

(a) AgI, silver iodide, a solid with antiseptic properties

(b) CaCO ₃ , calcium carbonate, the active ingredient in many over-the-counter chewable antacids

(c) Mg(OH) ₂ , magnesium hydroxide, the active ingredient in Milk of Magnesia

(d) Mg(NH ₄ )PO ₄ , magnesium ammonium phosphate, an essentially insoluble substance used in tests for magnesium

(e) Ca ₅ (PO ₄ ) ₃ OH, the mineral apatite, a source of phosphate for fertilizers

(Hint: When determining how to break (d) and (e) up into ions, refer to the list of polyatomic ions in the section on chemical nomenclature.)

Solution

(a) $\text{AgI}(s)\rightleftharpoons {\text{Ag}}^{\text{+}}(aq)+{\text{I}}^{\text{−}}(aq)K_{\text{sp}}={[\text{Ag}}^{\text{+}}][{\text{I}}^{\text{−}}]$

(b) ${\text{CaCO}}_3(s)\rightleftharpoons {\text{Ca}}^{\text{2+}}(aq)+{\text{CO}}_3{}^{\text{2−}}(aq)K_{\text{sp}}=[{\text{Ca}}^{\text{2+}}{\left]\right[\text{CO}}_3{}^{\text{2−}}]$

(c) ${\text{Mg}(\text{OH})}_2(s)\rightleftharpoons {\text{Mg}}^{\text{2+}}(aq)+{\text{2OH}}^{\text{−}}(aq)K_{\text{sp}}=[{\text{Mg}}^{\text{2+}}]{[{\text{OH}}^{\text{−}}]}^2$

(d) $\text{Mg}({\text{NH}}_{\text{4}}){\text{PO}}_{\text{4}}(s)\rightleftharpoons {\text{Mg}}^{\text{2+}}(aq)+{\text{NH}}_4{}^{\text{+}}(aq)+{\text{PO}}_4{}^{\text{3−}}(aq)K_{\text{sp}}=[{\text{Mg}}^{\text{2+}}{][\text{NH}}_4{}^{\text{+}}]{[\text{PO}}_4{}^{\text{3−}}]$

(e) ${\text{Ca}}_5({\text{PO}}_4)3\text{OH}(s)\rightleftharpoons {\text{5Ca}}^{\text{2+}}(aq)+{\text{3PO}}_4{}^{\text{3−}}(aq)+{\text{OH}}^{\text{−}}(aq)K_{\text{sp}}={{[\text{Ca}}^{\text{2+}}]}^5[\text{P}{\text{O}}_4{}^{\text{3−}}{]}^3[{\text{OH}}^{\text{−}}]$

Check your learning

Write the ionic equation for the dissolution and the solubility product for each of the following slightly soluble compounds:

(a) BaSO ₄

(b) Ag ₂ SO ₄

(c) Al(OH) ₃

(d) Pb(OH)Cl

Got questions? Get instant answers now!

Calculation of K _sp From equilibrium concentrations

We began the chapter with an informal discussion of how the mineral fluorite ( [link] ) is formed. Fluorite, CaF ₂ , is a slightly soluble solid that dissolves according to the equation:

The concentration of Ca ²⁺ in a saturated solution of CaF ₂ is 2.15 $\times$ 10 ^–4 M ; therefore, that of F ^– is 4.30 $\times$ 10 ^–4 M , that is, twice the concentration of Ca ²⁺ . What is the solubility product of fluorite?

Solution

First, write out the K _sp expression, then substitute in concentrations and solve for K _sp :

A saturated solution is a solution at equilibrium with the solid. Thus:

As with other equilibrium constants, we do not include units with K _sp .

Check your learning

In a saturated solution that is in contact with solid Mg(OH) ₂ , the concentration of Mg ²⁺ is 1.31 $\times$ 10 ^–4 M . What is the solubility product for Mg(OH) ₂ ?

Got questions? Get instant answers now!