AP Chemistry 7.11-7.12: Solubility Equilibria and Common-Ion Effect

Solubility equilibrium

A dynamic equilibrium in which a sparingly soluble solid dissolves and dissolved ions precipitate at equal rates. For example, $AgCl(s) \rightleftharpoons Ag^+(aq)+Cl^-(aq)$.

What conditions produce a saturated solution for a sparingly soluble ionic compound?

A saturated solution is in equilibrium with excess undissolved solid. At equilibrium, the rates of dissolution and precipitation are equal, so the dissolved-ion concentrations remain constant.

Solubility product constant, $K_{sp}$

The equilibrium constant for the dissolution of a sparingly soluble ionic solid. Only aqueous ions appear in the expression; the undissolved solid is omitted.

How is the $K_{sp}$ expression written for a dissolution reaction?

For $M_pX_q(s)\rightleftharpoons pM^{m+}(aq)+qX^{n-}(aq)$, $K_{sp}=[M^{m+}]^p[X^{n-}]^q$. The exponents are the stoichiometric coefficients of the dissolved ions.

Write the dissolution equation and $K_{sp}$ expression for $Mg(OH)_2$.

$Mg(OH)_2(s)\rightleftharpoons Mg^{2+}(aq)+2OH^-(aq)$; $K_{sp}=[Mg^{2+}][OH^-]^2$.

Write the dissolution equation and $K_{sp}$ expression for $Ag_2SO_4$.

$Ag_2SO_4(s)\rightleftharpoons 2Ag^+(aq)+SO_4^{2-}(aq)$; $K_{sp}=[Ag^+]^2[SO_4^{2-}]$.

Write the dissolution equation and $K_{sp}$ expression for hydroxylapatite, $Ca_5(PO_4)_3OH$.

$Ca_5(PO_4)_3OH(s)\rightleftharpoons 5Ca^{2+}(aq)+3PO_4^{3-}(aq)+OH^-(aq)$; $K_{sp}=[Ca^{2+}]^5[PO_4^{3-}]^3[OH^-]$.

What is molar solubility, and how does it relate to ion concentrations?

Molar solubility is the number of moles of compound that dissolve per liter of saturated solution. Ion concentrations are determined by multiplying the molar solubility by each ion's stoichiometric coefficient.

How can $K_{sp}$ be calculated when the molar solubility of $CaF_2$ is $s$?

From $CaF_2(s)\rightleftharpoons Ca^{2+}+2F^-$, $[Ca^{2+}]=s$ and $[F^-]=2s$, so $K_{sp}=s(2s)^2=4s^3$.

How can the molar solubility of a 1:1 salt such as $AgI$ be found from $K_{sp}$?

For $AgI(s)\rightleftharpoons Ag^++I^-$, if the molar solubility is $s$, then $K_{sp}=s^2$ and $s=\sqrt{K_{sp}}$, assuming no other source of either ion is present.

How can the molar solubility of $Ca(OH)_2$ be calculated from its $K_{sp}$?

Let the molar solubility be $s$. Because $[Ca^{2+}]=s$ and $[OH^-]=2s$, $K_{sp}=s(2s)^2=4s^3$, so $s=\sqrt[3]{K_{sp}/4}$.

A saturated $CaF_2$ solution has $[Ca^{2+}]=2.15\times10^{-4}\,M$. What is $K_{sp}$?

The fluoride concentration is $[F^-]=2(2.15\times10^{-4})=4.30\times10^{-4}\,M$. Therefore, $K_{sp}=[Ca^{2+}][F^-]^2=3.98\times10^{-11}$.

How is a solubility given in grams per liter converted for use in a $K_{sp}$ calculation?

Convert grams per liter to moles per liter using molar mass: $s\,(M)=\text{g/L}\times(1\,\text{mol}/\text{molar mass})$. Then use the dissolution stoichiometry to determine ion concentrations and substitute them into the $K_{sp}$ expression.

A solid has a very small $K_{sp}$. Does that always mean its molar solubility is smaller than that of another solid with a larger $K_{sp}$?

Not necessarily. The relationship between $K_{sp}$ and molar solubility depends on dissolution stoichiometry; different ion coefficients produce different powers of $s$ in the $K_{sp}$ expression.

Ion product, $Q_{sp}$

The reaction quotient calculated using the current aqueous-ion concentrations for a dissolution equilibrium. It has the same form as the $K_{sp}$ expression but describes a solution that may not yet be at equilibrium.

How does comparison of $Q_{sp}$ and $K_{sp}$ predict precipitation?

If $Q_{sp}<K_{sp}$, the solution is unsaturated and no precipitation occurs. If $Q_{sp}=K_{sp}$, the solution is saturated and at equilibrium. If $Q_{sp}>K_{sp}$, the solution is supersaturated and precipitation occurs until $Q_{sp}$ returns to $K_{sp}$.

Why must concentrations be recalculated after mixing two solutions before evaluating $Q_{sp}$?

Mixing changes the total volume and therefore dilutes each solute. For each ion, use $M_1V_1=M_2V_2$ or calculate moles divided by the combined volume before substituting into $Q_{sp}$.

Equal volumes of $2.0\times10^{-4}\,M$ $AgNO_3$ and $2.0\times10^{-4}\,M$ $NaCl$ are mixed. Given $K_{sp}(AgCl)=1.6\times10^{-10}$, will precipitation occur?

Each ion is diluted to $1.0\times10^{-4}\,M$. Thus $Q_{sp}=(1.0\times10^{-4})(1.0\times10^{-4})=1.0\times10^{-8}>K_{sp}$, so $AgCl$ precipitates.

What ion concentration marks the beginning of precipitation?

At the onset of precipitation, the solution is just saturated, so $Q_{sp}=K_{sp}$. For a salt producing ions in a 1:1 ratio, the threshold concentration of one ion is found by setting the product of the two ion concentrations equal to $K_{sp}$.

For $CaC_2O_4(s)\rightleftharpoons Ca^{2+}+C_2O_4^{2-}$ with $K_{sp}=1.96\times10^{-9}$, what oxalate concentration initiates precipitation when $[Ca^{2+}]=2.2\times10^{-3}\,M$?

Set $Q_{sp}=K_{sp}$: $[C_2O_4^{2-}]=\frac{1.96\times10^{-9}}{2.2\times10^{-3}}=8.9\times10^{-7}\,M$.

Selective precipitation

A separation method in which a precipitating ion is added gradually so that different dissolved ions form precipitates at different concentrations of the precipitating ion. The compound that reaches $Q_{sp}=K_{sp}$ first precipitates first.

How can the order of precipitation of two salts with a common precipitating ion be determined?

Calculate the concentration of the added ion required for each salt to begin precipitating by setting $Q_{sp}=K_{sp}$. The salt requiring the lower concentration of the added ion precipitates first; initial concentrations of the other ions must be included.

A solution contains $0.00010\,M$ $Br^-$ and $0.10\,M$ $Cl^-$. Given $K_{sp}(AgBr)=5.0\times10^{-13}$ and $K_{sp}(AgCl)=1.6\times10^{-10}$, which precipitates first as $Ag^+$ is added?

$AgCl$ begins at $[Ag^+]=\frac{1.6\times10^{-10}}{0.10}=1.6\times10^{-9}\,M$, while $AgBr$ begins at $\frac{5.0\times10^{-13}}{0.00010}=5.0\times10^{-9}\,M$. Therefore, $AgCl$ precipitates first despite its larger $K_{sp}$.

Common-ion effect on solubility

The decrease in the solubility of an ionic compound caused by adding a soluble substance that supplies one of the compound's ions. The added ion shifts the dissolution equilibrium toward the solid and increases precipitation.

What happens when $MgCl_2$ is added to a saturated solution of $Mg(OH)_2$?

The added $Mg^{2+}$ is a common ion, so the equilibrium $Mg(OH)_2(s)\rightleftharpoons Mg^{2+}+2OH^-$ shifts left. More solid forms, $[OH^-]$ decreases, and the equilibrium $[Mg^{2+}][OH^-]^2$ remains equal to $K_{sp}$.

What happens when $KOH$ is added to a saturated solution of $Mg(OH)_2$?

The added $OH^-$ is a common ion and shifts the equilibrium left. More $Mg(OH)_2$ precipitates, and the equilibrium concentration of $Mg^{2+}$ decreases.

What is the effect of adding an inert soluble electrolyte, such as $NaNO_3$, to a saturated solution of $Mg(OH)_2$?

Because $Na^+$ and $NO_3^-$ do not appear in the $Mg(OH)_2$ dissolution equilibrium, they do not cause a common-ion shift. In the ideal AP Chemistry treatment, there is no appreciable change in solubility or equilibrium ion concentrations.

What is the effect of adding more solid solute to a saturated solution?

The amount of undissolved solid increases, but the dissolved-ion concentrations do not change as long as the solution was already saturated and temperature remains constant. The solid does not appear in the $K_{sp}$ expression.

How does a common ion change the calculation of molar solubility?

Use the initial common-ion concentration plus the change from dissolution in an ICE setup. For example, for $CdS(s)\rightleftharpoons Cd^{2+}+S^{2-}$ in $0.010\,M$ $Cd^{2+}$, $K_{sp}=(0.010+s)(s)$; often $s$ is small enough that $0.010+s\approx0.010$.

Why does adding a common ion generally reduce the solubility of a sparingly soluble salt?

The common ion increases the reaction quotient for dissolution, making $Q_{sp}>K_{sp}$. The system responds by shifting toward the solid, lowering the amount that remains dissolved.

How can precipitation be used to remove phosphate from wastewater?

Adding $Ca(OH)_2$ supplies $Ca^{2+}$ and $OH^-$, which can precipitate phosphate as hydroxylapatite: $5Ca^{2+}+3PO_4^{3-}+OH^-\rightleftharpoons Ca_5(PO_4)_3OH(s)$. The precipitate can then be removed by filtration.

How can $K_{sp}$ be used to determine the pH needed to remove a metal ion as a hydroxide precipitate?

Set the desired metal-ion concentration and $Q_{sp}=K_{sp}$ to solve for $[OH^-]$. Then calculate $pOH=-\log[OH^-]$ and, at $25\,^{\circ}\mathrm{C}$, use $pH=14.00-pOH$.

For $Mn(OH)_2$ with $K_{sp}=2.0\times10^{-13}$, what pH is required for $[Mn^{2+}]=1.8\times10^{-6}\,M$ at equilibrium?

From $K_{sp}=[Mn^{2+}][OH^-]^2$, $[OH^-]=\sqrt{(2.0\times10^{-13})/(1.8\times10^{-6})}=3.3\times10^{-4}\,M$. Thus $pOH=3.48$ and $pH=10.52$ at $25\,^{\circ}\mathrm{C}$.