AP Chemistry 7.13-7.14: pH and Solubility, Free Energy of Dissolution

A compound $AB(s)$ has a molar solubility $s$ and dissociates into $A^+$ and $B^-$. What is its $K_{sp}$?

Because $[A^+]=[B^-]=s$, $K_{sp}=s^2$.

Reaction quotient for a solubility equilibrium, $Q_{sp}$

$Q_{sp}$ has the same concentration-based form as $K_{sp}$, but it uses the ion concentrations at a particular moment rather than necessarily equilibrium concentrations.

How does comparing $Q_{sp}$ with $K_{sp}$ 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 $Q_{sp}>K_{sp}$, precipitation occurs until the ion product decreases to $K_{sp}$.

Why must concentrations be adjusted when equal volumes of two ionic solutions are mixed?

The total volume doubles, so each solute concentration is reduced by a factor of two: $[\text{ion}]_{mix}=\frac{M_iV_i}{V_{total}}$. These post-mixing concentrations must be used to calculate $Q_{sp}$.

Equal volumes of $2.0\times10^{-4}$ M $AgNO_3$ and $2.0\times10^{-4}$ M $NaCl$ are mixed. Will $AgCl$ precipitate if $K_{sp}=1.6\times10^{-10}$?

After mixing, $[Ag^+]=[Cl^-]=1.0\times10^{-4}$ M, so $Q_{sp}=1.0\times10^{-8}$. Since $Q_{sp}>K_{sp}$, $AgCl$ precipitates.

At what condition does precipitation of a sparingly soluble salt begin?

Precipitation begins when the ion concentrations reach $Q_{sp}=K_{sp}$. The concentration of the ion being added at this point is found by solving the $K_{sp}$ expression with the other ion concentrations known.

A solution contains $[Ca^{2+}]=2.2\times10^{-3}$ M and $K_{sp}(CaC_2O_4)=1.96\times10^{-9}$. What oxalate concentration initiates precipitation?

At the onset, $K_{sp}=[Ca^{2+}][C_2O_4^{2-}]$. Therefore, $[C_2O_4^{2-}]=\frac{1.96\times10^{-9}}{2.2\times10^{-3}}=8.9\times10^{-7}$ M.

How can $K_{sp}$ be used to calculate the pH required to reduce a metal-ion concentration by precipitating its hydroxide?

Use $K_{sp}=[M^{n+}][OH^-]^x$ to solve for the required $[OH^-]$, then calculate $pOH=-\log[OH^-]$ and $pH=14.00-pOH$ at 25 °C.

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

$[OH^-]=\sqrt{K_{sp}/[Mn^{2+}]}=3.3\times10^{-4}$ M, giving $pOH=3.48$ and $pH=10.52$.

How does pH affect the solubility of salts containing basic anions?

Lowering pH increases solubility when the anion is basic, because $H^+$ reacts with the anion and removes it from the dissolution equilibrium. The equilibrium shifts right to replace the consumed anion, causing more solid to dissolve.

How is the molar solubility of $Ca(OH)_2$ related to its $K_{sp}$?

For $Ca(OH)_2(s)\rightleftharpoons Ca^{2+}+2OH^-$, $[Ca^{2+}]=s$ and $[OH^-]=2s$, so $K_{sp}=4s^3$ and $s=(K_{sp}/4)^{1/3}$.

How can a solubility in grams per liter be converted into the ion concentrations needed for a $K_{sp}$ calculation?

Convert using $s\,(mol/L)=\frac{\text{solubility }(g/L)}{\text{molar mass }(g/mol)}$, then use the dissolution stoichiometry to determine each ion concentration before substituting into the $K_{sp}$ expression.

Solubility equilibrium

A dynamic equilibrium in which a sparingly soluble solid dissolves and dissolved ions precipitate at equal rates. At equilibrium, the concentrations of dissolved species remain constant while both processes continue.

Saturated solution

A solution that contains the maximum amount of dissolved solute possible at a given temperature and is in equilibrium with excess undissolved solute.

How is the solubility-product constant, $K_{sp}$, defined for an ionic solid?

For a dissolution reaction, $K_{sp}$ is the equilibrium constant written using the molar concentrations of dissolved ions; the undissolved solid is omitted. Each ion concentration is raised to the power of its stoichiometric coefficient.

Writing a $K_{sp}$ expression for $M_pX_q(s)$

For $M_pX_q(s)\rightleftharpoons pM^{m+}(aq)+qX^{n-}(aq)$, the expression is $K_{sp}=[M^{m+}]^p[X^{n-}]^q$.

Why does an undissolved ionic solid not appear in a $K_{sp}$ expression?

Pure solids have constant activity, so their concentration is not included in equilibrium-constant expressions. Only aqueous ions and gases appear.

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 $Ca_5(PO_4)_3OH$.

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

How does molar solubility differ from mass solubility?

Molar solubility is the moles of compound that dissolve per liter of saturated solution, in $mol\,L^{-1}$ or M. Mass solubility is typically expressed in grams per liter and must be converted using molar mass before applying $K_{sp}$.

For $CaF_2(s)\rightleftharpoons Ca^{2+}+2F^-$, how are ion concentrations related to molar solubility $s$?

$[Ca^{2+}]=s$ and $[F^-]=2s$. Therefore, $K_{sp}=s(2s)^2=4s^3$.

For $M_pX_q(s)\rightleftharpoons pM+qX$, how are equilibrium ion concentrations related to molar solubility?

If no other reactions affect the ions, $[M]=ps$ and $[X]=qs$, where $s$ is the molar solubility. Thus, $K_{sp}=(ps)^p(qs)^q$ when the initial ion concentrations are negligible.

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

The stoichiometry gives $[F^-]=2(2.15\times10^{-4})=4.30\times10^{-4}$ M. Thus, $K_{sp}=[Ca^{2+}][F^-]^2=3.98\times10^{-11}$.

How is the molar solubility of a 1:1 salt such as $CuBr$ determined from $K_{sp}$?

For $CuBr(s)\rightleftharpoons Cu^++Br^-$, let each ion concentration be $s$. Then $K_{sp}=s^2$, so $s=\sqrt{K_{sp}}$; for $K_{sp}=6.3\times10^{-9}$, $s=7.9\times10^{-5}$ M.

Why are salts containing $OH^-$, $CO_3^{2-}$, $PO_4^{3-}$, or other basic anions generally more soluble in acidic solution?

Acid protonates the basic anion, forming species such as $H_2O$, $HCO_3^-$, $H_2CO_3$, or phosphate protonation products. Because the free anion concentration decreases, the ion product falls below $K_{sp}$ and additional solid dissolves.

How does increasing pH affect the solubility of a metal hydroxide?

Increasing pH increases $[OH^-]$, a common ion for the hydroxide dissolution equilibrium. The equilibrium shifts toward the solid, so the hydroxide becomes less soluble and more precipitates.

Selective precipitation

A separation method in which a reagent ion is added gradually so that different ions precipitate as their respective $K_{sp}$ conditions are reached. For target ions at equal concentrations, the salt with the smaller $K_{sp}$ generally precipitates first, but unequal ion concentrations can reverse that order.

How is the order of precipitation determined when $Ag^+$ is gradually added to a solution containing $Cl^-$ and $Br^-$?

Calculate the silver-ion concentration required for each salt to reach saturation: $[Ag^+]_{threshold}=K_{sp}/[X^-]$. The halide whose salt reaches $Q_{sp}=K_{sp}$ at the lower $[Ag^+]$ precipitates first.

A solution contains $0.10$ M $Cl^-$ and $1.0\times10^{-4}$ M $Br^-$. Given $K_{sp}(AgCl)=1.6\times10^{-10}$ and $K_{sp}(AgBr)=5.0\times10^{-13}$, which precipitates first?

For AgCl, precipitation begins at $[Ag^+]=1.6\times10^{-9}$ M. For AgBr, it begins at $5.0\times10^{-9}$ M, so AgCl precipitates first despite having the larger $K_{sp}$ because chloride is much more concentrated.

Common-ion effect on solubility

The decrease in the solubility of an ionic compound when a solution already contains one of the ions produced by its dissolution. The added common ion shifts the dissolution equilibrium toward the solid and reduces the amount that dissolves.

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

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

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

The added $OH^-$ is a common ion, so more $Mg(OH)_2$ forms. The equilibrium concentration of $Mg^{2+}$ decreases, while the hydroxide concentration increases because of the added KOH.

What happens when an inert soluble electrolyte such as $NaNO_3$ is added to a saturated $Mg(OH)_2$ solution?

Neither ion is part of the $Mg(OH)_2$ dissolution equilibrium, so no appreciable shift or change in solubility is expected under the ideal-concentration treatment used for AP Chemistry.

What happens when additional solid is added to a saturated solution already at solubility equilibrium?

The quantity of undissolved solid increases, but the equilibrium concentrations of the dissolved ions do not change, provided the solution remains saturated and the added solid does not introduce a common ion.

How does adding a common ion affect $Q_{sp}$ immediately?

Adding a common ion increases the ion product, making $Q_{sp}>K_{sp}$ momentarily. The resulting precipitation removes dissolved ions until $Q_{sp}$ again equals $K_{sp}$.

How can precipitation remove phosphate from wastewater?

Adding lime supplies $Ca^{2+}$ and makes the solution basic, allowing hydroxylapatite to precipitate: $5Ca^{2+}+3PO_4^{3-}+OH^-\rightleftharpoons Ca_5(PO_4)_3OH(s)$. The solid can then be filtered out; the water may be returned toward neutral pH using $CO_2$.

What is the thermodynamic relationship between standard free energy and $K_{sp}$?

For the dissolution reaction, $\Delta G^\circ=-RT\ln K_{sp}$. A larger $K_{sp}$ corresponds to a more negative $\Delta G^\circ$ and a more thermodynamically favorable standard-state dissolution.

What does the sign of $\Delta G^\circ$ imply about $K_{sp}$?

If $K_{sp}>1$, then $\Delta G^\circ<0$; if $K_{sp}<1$, then $\Delta G^\circ>0$; and if $K_{sp}=1$, then $\Delta G^\circ=0$. A small $K_{sp}$ indicates that dissolution is not favored under standard-state conditions.

How are the actual free energy change and the solubility reaction quotient related?

For a dissolution reaction, $\Delta G=\Delta G^\circ+RT\ln Q_{sp}$. At equilibrium, $Q_{sp}=K_{sp}$ and $\Delta G=0$. If $Q_{sp}<K_{sp}$, dissolution is favored; if $Q_{sp}>K_{sp}$, precipitation is favored.