Exercises — Solubility product Ksp — common-ion suppression, selective precipitation

WHAT we do: split each solid into its ions, then multiply the ion concentrations each raised to its own stoichiometric number. WHY: $K_{sp}$ is just the equilibrium constant of dissolution, and the pure solid has activity $1$ so it never appears.

(a) $\text{AgBr(s)}\rightleftharpoons {\text{Ag}}^{+}+{\text{Br}}^{-}$ $K_{sp}=[{\text{Ag}}^{+}][{\text{Br}}^{-}]$ (b) ${\text{PbCl}}_2\text{(s)}\rightleftharpoons {\text{Pb}}^{2+}+2{\text{Cl}}^{-}$ $K_{sp}=[{\text{Pb}}^{2+}][{\text{Cl}}^{-}{]}^2$ (c) ${\text{Ca}}_3({\text{PO}}_4{)}_2\text{(s)}\rightleftharpoons 3{\text{Ca}}^{2+}+2{\text{PO}}_4^{3-}$ $K_{sp}=[{\text{Ca}}^{2+}{]}^3[{\text{PO}}_4^{3-}{]}^2$

Figure 2 shows why the exponents appear: each formula unit shatters into a fixed head-count of ions (one, two, or three of each), and that head-count becomes the power.

Compare the "current reading" $Q$ with the "red line" $K_{sp}$. $Q=5\times 10^{-9}>1.8\times 10^{-10}=K_{sp}$ Since $Q>K_{sp}$, the solution is supersaturated — a precipitate forms until $Q$ falls back to $K_{sp}$.

The three-way rule (all cases):

• $Q<K_{sp}$ ⇒ unsaturated: no solid, more can still dissolve.
• $Q=K_{sp}$ ⇒ saturated equilibrium: the solution sits exactly on the red line. Here there is no net change — solid and ions coexist, dissolving and precipitating at equal rates. Nothing new precipitates and nothing extra dissolves; the system is already at rest.
• $Q>K_{sp}$ ⇒ supersaturated: precipitation runs until $Q$ drops back to $K_{sp}$.

The boundary case $Q=K_{sp}$ is exactly the "just begins" condition we exploit in 3.3 and 4.1.

WHAT: let $s$ = mol/L of AgBr that dissolves. WHY $s$ for both ions: the ratio is $1:1$, so each formula unit releases one ${\text{Ag}}^{+}$ and one ${\text{Br}}^{-}$. $[{\text{Ag}}^{+}]=s,[{\text{Br}}^{-}]=s$ $K_{sp}=s\cdot s=s^2\Rightarrow s=\sqrt{5.0\times 10^{-13}}=7.07\times 10^{-7}\text{mol/L}$

WHAT — set the ion concentrations. Each dissolved ${\text{PbCl}}_2$ gives one ${\text{Pb}}^{2+}$ and two ${\text{Cl}}^{-}$, so if $s$ mol/L dissolve, $[{\text{Pb}}^{2+}]=s$ and $[{\text{Cl}}^{-}]=2s$. WHY $2s$: the balanced equation releases two chlorides per formula unit, so their concentration grows twice as fast as the salt's. $[{\text{Pb}}^{2+}]=s,[{\text{Cl}}^{-}]=2s$ WHAT — substitute into $K_{sp}$; WHY: at saturation the ion product is $K_{sp}$, so plugging the $s$-expressions in turns the equilibrium condition into one equation in $s$. $K_{sp}=s(2s{)}^2=4s^3$ WHAT — solve; WHY the cube root: the unknown appears as $s^3$, so undoing the cube isolates $s$. $s^3=\frac{1.6\times 10^{-5}}{4}=4.0\times 10^{-6}\Rightarrow s=(4.0\times 10^{-6}{)}^{1/3}=1.587\times 10^{-2}\text{mol/L}$

WHAT — read the stoichiometry backwards; WHY: this time we are given $s$ and want $K_{sp}$, so we translate the measured solubility into ion concentrations first. ${\text{Mg(OH)}}_2\rightleftharpoons {\text{Mg}}^{2+}+2{\text{OH}}^{-}$ gives $[{\text{Mg}}^{2+}]=s$ and $[{\text{OH}}^{-}]=2s$ (two hydroxides per formula unit, same reason as 2.2). WHAT — build $K_{sp}$; WHY: $K_{sp}$ is by definition the ion product at saturation, and the measured $s$ is the saturation point, so we just evaluate it. $K_{sp}=s(2s{)}^2=4s^3=4(1.65\times 10^{-4}{)}^3=1.80\times 10^{-11}$

Notation first: we write the new solubility as $s^{\prime }$ (read "$s$-prime") purely to distinguish it from the pure-water value $s$ of Exercise 2.1 — same meaning (mol/L of AgBr that dissolves), just under the new common-ion conditions. The prime is a label, not a new quantity. WHAT: $\text{KBr}$ is fully soluble, so it floods the solution with ${\text{Br}}^{-}$ before any AgBr dissolves. Then: $[{\text{Ag}}^{+}]=s^{\prime },[{\text{Br}}^{-}]=0.10+s^{\prime }$ WHY the approximation: AgBr is so stingy that $s^{\prime }\ll 0.10$, so the ${\text{Br}}^{-}$ it adds is a rounding error on the $0.10\text{M}$ already there — hence $[{\text{Br}}^{-}]\approx 0.10$. $K_{sp}=s^{\prime }(0.10)\Rightarrow s^{\prime }=\frac{5.0\times 10^{-13}}{0.10}=5.0\times 10^{-12}\text{mol/L}$ Check: $5.0\times 10^{-12}\ll 0.10$ ✓. Compared to pure water ($7.07\times 10^{-7}$), solubility drops by a factor of about $1.4\times 10^5$ — the common ion crushes it.

Figure 3 shows this suppression on a bar chart: the pure-water solubility towers over the common-ion solubility, driven down by the flood of shared ${\text{Br}}^{-}$.

WHAT first — dilution. Mixing equal volumes halves every concentration, because each ion now spreads through $200\text{mL}$ instead of $100\text{mL}$. WHY: moles are conserved, volume doubles, so concentration halves. $[{\text{Ag}}^{+}]=\frac{0.010}{2}=5.0\times 10^{-3}\text{M},[{\text{Cl}}^{-}]=5.0\times 10^{-3}\text{M}$ Now compute the current reading $Q$: $Q=[{\text{Ag}}^{+}][{\text{Cl}}^{-}]=(5.0\times 10^{-3}{)}^2=2.5\times 10^{-5}$ $Q=2.5\times 10^{-5}\gg K_{sp}=1.8\times 10^{-10}$ $Q>K_{sp}$ ⇒ yes, AgCl precipitates.

Figure 4 draws the mixing: two $100\text{mL}$ beakers pour into one $200\text{mL}$ beaker, and the concentration bars halve — the visual reason $Q$ must use the diluted numbers.

Precipitation begins the instant $Q=K_{sp}$ (the boundary case from 1.2), so set them equal and solve for the unknown ion. $K_{sp}=[{\text{Mg}}^{2+}][{\text{OH}}^{-}{]}^2\Rightarrow [{\text{OH}}^{-}]=\sqrt{\frac{K_{sp}}{[{\text{Mg}}^{2+}]}}=\sqrt{\frac{1.8\times 10^{-11}}{0.010}}$ $[{\text{OH}}^{-}]=\sqrt{1.8\times 10^{-9}}=4.24\times 10^{-5}\text{M}$ Then $\text{pOH}=-\log (4.24\times 10^{-5})=4.37$, so (see pH and pOH Calculations): $\text{pH}=14-4.37=9.63$

WHAT — find each salt's trigger; WHY: precipitation starts the moment $Q=K_{sp}$ (from 1.2), so for each salt we solve for the $[{\text{Ag}}^{+}]$ that first reaches its own red line. Whichever red line is reached at the lower $[{\text{Ag}}^{+}]$ trips first, because we are raising $[{\text{Ag}}^{+}]$ gradually from zero.

For AgCl (1:1): $[{\text{Ag}}^{+}{]}_{\text{needed}}=\frac{K_{sp}}{[{\text{Cl}}^{-}]}=\frac{1.8\times 10^{-10}}{0.010}=1.8\times 10^{-8}\text{M}$ For ${\text{Ag}}_2{\text{CrO}}_4$ (2:1) — WHY the square: the balanced dissolution releases two ${\text{Ag}}^{+}$ per formula unit, so $[{\text{Ag}}^{+}]$ enters $K_{sp}$ squared, and undoing that needs a square root: $K_{sp}=[{\text{Ag}}^{+}{]}^2[{\text{CrO}}_4^{2-}]\Rightarrow [{\text{Ag}}^{+}{]}_{\text{needed}}=\sqrt{\frac{1.1\times 10^{-12}}{0.010}}=\sqrt{1.1\times 10^{-10}}=1.049\times 10^{-5}\text{M}$ AgCl needs only $1.8\times 10^{-8}\text{M}$ vs $1.049\times 10^{-5}\text{M}$ for the chromate — AgCl precipitates first (~580× sooner). This is the basis of the Mohr titration in Qualitative Inorganic Analysis.

WHAT — fix the $[{\text{Ag}}^{+}]$ at the second salt's trigger; WHY: the chromate begins the instant $[{\text{Ag}}^{+}]=1.049\times 10^{-5}\text{M}$ (from 4.1), and at that same instant the chloride still in solution must obey AgCl's own $K_{sp}$ — because AgCl is already saturated and sitting on its red line. So we read the leftover chloride straight off that equation. $[{\text{Cl}}^{-}{]}_{\text{left}}=\frac{K_{sp}(\text{AgCl})}{[{\text{Ag}}^{+}]}=\frac{1.8\times 10^{-10}}{1.049\times 10^{-5}}=1.716\times 10^{-5}\text{M}$ WHAT — turn into a fraction; WHY: dividing the leftover by the original $0.010\text{M}$ tells us the share of chloride that has not yet precipitated — the measure of how clean the cut is. $\frac{1.716\times 10^{-5}}{0.010}=1.716\times 10^{-3}=0.172\%$ Over $99.8\%$ of the chloride is already down as AgCl before the second salt begins — a clean separation.

Step 1 — dilute; WHY: equal volumes mixed means every species now occupies $100\text{mL}$ instead of $50\text{mL}$, so each concentration halves (moles conserved, volume doubled). $[{\text{Pb}}^{2+}{]}_0=\frac{0.20}{2}=0.10\text{M},[{\text{Cl}}^{-}{]}_0=\frac{0.20}{2}=0.10\text{M}$ Step 2 — test with $Q$; WHY: we compute the current ion product to see if it exceeds the red line (the 1.2 rule). $Q=[{\text{Pb}}^{2+}][{\text{Cl}}^{-}{]}^2=(0.10)(0.10{)}^2=1.0\times 10^{-3}\gg 1.6\times 10^{-5}=K_{sp}$ $Q>K_{sp}$ ⇒ precipitation occurs. (a) shown.

Step 3 — react to completion; WHY the limiting ion: each ${\text{Pb}}^{2+}$ consumes two ${\text{Cl}}^{-}$, so $0.10\text{M}$ chloride can only take down $0.05\text{M}$ lead — chloride runs out first, making it the limiting reagent. After ${\text{Pb}}^{2+}+2{\text{Cl}}^{-}\to {\text{PbCl}}_2$: $[{\text{Pb}}^{2+}{]}_{\text{excess}}=0.10-0.05=0.05\text{M}$ Step 4 — justify $[{\text{Cl}}^{-}]\approx 0$; WHY it is safe: the tiny amount of ${\text{PbCl}}_2$ that redissolves is capped by equilibrium against the large lead excess. With $[{\text{Pb}}^{2+}]\approx 0.05\text{M}$ the redissolved chloride is $[{\text{Cl}}^{-}]=\sqrt{\frac{K_{sp}}{[{\text{Pb}}^{2+}]}}=\sqrt{\frac{1.6\times 10^{-5}}{0.05}}=\sqrt{3.2\times 10^{-4}}=1.79\times 10^{-2}\text{M}$ This $1.79\times 10^{-2}\text{M}$ is much smaller than the $\approx 0.05\text{M}$ of chloride that precipitated as solid, so treating the post-reaction chloride as "nearly all gone" barely disturbs the lead balance — the excess lead stays $\approx 0.05\text{M}$. (b) answer: $\boxed{[{\text{Pb}}^{2+}]\approx 0.05\text{M}}$

Lower edge — Fe removed to $10^{-6}$ M; WHY: we ask what $[{\text{OH}}^{-}]$ pushes iron down to the target $10^{-6}\text{M}$, using Fe's own $K_{sp}$ (three hydroxides ⇒ cube): $[{\text{OH}}^{-}{]}^3=\frac{2.6\times 10^{-39}}{10^{-6}}=2.6\times 10^{-33}\Rightarrow [{\text{OH}}^{-}]=(2.6\times 10^{-33}{)}^{1/3}=1.375\times 10^{-11}\text{M}$ $\text{pOH}=-\log (1.375\times 10^{-11})=10.86\Rightarrow \text{pH}=14-10.86=3.14$ So above pH 3.14 iron is essentially gone.

Upper edge — Mg must NOT start; WHY: we find the $[{\text{OH}}^{-}]$ at which Mg first reaches its red line (the "just begins" condition from 1.2); staying below it keeps Mg dissolved. $[{\text{OH}}^{-}]=\sqrt{\frac{K_{sp}}{[{\text{Mg}}^{2+}]}}=\sqrt{\frac{1.8\times 10^{-11}}{0.010}}=\sqrt{1.8\times 10^{-9}}=4.24\times 10^{-5}\text{M}$ $\text{pOH}=-\log (4.24\times 10^{-5})=4.37\Rightarrow \text{pH}=14-4.37=9.63$ So Mg stays dissolved below pH 9.63.

Window: hold the pH between $\approx 3.1$ and $\approx 9.6$ (e.g. an acetate buffer near pH 5) to drop Fe(OH)₃ cleanly while keeping Mg²⁺ in solution. This is precisely the Group-III hydroxide separation in Qualitative Inorganic Analysis, and the thermodynamic direction is fixed by $\Delta G<0$ (see Gibs Free Energy and Spontaneity).

Why $K_{sp}$ alone is not enough here; WHAT is new: the naive picture says "more ${\text{OH}}^{-}$ ⇒ more precipitate" (common-ion, Le Chatelier). But aluminium has a second channel — the solid can react with the excess base to form a soluble complex. So total dissolved Al is the sum of two routes.

Route 1 — free ion via $K_{sp}$: $[{\text{Al}}^{3+}]=\frac{K_{sp}}{[{\text{OH}}^{-}{]}^3}=\frac{3.0\times 10^{-34}}{(1.0{)}^3}=3.0\times 10^{-34}\text{M}$ Vanishingly small — the common-ion route says aluminium is "gone".

Route 2 — the complex via $K_c$; WHY: the solid dissolving as ${\text{Al(OH)}}_4^{-}$ obeys its own equilibrium, and with the pure solid at activity $1$: $K_c=\frac{[{\text{Al(OH)}}_4^{-}]}{[{\text{OH}}^{-}]}\Rightarrow [{\text{Al(OH)}}_4^{-}]=K_c[{\text{OH}}^{-}]=40\times 1.0=40\text{M (formally)}$ Read this as "the complex route is enormously favoured" — in practice all the solid dissolves long before $40\text{M}$ is approached; the number tells us there is effectively no solid left.

(a) Total dissolved aluminium $\approx 3.0\times 10^{-34}+40\approx 40\text{M}$ formally — dominated entirely by the complex. (b) The ${\text{Al(OH)}}_4^{-}$ term outweighs the free-ion term by ~35 orders of magnitude. Lesson: never wash an amphoteric hydroxide precipitate (${\text{Al}}^{3+}$, ${\text{Zn}}^{2+}$, ${\text{Pb}}^{2+}$, ${\text{Cr}}^{3+}$) with excess strong base — the very ion you tried to precipitate re-dissolves as a complex, and $K_{sp}$ reasoning silently fails because a second equilibrium has opened. This is the amphoteric trap of Group-III analysis in Qualitative Inorganic Analysis.

WHY the ideal picture fails; WHAT changes: ${\text{KNO}}_3$ shares no ion, so common-ion reasoning predicts no effect. But the true $K_{sp}$ is built from activities $a=\gamma [\text{ion}]$, not concentrations. The real constant is $K_{sp}=a_{\text{Ag}}{a}_{\text{Cl}}={\gamma }_{\pm }^2[{\text{Ag}}^{+}][{\text{Cl}}^{-}]$ Solving for the concentration product (what actually dissolves): $[{\text{Ag}}^{+}][{\text{Cl}}^{-}]=\frac{K_{sp}}{{\gamma }_{\pm }^2}=\frac{1.8\times 10^{-10}}{(0.75{)}^2}=3.2\times 10^{-10}$ With $[{\text{Ag}}^{+}]=[{\text{Cl}}^{-}]=s$: $s=\sqrt{3.2\times 10^{-10}}=1.789\times 10^{-5}\text{M}$ Direction and lesson: $s$ rose from $1.34\times 10^{-5}$ to $1.79\times 10^{-5}\text{M}$ — about $33\%$ more soluble, even though we added a salt with no common ion. WHY it rises: the ionic atmosphere of ${\text{K}}^{+}/{\text{NO}}_3^{-}$ shields each ${\text{Ag}}^{+}$ and ${\text{Cl}}^{-}$, lowering their effective (activity) concentration; to keep the true activity product at $K_{sp}$, more solid must dissolve. This is the salting-in / inert-salt effect — the concrete pay-off of the activity caveat opened at the top of the page. More in Ionic Equilibria in Solutions.