Corrected exercises: solubility and the solubility product (202-SN2-RE)

$AgCl(s)\rightleftharpoons Ag^{+}+Cl^{-}$, so $K_{sp}=\left[Ag^{+}\right]\left[Cl^{-}\right]$. $PbI_{2}(s)\rightleftharpoons Pb^{2+}+2I^{-}$, so $K_{sp}=\left[Pb^{2+}\right]\left[I^{-}\right]^{2}$. $Ca_{3}(PO_{4})_{2}(s)\rightleftharpoons 3Ca^{2+}+2PO_{4}^{3-}$, so $K_{sp}=\left[Ca^{2+}\right]^{3}\left[PO_{4}^{3-}\right]^{2}$. The solid never appears in the expression: its activity is 1.

At saturation, $\left[Ag^{+}\right]=\left[Cl^{-}\right]=s$, so $K_{sp}=s^{2}$ and $s=\sqrt{1.8\times 10^{-10}}=1.34\times 10^{-5}$ mol/L. In grams: $1.34\times 10^{-5}\times 143.3=1.92\times 10^{-3}$ g/L, about 1.9 mg per litre. Silver chloride is indeed an “insoluble” salt in the sense of the solubility tables, but it is not strictly insoluble.

For $PbI_{2}$: $\left[Pb^{2+}\right]=s$ and $\left[I^{-}\right]=2s$, so $K_{sp}=s(2s)^{2}=4s^{3}$. Then $s=\sqrt[3]{\dfrac{7.1\times 10^{-9}}{4}}=1.21\times 10^{-3}$ mol/L. For $Ca_{3}(PO_{4})_{2}$: $\left[Ca^{2+}\right]=3s$ and $\left[PO_{4}^{3-}\right]=2s$, so $K_{sp}=(3s)^{3}(2s)^{2}=27s^{3}\cdot 4s^{2}=108s^{5}$. Then $s=\sqrt[5]{\dfrac{2.0\times 10^{-29}}{108}}=7.14\times 10^{-7}$ mol/L. Watch the coefficient: forgetting the 108 changes the result by an order of magnitude.

$Ag_{2}CrO_{4}(s)\rightleftharpoons 2Ag^{+}+CrO_{4}^{2-}$, so $K_{sp}=(2s)^{2}(s)=4s^{3}$ and $s=\sqrt[3]{\dfrac{1.1\times 10^{-12}}{4}}=6.50\times 10^{-5}$ mol/L. Now $AgCl$ only reaches $1.34\times 10^{-5}$ mol/L: silver chromate is almost FIVE TIMES MORE soluble, even though its $K_{sp}$ is 160 times smaller. The reason is stoichiometry: $Ag_{2}CrO_{4}$ releases three ions per formula unit dissolved, so $K_{sp}$ varies as $s^{3}$ and not as $s^{2}$, which squashes the numerical value of the constant. RULE: two $K_{sp}$ values can be compared directly only if the two salts have the same general formula.

The common ion is $Cl^{-}$, held at 0.10 mol/L by the $NaCl$, which is completely dissociated. At saturation, $\left[Ag^{+}\right]=s$ and $\left[Cl^{-}\right]\approx 0.10$. Then $K_{sp}=s\times 0.10$, so $s=\dfrac{1.8\times 10^{-10}}{0.10}=1.8\times 10^{-9}$ mol/L. In pure water, $s$ was $1.34\times 10^{-5}$ mol/L: the solubility is divided by about 7500.

The equilibrium is $AgCl(s)\rightleftharpoons Ag^{+}+Cl^{-}$. Adding $Cl^{-}$ raises the concentration of a product; the system responds by consuming that product, that is, by shifting the equilibrium to the LEFT, toward the formation of solid. Less salt dissolves, so the solubility decreases. Note that $K_{sp}$ itself has not changed: only the split between the two ions has.

In pure water: $CaF_{2}(s)\rightleftharpoons Ca^{2+}+2F^{-}$, so $K_{sp}=4s^{3}$ and $s=\sqrt[3]{\dfrac{3.9\times 10^{-11}}{4}}=2.14\times 10^{-4}$ mol/L. In 0.050 mol/L $Ca(NO_{3})_{2}$, the common ion is $Ca^{2+}$: $\left[Ca^{2+}\right]\approx 0.050$ and $\left[F^{-}\right]=2s$. Then $K_{sp}=0.050\times(2s)^{2}=0.20s^{2}$, so $s=\sqrt{\dfrac{3.9\times 10^{-11}}{0.20}}=1.40\times 10^{-5}$ mol/L: about 15 times lower.

The dissolution contributes $s=1.8\times 10^{-9}$ mol/L of $Cl^{-}$, to be compared with the 0.10 mol/L already present. The contribution is $1.8\times 10^{-6}$% of the total: completely negligible next to the precision of the data. The approximation “$\left[\text{common ion}\right]\approx$ concentration added” is therefore legitimate, and it is legitimate whenever $s$ is small compared with the imposed concentration. It would stop being so if the common ion were very dilute, in which case the full equation would have to be solved.

The total volume is $100+100=200$ mL, so each solution is diluted by a factor of 2. $\left[Pb^{2+}\right]=\dfrac{0.010\times 0.100}{0.200}=5.0\times 10^{-3}$ mol/L, and likewise $\left[Cl^{-}\right]=5.0\times 10^{-3}$ mol/L. The spectator ions $NO_{3}^{-}$ and $Na^{+}$ play no part.

$Q=\left[Pb^{2+}\right]\left[Cl^{-}\right]^{2}=5.0\times 10^{-3}\times\left(5.0\times 10^{-3}\right)^{2}=1.25\times 10^{-7}$. Since $Q=1.25\times 10^{-7}<K_{sp}=1.7\times 10^{-5}$, the solution is UNSATURATED: there is no precipitate, everything stays dissolved.

With 0.20 mol/L, after dilution by 2: $\left[Pb^{2+}\right]=\left[Cl^{-}\right]=0.10$ mol/L. Then $Q=0.10\times(0.10)^{2}=1.0\times 10^{-3}$. This time $Q>K_{sp}$: $PbCl_{2}$ PRECIPITATES, and it keeps doing so until the product of the concentrations falls back to $1.7\times 10^{-5}$.

Ignoring the dilution, the student computes $Q=0.010\times(0.010)^{2}=1.0\times 10^{-6}$, eight times the true value. Here the conclusion does NOT change, because $1.0\times 10^{-6}$ is still below $K_{sp}$: the error goes unnoticed. That is precisely what makes it dangerous. With an exponent of 2 on $\left[Cl^{-}\right]$, forgetting a dilution by 2 distorts $Q$ by a factor of $2^{3}=8$, and it only takes a true $Q$ slightly below $K_{sp}$ for the conclusion to flip. RULE: recalculate every concentration on the TOTAL volume before writing $Q$.

Precipitation begins when $Q$ reaches $K_{sp}$. For $AgCl$: $\left[Ag^{+}\right]\left[Cl^{-}\right]=K_{sp}$, so $\left[Ag^{+}\right]=\dfrac{1.8\times 10^{-10}}{0.010}=1.8\times 10^{-8}$ mol/L.

For $Ag_{2}CrO_{4}$: $\left[Ag^{+}\right]^{2}\left[CrO_{4}^{2-}\right]=K_{sp}$, so $\left[Ag^{+}\right]^{2}=\dfrac{1.1\times 10^{-12}}{0.010}=1.1\times 10^{-10}$ and $\left[Ag^{+}\right]=\sqrt{1.1\times 10^{-10}}=1.05\times 10^{-5}$ mol/L. The exponent 2 is the only difficulty: it is the one that gets forgotten, and it changes the result by several orders of magnitude.

$AgCl$ needs $1.8\times 10^{-8}$ mol/L, $Ag_{2}CrO_{4}$ needs $1.05\times 10^{-5}$ mol/L, nearly 600 times more. So $AgCl$, silver chloride, precipitates FIRST, even though it has the larger $K_{sp}$ of the two: once again, stoichiometry decides.

The second salt starts to precipitate when $\left[Ag^{+}\right]=1.05\times 10^{-5}$ mol/L. At that moment, the remaining chloride still satisfies the $AgCl$ equilibrium: $\left[Cl^{-}\right]=\dfrac{K_{sp}}{\left[Ag^{+}\right]}=\dfrac{1.8\times 10^{-10}}{1.05\times 10^{-5}}=1.72\times 10^{-5}$ mol/L. Relative to the starting 0.010 mol/L, $0.17$% of the chloride remains, so more than 99.8% was precipitated before the chromate began. The separation is excellent. This is the principle of the Mohr method: the chromate serves as the indicator, and the appearance of the red $Ag_{2}CrO_{4}$ precipitate signals that all the chloride has been consumed.

$s=\frac{2.40\times 10^{-3}}{233.40}=1.028\times 10^{-5}$ mol/L. The dissolution $BaSO_{4}(s)\rightleftharpoons Ba^{2+}+SO_{4}^{2-}$ gives one ion of each kind per formula unit dissolved, so $K_{sp}=s^{2}=(1.028\times 10^{-5})^{2}=1.06\times 10^{-10}$.

$s=\frac{2.16\times 10^{-2}}{331.74}=6.51\times 10^{-5}$ mol/L. Here $Ag_{2}CrO_{4}(s)\rightleftharpoons 2Ag^{+}+CrO_{4}^{2-}$: each formula unit dissolved releases TWO silver ions, so $[Ag^{+}]=2s$ and $[CrO_{4}^{2-}]=s$. $K_{sp}=(2s)^{2}(s)=4s^{3}=4(6.51\times 10^{-5})^{3}=1.10\times 10^{-12}$. Forgetting the coefficient 2 (in the concentration AND in the exponent) is the most common error of the chapter: here it throws the result off by a factor of 4.

If heating dissolves more, the dissolution equilibrium shifts to the right when heat is supplied: by Le Chatelier, the dissolution consumes heat, so it is ENDOTHERMIC, $\Delta H_{diss}>0$. And since the equilibrium composition really changes, it is the CONSTANT itself that has changed: $K_{sp}$ increases with temperature. This is the only disturbance that alters a $K_{sp}$; all the others (common ion, pH, complexing agent) shift the equilibrium without touching the constant.

Nothing changes: the ion concentrations remain those of saturation. The expression $K_{sp}=[Ba^{2+}][SO_{4}^{2-}]$ contains no term for the solid, because the activity of a pure solid is 1 whatever the amount present. The excess solid is only a reservoir: it guarantees saturation, it does not strengthen it. One spoonful or a kilogram gives the same solution.

The free barium ion concentration is that of saturation: $1.03\times 10^{-5}$ mol/L, about 1.4 mg/L, a tiny amount, and above all CAPPED by $K_{sp}$ whatever the mass swallowed. The patient swallows hundreds of grams of solid, but the equilibrium forbids more than a few milligrams per litre from going into solution: the poison stays locked in a solid that the body does not dissolve. The safety principle rests entirely on the smallness of $K_{sp}$, which is why the mix-up in a pharmacy between barium sulfate and barium CARBONATE, far more soluble and soluble in acid on top of that, has killed before.

$Mg(OH)_{2}(s)\rightleftharpoons Mg^{2+}+2OH^{-}$, so $K_{sp}=s(2s)^{2}=4s^{3}$ and $s=\sqrt[3]{\dfrac{5.6\times 10^{-12}}{4}}=1.12\times 10^{-4}$ mol/L. Then $\left[OH^{-}\right]=2s=2.24\times 10^{-4}$ mol/L, so $pOH=-\log\left(2.24\times 10^{-4}\right)=3.65$ and $pH=14-3.65=10.35$. A saturated suspension is therefore clearly basic.

At pH 9.00, the buffer imposes $pOH=5.00$, so $\left[OH^{-}\right]=1.0\times 10^{-5}$ mol/L, a value FIXED independently of the dissolution. Then $s=\left[Mg^{2+}\right]=\dfrac{K_{sp}}{\left[OH^{-}\right]^{2}}=\dfrac{5.6\times 10^{-12}}{\left(1.0\times 10^{-5}\right)^{2}}=5.6\times 10^{-2}$ mol/L. The solubility is multiplied by about 500. Lowering the pH from 10.35 to 9.00 is therefore enough to take magnesium from traces to a sizeable concentration. On a graph of $\log s$ against pH, the line has a slope of $-2$: the exponent 2 carried by $\left[OH^{-}\right]$ amplifies everything, and a single pH unit divides the solubility by 100.

Two coupled equilibria are at work: $CaF_{2}(s)\rightleftharpoons Ca^{2+}+2F^{-}$ and $F^{-}+H^{+}\rightleftharpoons HF$. In acidic medium, $H^{+}$ consumes the $F^{-}$ ions to form $HF$, which is a WEAK acid and therefore barely dissociated. The disappearance of $F^{-}$ shifts the first equilibrium to the right, and more $CaF_{2}$ dissolves. This is Le Chatelier again, this time by removing a product.

No. The anion is $Cl^{-}$, the conjugate base of $HCl$, which is a STRONG acid: $Cl^{-}$ therefore has no tendency to pick up $H^{+}$, and no second equilibrium removes chloride ions. The solubility of $AgCl$ is practically insensitive to pH. RULE: the pH effect exists only if the anion of the salt is an appreciable base, that is, if it is the conjugate base of a weak acid.

In the bottle, the water is nearly neutral and $Mg(OH)_{2}$ is very sparingly soluble there: the product stays a stable white suspension, with no marked metallic taste and no risk of overdose. In the stomach, the pH drops to about 1 to 2: the $H^{+}$ ions neutralize the $OH^{-}$ according to $H^{+}+OH^{-}\rightarrow H_{2}O$, which removes a product from the dissolution equilibrium and makes the solid dissolve as it goes. The solid thus acts as a RESERVE that is released only in the presence of acid, and that stops by itself once the acid is neutralized. This is exactly the behaviour wanted from an antacid: self-regulated by pH.

$Ca^{2+}+CO_{3}^{2-}\rightarrow CaCO_{3}(s)$, with the dissolution equilibrium written $CaCO_{3}(s)\rightleftharpoons Ca^{2+}+CO_{3}^{2-}$ and $K_{sp}=\left[Ca^{2+}\right]\left[CO_{3}^{2-}\right]$.

Precipitation begins when $Q$ reaches $K_{sp}$: $\left[CO_{3}^{2-}\right]=\dfrac{K_{sp}}{\left[Ca^{2+}\right]}=\dfrac{4.5\times 10^{-9}}{2.0\times 10^{-3}}=2.25\times 10^{-6}$ mol/L. A minute amount is enough to start the deposit, which explains why limescale forms spontaneously in natural waters loaded with carbonate.

At final equilibrium, the product must still equal $K_{sp}$, but with the calcium lowered: $\left[CO_{3}^{2-}\right]=\dfrac{4.5\times 10^{-9}}{5.0\times 10^{-5}}=9.0\times 10^{-5}$ mol/L. About 40 times more carbonate is therefore needed than merely to start the precipitation: the last of the calcium is always the most expensive to remove, and this is a general property of solubility equilibria.

The calcium removed is $2.0\times 10^{-3}-5.0\times 10^{-5}=1.95\times 10^{-3}$ mol/L, that is $\dfrac{1.95\times 10^{-3}}{2.0\times 10^{-3}}=97.5$% of the initial amount. The process is therefore very efficient, but it can never reach 100%: as long as solid remains in contact with the water, the product of the concentrations stays equal to $K_{sp}$, and a trace of calcium necessarily remains in solution.

Dissolution: $AgCl(s)\rightleftharpoons Ag^{+}+Cl^{-}$ ($K_{sp}$). Complexation: $Ag^{+}+2NH_{3}\rightleftharpoons Ag(NH_{3})_{2}^{+}$ ($K_{f}$). Sum: $AgCl(s)+2NH_{3}\rightleftharpoons Ag(NH_{3})_{2}^{+}+Cl^{-}$. Adding two reactions multiplies their constants, so $K=K_{sp}K_{f}$.