Exercises — Common ion effect

(a) ${\text{CaF}}_2\rightleftharpoons {\text{Ca}}^{2+}+2{\text{F}}^{-}$. NaF supplies extra ${\text{F}}^{-}$ — the common ion is ${\text{F}}^{-}$. Equilibrium shifts left, so solubility goes down.

(b) $\text{AgCl}\rightleftharpoons {\text{Ag}}^{+}+{\text{Cl}}^{-}$. ${\text{KNO}}_3$ gives ${\text{K}}^{+}$ and ${\text{NO}}_3^{-}$ — neither appears in the AgCl equilibrium. No common ion, so to a first approximation solubility is unchanged.

(c) ${\text{BaSO}}_4\rightleftharpoons {\text{Ba}}^{2+}+{\text{SO}}_4^{2-}$. ${\text{Na}}_2{\text{SO}}_4$ supplies ${\text{SO}}_4^{2-}$ — common ion is ${\text{SO}}_4^{2-}$. Solubility goes down.

Dissolving: ${\text{Ag}}_2{\text{CrO}}_4\text{(s)}\rightleftharpoons 2{\text{Ag}}^{+}+{\text{CrO}}_4^{2-}$.

$K_{sp}=[{\text{Ag}}^{+}{]}^2[{\text{CrO}}_4^{2-}]$

In pure water, if $s$ mol of salt dissolves per litre it makes $2s$ of ${\text{Ag}}^{+}$ and $s$ of ${\text{CrO}}_4^{2-}$:

$K_{sp}=(2s{)}^2(s)=4s^3⟹s={\left(\frac{K_{sp}}{4}\right)}^{1/3}$

Why the $2$ inside the square? The coefficient $2$ multiplies the concentration ($[{\text{Ag}}^{+}]=2s$) and the exponent is $2$ because the ion appears twice in the product. Both effects are separate; do not merge them.

${\text{CaF}}_2\rightleftharpoons {\text{Ca}}^{2+}+2{\text{F}}^{-}$, so $[{\text{Ca}}^{2+}]=s$, $[{\text{F}}^{-}]=2s$.

$K_{sp}=s(2s{)}^2=4s^3$ $s={\left(\frac{3.9\times 10^{-11}}{4}\right)}^{1/3}={\left(9.75\times 10^{-12}\right)}^{1/3}=2.14\times 10^{-4}\text{M}$

NaF gives $[{\text{F}}^{-}]=0.10\text{M}$ before ${\text{CaF}}_2$ reacts. Let new solubility $=s^{\prime }$. $[{\text{Ca}}^{2+}]=s^{\prime },[{\text{F}}^{-}]=0.10+2s^{\prime }$ Expect $s^{\prime }$ tiny, so $0.10+2s^{\prime }\approx 0.10$: $K_{sp}=s^{\prime }(0.10{)}^2⟹s^{\prime }=\frac{3.9\times 10^{-11}}{0.01}=3.9\times 10^{-9}\text{M}$ Check: $2s^{\prime }=7.8\times 10^{-9}\ll 0.10$ ✓ (well under $5\%$). Solubility fell from $2.14\times 10^{-4}$ to $3.9\times 10^{-9}$ — about a 55 000× drop.

${\text{PbI}}_2\rightleftharpoons {\text{Pb}}^{2+}+2{\text{I}}^{-}$. KI gives $[{\text{I}}^{-}]=0.010$. $[{\text{Pb}}^{2+}]=s^{\prime },[{\text{I}}^{-}]=0.010+2s^{\prime }$ Shortcut (assume $0.010+2s^{\prime }\approx 0.010$): $K_{sp}=s^{\prime }(0.010{)}^2⟹s^{\prime }=\frac{7.1\times 10^{-9}}{1.0\times 10^{-4}}=7.1\times 10^{-5}\text{M}$ Validity check: the extra iodide from dissolution is $2s^{\prime }=1.42\times 10^{-4}$. Compare to $0.010$: that is $1.42\%$. Under $5\%$, so the approximation is valid here. ✓

Contrast with the parent's PbCl₂ example, where the common ion was the cation and the dissolved salt made a lot of the other ion — there the approximation failed. The lesson: it is the size of $s^{\prime }$ relative to $c$ that decides, not the wording of the problem.

Add ${\text{Pb}}^{2+}$ ($c=0.010$): $[{\text{Pb}}^{2+}]\approx 0.010$, $[{\text{I}}^{-}]=2s^{\prime }$. $K_{sp}=0.010(2s^{\prime }{)}^2⟹(s^{\prime }{)}^2=\frac{7.1\times 10^{-9}}{0.04}=1.775\times 10^{-7}$ $s^{\prime }=4.21\times 10^{-4}\text{M}$ Add ${\text{I}}^{-}$ ($c=0.010$): from 3.1, $s^{\prime }=7.1\times 10^{-5}\text{M}$.

Conclusion: adding ${\text{I}}^{-}$ suppresses more ($7.1\times 10^{-5}$ vs $4.21\times 10^{-4}$, roughly $6\times$ stronger).

Why? ${\text{I}}^{-}$ appears squared in $K_{sp}$. Forcing up a squared term costs the equilibrium far more, so the salt must dissolve less to keep $K_{sp}$ constant. Attack the ion with the higher exponent.

${\text{Ag}}^{+}$ needed to start each precipitate (set reaction quotient $=K_{sp}$):

• AgCl: $[{\text{Ag}}^{+}]=\frac{K_{sp}}{[{\text{Cl}}^{-}]}=\frac{1.8\times 10^{-10}}{0.020}=9.0\times 10^{-9}\text{M}$
• ${\text{Ag}}_2{\text{CrO}}_4$: $[{\text{Ag}}^{+}]=\sqrt{\frac{K_{sp}}{[{\text{CrO}}_4^{2-}]}}=\sqrt{\frac{1.1\times 10^{-12}}{0.020}}=\sqrt{5.5\times 10^{-11}}=7.42\times 10^{-6}\text{M}$

AgCl needs far less ${\text{Ag}}^{+}$, so AgCl precipitates first.

$[{\text{Cl}}^{-}]$ left when ${\text{Ag}}_2{\text{CrO}}_4$ just starts (i.e. when $[{\text{Ag}}^{+}]=7.42\times 10^{-6}$): $[{\text{Cl}}^{-}]=\frac{K_{sp}(\text{AgCl})}{[{\text{Ag}}^{+}]}=\frac{1.8\times 10^{-10}}{7.42\times 10^{-6}}=2.43\times 10^{-5}\text{M}$ So by the time chromate begins to drop, chloride is down to $2.43\times 10^{-5}\text{M}$ — over $99.8\%$ of the chloride is already removed. Clean separation.

${\text{CH}}_3\text{COOH}\rightleftharpoons {\text{H}}^{+}+{\text{CH}}_3{\text{COO}}^{-}$. $K_a=\frac{[{\text{H}}^{+}][{\text{CH}}_3{\text{COO}}^{-}]}{[{\text{CH}}_3\text{COOH}]}$ With common ion (acetate $\approx 0.10$, acid $\approx 0.10$): $[{\text{H}}^{+}]=K_a\frac{[{\text{CH}}_3\text{COOH}]}{[{\text{CH}}_3{\text{COO}}^{-}]}=1.8\times 10^{-5}\cdot \frac{0.10}{0.10}=1.8\times 10^{-5}\text{M}$ Acid alone ($0.10\text{M}$): $[{\text{H}}^{+}]=\sqrt{K_a\cdot 0.10}=\sqrt{1.8\times 10^{-6}}=1.34\times 10^{-3}\text{M}$. Adding the common ion cut $[{\text{H}}^{+}]$ from $1.34\times 10^{-3}$ to $1.8\times 10^{-5}$ — a 74× suppression of ionisation. Same physics as solubility suppression, applied to a weak acid.

${\text{PbCl}}_2\rightleftharpoons {\text{Pb}}^{2+}+2{\text{Cl}}^{-}$. NaCl gives $[{\text{Cl}}^{-}]=0.10$. $[{\text{Pb}}^{2+}]=s^{\prime },[{\text{Cl}}^{-}]=0.10+2s^{\prime }$ Shortcut ($0.10+2s^{\prime }\approx 0.10$): $s^{\prime }=\frac{K_{sp}}{(0.10{)}^2}=\frac{1.7\times 10^{-5}}{0.010}=1.7\times 10^{-3}\text{M}$ Check: dissolution adds $2s^{\prime }=3.4\times 10^{-3}$, which is $3.4\%$ of $0.10$. Under $5\%$, so the shortcut is valid ✓. $s^{\prime }\approx 1.7\times 10^{-3}\text{M}$.

Note how differently this behaves from the parent's example where ${\text{Pb}}^{2+}$ was the common ion: there the chloride (squared, from dissolution) grew large and broke the approximation. Which ion you add changes everything — attacking the squared ion (as here) suppresses hardest and keeps $s^{\prime }$ small.

${\text{Mg(OH)}}_2\rightleftharpoons {\text{Mg}}^{2+}+2{\text{OH}}^{-}$, so $K_{sp}=[{\text{Mg}}^{2+}][{\text{OH}}^{-}{]}^2$. At $\text{pH}=9.00$: $[{\text{H}}^{+}]=10^{-9}$, so $[{\text{OH}}^{-}]=\frac{10^{-14}}{10^{-9}}=10^{-5}\text{M}$ (fixed by buffer). $[{\text{Mg}}^{2+}]=\frac{K_{sp}}{[{\text{OH}}^{-}{]}^2}=\frac{5.6\times 10^{-12}}{(10^{-5}{)}^2}=\frac{5.6\times 10^{-12}}{10^{-10}}=5.6\times 10^{-2}\text{M}$ Since each dissolved ${\text{Mg(OH)}}_2$ gives one ${\text{Mg}}^{2+}$, solubility $=[{\text{Mg}}^{2+}]=5.6\times 10^{-2}\text{M}$.

The insight: here a higher ${\text{OH}}^{-}$ (more basic) would lower solubility (common ion), and the buffer lets us dial $[{\text{OH}}^{-}]$ to any value we want. At $\text{pH}=9$ the hydroxide is low enough that ${\text{Mg(OH)}}_2$ is quite soluble. This is exactly how ${\text{Mg(OH)}}_2$ is dissolved by acid: acid destroys ${\text{OH}}^{-}$, forcing dissolution — the anti-common-ion effect.