13/14 Acid–Base and Solubility Equilibria

13/14 Acid–Base and Solubility Equilibria

Acid–base and solubility chemistry are applications of chemical equilibrium. A forward process and its reverse process continue simultaneously; at equilibrium, they occur at equal rates. Equilibrium calculations connect molecular reactions to measurable quantities such as concentration, $\mathrm{pH}$, titrant volume, and precipitate formation.

Unless otherwise stated, aqueous solutions are assumed to be at $25\,^{\circ}\mathrm{C}$, where

A useful strategy is to identify the reacting species, write the appropriate equilibrium or stoichiometric relationship, solve using the controlling approximation, and then check whether the result is chemically reasonable.

Takeaway: Equilibrium does not mean that reactions stop; it means that opposing reaction rates are equal.

Three Theories of Acids and Bases

Acid–base definitions become broader as the theory changes.

• An Arrhenius acid produces $\mathrm{H_3O^+}$ in water, while an Arrhenius base produces $\mathrm{OH^-}$. For example, $\mathrm{HCl}$ reacts with water to form $\mathrm{H_3O^+}$, and $\mathrm{NaOH}$ dissociates to form $\mathrm{OH^-}$.

• A Brønsted–Lowry acid donates a proton, and a Brønsted–Lowry base accepts a proton. In $\mathrm{NH_3+H_2O\rightleftharpoons NH_4^++OH^-}$, ammonia accepts a proton and is the base, while water donates a proton and is the acid.

• A Lewis acid accepts an electron pair, and a Lewis base donates an electron pair. In $\mathrm{BF_3+NH_3\rightarrow F_3B\leftarrow NH_3}$, $\mathrm{BF_3}$ is the Lewis acid and $\mathrm{NH_3}$ is the Lewis base.

A conjugate acid–base pair differs by one proton. For the general reaction

$\mathrm{HA/A^-}$ and $\mathrm{B/HB^+}$ are conjugate pairs. Water is amphiprotic because it can either donate or accept a proton.

Takeaway: Use Arrhenius definitions for ion production in water, Brønsted–Lowry definitions for proton transfer, and Lewis definitions for electron-pair transfer.

Water, pH, and pOH

Water undergoes slight self-ionization:

The acidity and basicity scales are defined by

At $25\,^{\circ}\mathrm{C}$,

A solution is acidic when $\mathrm{[H_3O^+]>[OH^-]}$, neutral when the concentrations are equal, and basic when $\mathrm{[H_3O^+]<[OH^-]}$. The neutral value $\mathrm{pH=7.00}$ applies specifically at $25\,^{\circ}\mathrm{C}$.

To convert concentration to pH, use the negative logarithm. For $\mathrm{[H_3O^+]=2.5\times10^{-3}\,M}$,

Then $\mathrm{pOH=14.00-2.60=11.40}$. Conversely, if $\mathrm{pH=9.20}$,

Because pH is logarithmic, its decimal places indicate the significant figures reported for the ion concentration.

Takeaway: Use $K_w$ to connect hydronium and hydroxide concentrations, and use $\mathrm{pH+pOH=14.00}$ at $25\,^{\circ}\mathrm{C}$.

Strong and Weak Acids and Bases

The distinction between strong and weak electrolytes determines whether an equilibrium calculation is needed.

A strong acid or base ionizes essentially completely. For a monoprotic strong acid such as $\mathrm{HCl}$,

For $\mathrm{Ba(OH)_2}$, each formula unit supplies two hydroxide ions:

A weak acid or base ionizes only partially. For a weak acid,

For a weak base,

A larger $K_a$ indicates a stronger acid, while a larger $K_b$ indicates a stronger base. For conjugate pairs at $25\,^{\circ}\mathrm{C}$,

For a weak acid with initial concentration $C$, an ICE table gives

where $x$ is the equilibrium hydronium concentration produced by ionization. If $x$ is less than about $5\%$ of $C$, use

For $0.100\,\mathrm{M}$ acetic acid with $K_a=1.8\times10^{-5}$, this gives $\mathrm{[H_3O^+]=1.34\times10^{-3}\,M}$ and $\mathrm{pH=2.87}$. The percent ionization is $1.34\%$, so the approximation is reasonable.

Takeaway: Strong species are treated primarily by stoichiometry; weak species require an equilibrium expression and a check of any approximation.

Salt Hydrolysis

Ions from weak acids and weak bases can react with water. This process is called hydrolysis.

The acetate ion, which is the conjugate base of acetic acid, reacts according to

Because hydroxide is produced, sodium acetate gives a basic solution. The ammonium ion, which is the conjugate acid of ammonia, reacts according to

Because hydronium is produced, ammonium chloride gives an acidic solution. A salt such as $\mathrm{NaCl}$, formed from a strong acid and a strong base, generally gives a nearly neutral solution because its ions are very weak acid–base reactants.

For a conjugate base and conjugate acid,

Takeaway: Determine whether a salt solution is acidic or basic by asking whether either ion is the conjugate partner of a weak acid or weak base.

Buffers and pH Control

A buffer contains appreciable amounts of a weak acid and its conjugate base, or a weak base and its conjugate acid. It resists large pH changes because added strong acid or base is consumed by one of the buffer components.

For an acidic buffer, added acid reacts mainly with the conjugate base:

Added base reacts mainly with the weak acid:

The Henderson–Hasselbalch equation is

When both components occupy the same solution, mole ratios can replace concentration ratios. Buffer effectiveness is greatest when $\mathrm{[A^-]=[HA]}$, because then $\mathrm{pH=p}K_a$. To design a buffer, choose a weak acid with a $\mathrm{p}K_a$ near the desired pH and then adjust the conjugate-base-to-acid ratio.

For a buffer containing $0.20$ mol of acetic acid and $0.30$ mol of acetate with $\mathrm{p}K_a=4.76$,

If $0.010$ mol of strong acid is added, the acetate amount becomes $0.290$ mol and the acid amount becomes $0.210$ mol. The new pH is approximately $4.90$, illustrating the buffer’s resistance to a large change.

Takeaway: A buffer changes the ratio of conjugate partners rather than allowing the added strong reagent to change the pH dramatically.

Acid–Base Titrations

A titration determines an unknown concentration by reacting an analyte with a solution of known concentration, called the titrant. The equivalence point occurs when the reactants have combined in stoichiometrically equivalent amounts. The endpoint is the observed indicator or instrumental signal and should be close to the equivalence point.

For a 1:1 reaction such as

For a strong-acid–strong-base titration, calculate pH from the initial strong reagent before titrant is added, use excess moles divided by total volume before or after equivalence, and expect approximately $\mathrm{pH=7.00}$ at equivalence at $25\,^{\circ}\mathrm{C}$.

For example, titrating $25.00\,\mathrm{mL}$ of $0.100\,\mathrm{M}$ $\mathrm{HCl}$ with $30.00\,\mathrm{mL}$ of $0.100\,\mathrm{M}$ $\mathrm{NaOH}$ leaves $0.000500\,\mathrm{mol}$ of excess hydroxide in $0.05500\,\mathrm{L}$. Thus $\mathrm{[OH^-]=9.09\times10^{-3}\,M}$, $\mathrm{pOH=2.04}$, and $\mathrm{pH=11.96}$.

Weak-acid–strong-base titrations contain a buffer region before equivalence. At the half-equivalence point, equal amounts of weak acid and conjugate base are present, so $\mathrm{pH=p}K_a$. At equivalence, the conjugate base hydrolyzes and the pH is greater than $7$. Weak-base–strong-acid titrations are analogous, but their half-equivalence point satisfies $\mathrm{pOH=p}K_b$ and their equivalence point is acidic.

Choose an indicator whose color-change interval overlaps the steep portion of the titration curve near equivalence.

Takeaway: Divide a titration into regions, identify the controlling reaction in each region, and use stoichiometry before equilibrium calculations whenever a strong reagent is in excess.

Solubility Equilibria and Precipitation

A sparingly soluble ionic solid establishes an equilibrium between the solid and its dissolved ions. For silver chloride,

The solubility-product constant is

For a general salt,

The stoichiometric coefficients determine the exponents. For $\mathrm{CaF_2}$, if the molar solubility is $s$, then $\mathrm{[Ca^{2+}]=s}$ and $\mathrm{[F^-]=2s}$, so

To predict precipitation after mixing solutions, calculate the ion product $Q_{sp}$ from concentrations immediately after dilution:

• If $Q_{sp}<K_{sp}$, the solution is unsaturated and precipitation is not expected.

• If $Q_{sp}=K_{sp}$, the solution is saturated and at equilibrium.

• If $Q_{sp}>K_{sp}$, precipitation is thermodynamically favored.

For $\mathrm{[Ag^+]=[Cl^-]=1.0\times10^{-4}\,M}$, $Q_{sp}=1.0\times10^{-8}$. With $K_{sp}(\mathrm{AgCl})=1.6\times10^{-10}$, $Q_{sp}>K_{sp}$, so silver chloride precipitates.

Takeaway: Always use post-mixing concentrations and compare $Q_{sp}$ directly with $K_{sp}$.

Factors Affecting Solubility and Measurement

Solubility can change when other chemical equilibria alter the concentration of a dissolved ion.

The common-ion effect decreases the solubility of a solid when an ion already present in its dissolution equilibrium is added. For silver chloride,

adding excess $\mathrm{Cl^-}$ shifts the equilibrium toward solid $\mathrm{AgCl}$.

Acid can increase the solubility of salts containing basic anions. Carbonate reacts with hydronium:

This removes carbonate from solution and shifts the dissolution of a carbonate such as calcium carbonate toward additional dissolution.

In selective precipitation, two ions are exposed to a common precipitating ion. The compound with the smaller effective solubility generally begins to precipitate first, allowing the ions to be separated if the conditions are suitable.

For reliable quantitative work, account for dilution after mixing, calibrate pH meters, read buret menisci at eye level, add titrant slowly near the endpoint, use replicate trials, and report appropriate significant figures and uncertainty.

Takeaway: Le Châtelier’s principle links acid effects and common-ion effects to changes in solubility, while careful measurement determines whether the predicted change can be observed quantitatively.