AP Chemistry 8.3-8.6: Weak Acid/Base Equilibria and Molecular Structure

Relative strength of an acid

Acid strength describes how extensively an acid ionizes in water. Strong acids ionize essentially completely, whereas weak acids ionize only partially.

Acid-ionization constant, $K_a$

For $HA(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+A^-(aq)$, $K_a=\frac{[H_3O^+][A^-]}{[HA]}$. Water is omitted because it is the solvent.

Why is liquid water omitted from the $K_a$ expression?

The concentration of a pure liquid solvent is effectively constant, so it is incorporated into the equilibrium constant rather than written in the expression.

How does the magnitude of $K_a$ compare acids of the same initial concentration?

A larger $K_a$ indicates greater ionization, a higher equilibrium concentration of $H_3O^+$, and a stronger acid.

What does an extremely large or immeasurable $K_a$ signify?

It signifies an acid that ionizes essentially completely in water. Such an acid is classified as strong, and the equilibrium concentration of $HA$ is approximately zero.

Rank $CH_3CO_2H$, $HNO_2$, and $HSO_4^-$ by increasing acid strength using their $K_a$ values: $1.8\times10^{-5}$, $4.6\times10^{-4}$, and $1.2\times10^{-2}$.

$CH_3CO_2H < HNO_2 < HSO_4^-$. The species with the largest $K_a$ is the strongest acid.

Percent ionization of a weak acid

The percent ionization is $\%\text{ ionization}=\frac{[H_3O^+]_{eq}}{[HA]_0}\times100\%$. For a 1:1 acid-ionization reaction, $[H_3O^+]_{eq}=[A^-]_{eq}$ when water's initial contribution is negligible.

How does weak-acid percent ionization usually change when the initial acid concentration increases?

It usually decreases as the initial concentration increases, even though the acid's $K_a$ remains constant.

A $0.125\,M$ weak acid solution has pH $2.09$. What is its percent ionization?

$[H_3O^+]=10^{-2.09}=8.1\times10^{-3}\,M$, so $\%\text{ ionization}=\frac{8.1\times10^{-3}}{0.125}\times100\%=6.5\%$.

Base-ionization constant, $K_b$

For $B(aq)+H_2O(l)\rightleftharpoons HB^+(aq)+OH^-(aq)$, $K_b=\frac{[HB^+][OH^-]}{[B]}$. A larger $K_b$ means a stronger base.

How does the magnitude of $K_b$ relate to hydroxide concentration for bases of the same initial concentration?

A stronger base has a larger $K_b$, ionizes more extensively, and produces a greater equilibrium concentration of $OH^-.

Percent ionization of a weak base

The percent ionization is $\%\text{ ionization}=\frac{[OH^-]_{eq}}{[B]_0}\times100\%$ for the reaction $B+H_2O\rightleftharpoons HB^++OH^-$.

How do a weak base's $K_b$ and initial concentration affect its percent ionization?

Percent ionization depends on both $K_b$ and the initial concentration. For a given weak base, it generally decreases as the initial concentration increases.

Rank $NO_2^-$, $CH_3CO_2^-$, and $NH_3$ by increasing base strength given $K_b=2.17\times10^{-11}$, $5.6\times10^{-10}$, and $1.8\times10^{-5}$, respectively.

$NO_2^- < CH_3CO_2^- < NH_3$. The increasing order follows the increasing values of $K_b$.

Relationship between the ionization constants of a conjugate acid-base pair

For conjugate pair $HA/A^-$, $K_aK_b=K_w$. At $25^\circ\mathrm{C}$, $K_w=1.0\times10^{-14}$, so $K_a=\frac{K_w}{K_b}$ and $K_b=\frac{K_w}{K_a}$.

How are the strengths of conjugate acid-base partners related?

They are inversely related: a stronger acid has a weaker conjugate base, and a stronger base has a weaker conjugate acid.

Calculate $K_a$ for $HNO_2$ if $K_b$ for its conjugate base, $NO_2^-$, is $2.17\times10^{-11}$.

$K_a=\frac{K_w}{K_b}=\frac{1.0\times10^{-14}}{2.17\times10^{-11}}=4.6\times10^{-4}$.

What is the conjugate-base strength of a strong acid?

It is negligible: because a strong acid has an extremely large $K_a$, its conjugate base has $K_b=K_w/K_a\approx0$.

What is the conjugate-acid strength of a strong base?

It is negligible: because a strong base has an extremely large $K_b$, its conjugate acid has $K_a=K_w/K_b\approx0$.

Leveling effect in water

Water limits the apparent strength of acids and bases. Any acid stronger than $H_3O^+$ transfers its proton completely to water, and any base stronger than $OH^-$ reacts completely with water; their strengths are therefore leveled to $H_3O^+$ or $OH^-$.

Why can’t the relative strengths of strong acids in water be distinguished by their aqueous $K_a$ values?

They all ionize essentially completely, so the principal acid species remaining in solution is $H_3O^+$. A less basic solvent can reveal differences among these acids.

How does the solvent affect the apparent strengths of HCl, HBr, and HI?

All three behave as strong acids in water because of the leveling effect. In ethanol, where leveling is less extensive, their relative strength is observed as $HCl<HBr<HI$.

Using equilibrium concentrations, how is $K_a$ determined for a weak acid?

Write the acid-ionization expression and substitute equilibrium concentrations: $K_a=\frac{[H_3O^+][A^-]}{[HA]}$. Initial concentrations are not used unless they equal the equilibrium values.

At equilibrium, a solution contains $[CH_3CO_2H]=0.0787\,M$ and $[H_3O^+]=[CH_3CO_2^-]=0.00118\,M$. What is $K_a$?

$K_a=\frac{(0.00118)(0.00118)}{0.0787}=1.77\times10^{-5}$.

Using equilibrium concentrations, how is $K_b$ determined for a weak base?

For $B+H_2O\rightleftharpoons HB^++OH^-$, use $K_b=\frac{[HB^+][OH^-]}{[B]}$ with equilibrium concentrations.

A weak base has equilibrium concentrations $[B]=0.050\,M$, $[HB^+]=5.0\times10^{-3}\,M$, and $[OH^-]=2.5\times10^{-3}\,M$. Find $K_b$.

$K_b=\frac{(5.0\times10^{-3})(2.5\times10^{-3})}{0.050}=2.5\times10^{-4}$.

How can pH data be used to find the $K_a$ of a weak monoprotic acid?

Convert pH to equilibrium hydronium concentration using $[H_3O^+]=10^{-\mathrm{pH}}$. Set the change in $HA$ equal to this concentration, construct the equilibrium concentrations, and substitute them into $K_a=\frac{[H_3O^+][A^-]}{[HA]}$.

A $0.0516\,M$ $HNO_2$ solution has pH $2.34$. What is its approximate $K_a$?

$[H_3O^+]=10^{-2.34}=0.0046\,M$. Thus $[NO_2^-]\approx0.0046\,M$, $[HNO_2]\approx0.0516-0.0046=0.0470\,M$, and $K_a\approx\frac{(0.0046)^2}{0.0470}=4.6\times10^{-4}$.

How can pH data be used to find the $K_b$ of a weak base?

First use $\mathrm{pOH}=14.00-\mathrm{pH}$ and $[OH^-]=10^{-\mathrm{pOH}}$. Treat this hydroxide concentration as the amount of base ionized, then use the equilibrium concentrations in $K_b=\frac{[HB^+][OH^-]}{[B]}$.

At $25^\circ\mathrm{C}$, how are pH and pOH related?

$\mathrm{pH}+\mathrm{pOH}=\mathrm{p}K_w=14.00$. Thus, $\mathrm{pH}=14.00-\mathrm{pOH}$.

ICE-table setup for a weak acid initially present at concentration $C$

For $HA\rightleftharpoons H_3O^++A^-$, the equilibrium concentrations are $[HA]=C-x$, $[H_3O^+]=x$, and $[A^-]=x$, neglecting initial hydronium when appropriate.

ICE-table setup for a weak base initially present at concentration $C$

For $B\rightleftharpoons HB^++OH^-$, the equilibrium concentrations are $[B]=C-x$, $[HB^+]=x$, and $[OH^-]=x$, neglecting initial hydroxide when appropriate.

When is the small-$x$ approximation valid in a weak-acid or weak-base equilibrium calculation?

If $x$ is less than about $5\%$ of the initial concentration, replace $C-x$ with $C$. After solving, check that $\frac{x}{C}\times100\%<5\%$.

What approximation commonly estimates hydronium concentration for a weak acid?

When the small-$x$ approximation is valid, $K_a\approx\frac{x^2}{C}$, so $[H_3O^+]=x\approx\sqrt{K_aC}$.

What approximation commonly estimates hydroxide concentration for a weak base?

When the small-$x$ approximation is valid, $K_b\approx\frac{x^2}{C}$, so $[OH^-]=x\approx\sqrt{K_bC}$.

How is the pH of a weak-base solution calculated after finding its equilibrium hydroxide concentration?

Find $[OH^-]$ from the ICE table or $[OH^-]\approx\sqrt{K_bC}$ when valid, calculate $\mathrm{pOH}=-\log[OH^-]$, then use $\mathrm{pH}=14.00-\mathrm{pOH}$ at $25^\circ\mathrm{C}$.

A $0.534\,M$ formic acid solution has $K_a=1.8\times10^{-4}$. Estimate $[H_3O^+]$ and pH.

$[H_3O^+]\approx\sqrt{(1.8\times10^{-4})(0.534)}=9.8\times10^{-3}\,M$. Therefore, $\mathrm{pH}=-\log(9.8\times10^{-3})\approx2.01$; the ionization is about $1.8\%$, so the approximation is valid.

How should a weak-acid or weak-base equilibrium be solved when the small-$x$ approximation fails?

Retain the $C-x$ term in the equilibrium expression and solve the resulting quadratic equation. Discard any negative concentration root as physically meaningless.

Why must the quadratic equation be used for a $0.50\,M$ $HSO_4^-$ solution with $K_a=1.2\times10^{-2}$?

The approximate solution gives $x\approx0.077\,M$, which is about $15\%$ of the initial concentration—greater than the $5\%$ criterion. The $C-x$ term cannot be neglected; solving exactly gives $[H_3O^+]\approx0.072\,M$ and pH $\approx1.14$.

Binary acid trend down a group

For hydrogen compounds of nonmetals in the same group, acidity generally increases down the group because the H–A bond becomes weaker and is easier to break. For halogen hydrides, $HF<HCl<HBr<HI$ in increasing acid strength.

How does the H–A bond-strength trend explain the acidity of group 16 hydrides?

Acidity increases down the group as the H–A bond weakens: $H_2O<H_2S<H_2Se<H_2Te$.

Binary acid trend across a period

Across a period, binary hydrogen compounds generally become more acidic as the electronegativity of the nonmetal increases, because the H–A bond becomes more polar. In the second period: $CH_4<NH_3<H_2O<HF$.

What is the acidity trend across the third period for binary hydrogen compounds?

Acidity increases with the electronegativity of the nonmetal: $SiH_4<PH_3<H_2S<HCl$.

General structure of an oxyacid

An oxyacid contains a central atom bonded to oxygen atoms, with at least one oxygen bearing hydrogen; it can be represented generally as $O_mE(OH)_n$.

How does the electronegativity of the central atom affect whether a ternary hydroxide compound is acidic or basic?

A low-electronegativity, metallic central atom favors ionic M–OH bonding and release of $OH^-$, producing a basic hydroxide. A high-electronegativity, nonmetallic central atom strengthens the E–O bond, weakens O–H, and favors proton release as an oxyacid.

How does the oxidation number of the central atom affect oxyacid strength?

For oxyacids with the same central atom, a higher oxidation number generally increases acidity by increasing the central atom’s electron-withdrawing effect and weakening the O–H bond. Thus, $H_2SO_4$ is more acidic than $H_2SO_3$, and $HNO_3$ is more acidic than $HNO_2$.

Amphoteric hydroxide

An amphoteric hydroxide can act as either an acid or a base, depending on the reactant. Aluminum hydroxide dissolves in strong base by donating a proton to $OH^-$ and dissolves in strong acid by accepting protons.

How does hydrated aluminum hydroxide react with strong base?

It acts as an acid by transferring a proton from a coordinated water molecule to hydroxide: $Al(H_2O)_3(OH)_3+OH^-\rightleftharpoons H_2O+[Al(H_2O)_2(OH)_4]^-$.

How does hydrated aluminum hydroxide react with strong acid?

It acts as a base by accepting protons from hydronium: $3H_3O^++Al(H_2O)_3(OH)_3\rightleftharpoons Al(H_2O)_6^{3+}+3H_2O$.