Exercises — Conjugate acid-base pairs

Remove one ${\text{H}}^{+}$; charge drops by $1$.

• ${\text{HNO}}_3\to {\text{NO}}_3^{-}$ (neutral $\to -1$)
• ${\text{H}}_2{\text{SO}}_4\to {\text{HSO}}_4^{-}$ (neutral $\to -1$)
• ${\text{HCO}}_3^{-}\to {\text{CO}}_3^{2-}$ ($-1\to -2$)

Notice the last one: both species are negative. A conjugate base is not required to be neutral or positive — it is just one proton lighter.

Add one ${\text{H}}^{+}$; charge rises by $1$.

• ${\text{NH}}_3\to {\text{NH}}_4^{+}$ (neutral $\to +1$)
• ${\text{OH}}^{-}\to {\text{H}}_2\text{O}$ ($-1\to 0$)
• ${\text{HPO}}_4^{2-}\to {\text{H}}_2{\text{PO}}_4^{-}$ ($-2\to -1$)

• Pair 1: $\text{HF}$ (acid) / ${\text{F}}^{-}$ (conjugate base) — differ by one ${\text{H}}^{+}$.
• Pair 2: ${\text{H}}_3{\text{O}}^{+}$ (conjugate acid) / ${\text{H}}_2\text{O}$ (base) — differ by one ${\text{H}}^{+}$.

The acid on the left ($\text{HF}$) always pairs with the base product on the right (${\text{F}}^{-}$), and the base on the left (${\text{H}}_2\text{O}$) pairs with the acid product (${\text{H}}_3{\text{O}}^{+}$). Cross-connect, never straight down.

Rearrange the workhorse: $K_b=\frac{K_w}{K_a}$. $K_b=\frac{1.0\times 10^{-14}}{6.8\times 10^{-4}}=1.47\times 10^{-11}$ Since $K_b\ll 1$, ${\text{F}}^{-}$ is a weak base — the conjugate of a (relatively) strong-ish weak acid is a weak base.

$K_a=\frac{K_w}{K_b}=\frac{1.0\times 10^{-14}}{2.0\times 10^{-5}}=5.0\times 10^{-10}$ Compare: $K_a(\text{HCN})=5.0\times 10^{-10}$ versus $K_a(\text{HF})=6.8\times 10^{-4}$. Since $5.0\times 10^{-10}<6.8\times 10^{-4}$, HCN is the weaker acid.

Start from $K_a{K}_b=K_w$. Take $-{\log }_{10}$ of both sides: $-{\log }_{10}(K_a{K}_b)=-{\log }_{10}{K}_w$ The log of a product is a sum, so the left side splits: $(-{\log }_{10}{K}_a)+(-{\log }_{10}{K}_b)=-{\log }_{10}(1.0\times 10^{-14})$ $\text{p}{K}_a+\text{p}{K}_b=14$ This is the same rule in "p" clothing — handy because you can just subtract from 14.

Acid strength = larger $K_a$. Order by size: $D(10^{-2})>B(6.8\times 10^{-4})>A(1.8\times 10^{-5})>C(4.9\times 10^{-10})$ Conjugate base strength is the reverse (inverse relationship $K_b=K_w/K_a$: bigger $K_a\Rightarrow$ smaller $K_b$): $C^{-}>A^{-}>B^{-}>D^{-}$ See the mirror image in the figure below — as the acid bars shrink, the conjugate-base bars grow.

Equilibrium favours the side with the weaker acid (weaker acids hold protons, so protons pile up there).

• Left has acid A ($K_a=1.8\times 10^{-5}$).
• Right has acid HB ($K_a=6.8\times 10^{-4}$). A is the weaker acid, so protons prefer to sit on A. Equilibrium lies to the left (reactants favoured). Quick check via $K=K_a(\text{A})/K_a(\text{HB})=\frac{1.8\times 10^{-5}}{6.8\times 10^{-4}}=0.026<1$ ✓ — indeed left-favoured.

As an acid (donates ${\text{H}}^{+}$): ${\text{HCO}}_3^{-}\rightleftharpoons {\text{H}}^{+}+{\text{CO}}_3^{2-}$ Conjugate base $={\text{CO}}_3^{2-}$. As a base (accepts ${\text{H}}^{+}$): ${\text{HCO}}_3^{-}+{\text{H}}^{+}\rightleftharpoons {\text{H}}_2{\text{CO}}_3$ Conjugate acid $={\text{H}}_2{\text{CO}}_3$. So ${\text{HCO}}_3^{-}$ sits in the middle of the ladder ${\text{H}}_2{\text{CO}}_3/{\text{HCO}}_3^{-}/{\text{CO}}_3^{2-}$, one proton above and one below.

When ${\text{HCO}}_3^{-}$ acts as a base, its conjugate acid is ${\text{H}}_2{\text{CO}}_3$, whose acid constant is $K_{a1}$. So use $K_{a1}$: $K_b({\text{HCO}}_3^{-})=\frac{K_w}{K_{a1}}=\frac{1.0\times 10^{-14}}{4.3\times 10^{-7}}=2.33\times 10^{-8}$ When ${\text{HCO}}_3^{-}$ acts as an acid, its constant is $K_{a2}=4.8\times 10^{-11}$. Compare: as a base $K_b=2.33\times 10^{-8}$; as an acid $K_a=4.8\times 10^{-11}$. Since $2.33\times 10^{-8}>4.8\times 10^{-11}$, the base behaviour dominates — a solution of ${\text{NaHCO}}_3$ is mildly basic. (This is exactly why baking soda solutions are slightly alkaline.)

(a) The proton ladder: ${\text{H}}_3{\text{PO}}_4\stackrel{-{\text{H}}^{+}}{\to }{\text{H}}_2{\text{PO}}_4^{-}\stackrel{-{\text{H}}^{+}}{\to }{\text{HPO}}_4^{2-}\stackrel{-{\text{H}}^{+}}{\to }{\text{PO}}_4^{3-}$ Pairs (acid / conjugate base): $({\text{H}}_3{\text{PO}}_4/{\text{H}}_2{\text{PO}}_4^{-})$, $({\text{H}}_2{\text{PO}}_4^{-}/{\text{HPO}}_4^{2-})$, $({\text{HPO}}_4^{2-}/{\text{PO}}_4^{3-})$. Each step lowers charge by $1$. See the ladder figure.

(b) ${\text{PO}}_4^{3-}$ accepting a proton becomes ${\text{HPO}}_4^{2-}$. The conjugate acid ${\text{HPO}}_4^{2-}$ is described by $K_{a3}$ (the acid that gives up the third proton). So: $K_b({\text{PO}}_4^{3-})=\frac{K_w}{K_{a3}}=\frac{1.0\times 10^{-14}}{4.2\times 10^{-13}}=2.38\times 10^{-2}$ That is a fairly large $K_b$ — ${\text{PO}}_4^{3-}$ is a decently strong base, consistent with $K_{a3}$ being the tiniest acid constant.

(c) Each anion's base strength uses the $K_a$ of its conjugate acid (the species one proton heavier):

• ${\text{H}}_2{\text{PO}}_4^{-}$: conjugate acid ${\text{H}}_3{\text{PO}}_4$, use $K_{a1}$ → $K_b=10^{-14}/7.5\times 10^{-3}=1.33\times 10^{-12}$.
• ${\text{HPO}}_4^{2-}$: conjugate acid ${\text{H}}_2{\text{PO}}_4^{-}$, use $K_{a2}$ → $K_b=10^{-14}/6.2\times 10^{-8}=1.61\times 10^{-7}$.
• ${\text{PO}}_4^{3-}$: from (b), $K_b=2.38\times 10^{-2}$. Ranking (strongest base first): ${\text{PO}}_4^{3-}>{\text{HPO}}_4^{2-}>{\text{H}}_2{\text{PO}}_4^{-}$ The more deprotonated (more negative) the anion, the hungrier it is for a proton — the stronger the base.

${\text{HPO}}_4^{2-}$ acting as a base accepts a proton to become ${\text{H}}_2{\text{PO}}_4^{-}$; its conjugate acid ${\text{H}}_2{\text{PO}}_4^{-}$ has acid constant $K_{a2}$. Therefore $K_b=K_w/K_{a2}$, and: $K_{a2}\times K_b=K_{a2}\times \frac{K_w}{K_{a2}}=K_w=1.0\times 10^{-14}✓$ The pair ${\text{H}}_2{\text{PO}}_4^{-}$/${\text{HPO}}_4^{2-}$ obeys the same $K_a{K}_b=K_w$ law as any simple pair — polyprotic acids are just several ordinary pairs stacked on a ladder.

L1: conjugates differ by one ${\text{H}}^{+}$ (charge $\pm 1$). ::: L2: $K_a{K}_b=K_w$, so $K_b=K_w/K_a$. L3: base strength runs opposite to acid strength. L4: for amphoteric ions, pick $K_a$ of the correct conjugate. L5: polyprotic = a ladder of ordinary pairs; match each base to the deprotonation that makes its conjugate acid.