Corrected exercises: acids and bases (202-SN2-RE)

Three pH units apart: $10^{3}=1000$ times more concentrated. The pH scale is LOGARITHMIC: each unit is a factor of 10, so a small difference in pH hides an enormous difference in concentration. This is why you never average two pH values.

HCl is a strong acid, completely dissociated: $[H_{3}O^{+}]=0.020$ mol/L. $pH=-\log(0.020)=1.70$.

NaOH is a strong base: $[OH^{-}]=0.0050$ mol/L, so $pOH=-\log(0.0050)=2.30$ and $pH=14-2.30=11.70$.

Diluting by 10 divides $[H_{3}O^{+}]$ by 10: $pH=-\log(0.0020)=2.70$, one unit higher. But the limit is the water itself: its autoionization already supplies $10^{-7}$ mol/L of $H_{3}O^{+}$ ions. When the dilution is extreme, the contribution of the acid becomes negligible next to that of the water and the pH approaches 7 FROM BELOW: diluting an acid will never make a base. Believing that a dilution by $10^{6}$ gives $1.70+6=7.70$ is the classic trap.

Pairs: $NH_{4}^{+}/NH_{3}$ (ammonia accepts a proton: it is the BASE) and $H_{2}O/OH^{-}$ (water donates a proton: it is the ACID). Here the water acts as an acid; facing an acid such as HCl, it would act as a base: it is amphoteric, a chameleon belonging to both camps.

As an acid: $HCO_{3}^{-}+H_{2}O\rightleftharpoons CO_{3}^{2-}+H_{3}O^{+}$ (it donates its proton). As a base: $HCO_{3}^{-}+H_{2}O\rightleftharpoons H_{2}CO_{3}+OH^{-}$ (it accepts one). The same ion belongs to two pairs: $H_{2}CO_{3}/HCO_{3}^{-}$ and $HCO_{3}^{-}/CO_{3}^{2-}$. It is the ion that buffers blood.

Acids, from strongest to weakest: $HF>CH_{3}COOH>HCN$ (the $K_{a}$ values decrease from $6.6\times 10^{-4}$ to $6.2\times 10^{-10}$). Conjugate bases: the REVERSE ranking, $CN^{-}>CH_{3}COO^{-}>F^{-}$. Rule: the stronger an acid, the weaker its conjugate base; an acid that gives up its proton easily leaves behind a base that has no urge to take it back.

To the RIGHT. The forward reaction sets the acid HF ($K_{a}=6.6\times 10^{-4}$) against the acid HCN ($K_{a}=6.2\times 10^{-10}$): the reaction proceeds from the stronger acid to the weaker acid, and the proton always goes down to the hungrier base (here $CN^{-}$). The ratio of the two $K_{a}$ values, about $10^{6}$, would in fact give the equilibrium constant of the reaction directly: strongly favoured.

ICE: $K_{a}=\frac{x^{2}}{0.10-x}\approx\frac{x^{2}}{0.10}$, so $x=\sqrt{(1.8\times 10^{-5})(0.10)}=1.3\times 10^{-3}$ mol/L. Test: $\frac{1.34\times 10^{-3}}{0.10}=1.3$% $<5$%: the approximation is valid. $pH=-\log(1.34\times 10^{-3})=2.87$.

$\frac{x}{C_{0}}\times 100=\frac{1.34\times 10^{-3}}{0.10}\times 100\approx 1.3$%. In concrete terms: out of 1000 dissolved acetic acid molecules, only 13 have given up their proton; the others stay intact. That is the whole difference from a strong acid, which is 100% ionized.

At 0.0010 mol/L, the approximation would give $x=\sqrt{1.8\times 10^{-8}}=1.34\times 10^{-4}$, that is 13% of $C_{0}$: the test fails. Quadratic: $x^{2}+1.8\times 10^{-5}x-1.8\times 10^{-8}=0$, hence $x=\frac{-1.8\times 10^{-5}+\sqrt{(1.8\times 10^{-5})^{2}+4(1.8\times 10^{-8})}}{2}\approx 1.25\times 10^{-4}$ mol/L. $pH\approx 3.90$ and percent ionization $\approx 13$%.

Diluting from 0.10 to 0.0010 mol/L, the ionization goes from 1.3% to 13%: dilution IONIZES MORE (in proportion), even though the pH rises. Le Chatelier explains it: the ionization $HA+H_{2}O\rightleftharpoons H_{3}O^{+}+A^{-}$ turns one dissolved species into two; dilution pushes the equilibrium toward the side with more dissolved particles, that is, toward ionization. A very dilute weak acid behaves almost like a strong acid: it is the favourite paradox of exams.

ICE on $NH_{3}+H_{2}O\rightleftharpoons NH_{4}^{+}+OH^{-}$: $x=\sqrt{K_{b}C_{0}}=\sqrt{(1.8\times 10^{-5})(0.20)}=1.9\times 10^{-3}$ mol/L, which is 0.95% of $C_{0}$: approximation justified. $pOH=-\log(1.90\times 10^{-3})=2.72$, so $pH=11.28$.

For a conjugate pair, $K_{a}\times K_{b}=K_{w}$: $K_{a}(NH_{4}^{+})=\frac{1.0\times 10^{-14}}{1.8\times 10^{-5}}=5.6\times 10^{-10}$. The strength on the base side and the weakness on the acid side are two faces of the same coin.

ACIDIC. The salt dissociates into $NH_{4}^{+}$ and $Cl^{-}$. The chloride ion, the conjugate base of a STRONG acid, is a spectator (no urge for a proton). The ammonium ion, on the other hand, is a weak acid ($K_{a}=5.6\times 10^{-10}$): it gives a few protons to the water and the pH drops below 7. General rule: the salt of a strong acid and a weak base is acidic.

Because they are weak: in both cases only a small fraction of the molecules ionizes (about 1% here). Vinegar at 0.10 mol/L gives a pH of 2.9 where a STRONG acid of the same concentration would give 1.0; ammonia gives 11.3 where a strong base would give 13.3. The gap of two units, a factor of 100 in ion concentration, is the signature of the weak electrolyte: the bottle contains a lot of acid or base, but the solution contains few ions.

First: $H_{2}CO_{3}+H_{2}O\rightleftharpoons H_{3}O^{+}+HCO_{3}^{-}$, $K_{a1}=\frac{[H_{3}O^{+}][HCO_{3}^{-}]}{[H_{2}CO_{3}]}$. Second: $HCO_{3}^{-}+H_{2}O\rightleftharpoons H_{3}O^{+}+CO_{3}^{2-}$, $K_{a2}=\frac{[H_{3}O^{+}][CO_{3}^{2-}]}{[HCO_{3}^{-}]}$.

$K_{a1}$ is about 9100 times larger than $K_{a2}$: the second ionization releases a negligible amount of $H_{3}O^{+}$ next to the first, and the whole pH depends on $K_{a1}$. With the usual approximation: $[H_{3}O^{+}]=\sqrt{K_{a1}C}=\sqrt{(4.3\times 10^{-7})(0.10)}=\sqrt{4.3\times 10^{-8}}=2.07\times 10^{-4}$ mol/L, hence pH $=3.68$. (5% test: $\frac{2.07\times 10^{-4}}{0.10}=0.21$%, comfortably satisfied.)

The first ionization produces as much $HCO_{3}^{-}$ as $H_{3}O^{+}$: $[HCO_{3}^{-}]=2.07\times 10^{-4}$ mol/L. Substituting into $K_{a2}=\frac{[H_{3}O^{+}][CO_{3}^{2-}]}{[HCO_{3}^{-}]}$, and since $[H_{3}O^{+}]\approx[HCO_{3}^{-}]$, these two terms cancel: $[CO_{3}^{2-}]=K_{a2}=4.7\times 10^{-11}$ mol/L. A remarkable result: the carbonate ion concentration of a diprotic acid alone in solution does NOT depend on the dilution; it is fixed by $K_{a2}$ alone.

Because the second ionization means pulling a positively charged proton off a species that is already NEGATIVELY charged ($HCO_{3}^{-}$). The electrostatic attraction to overcome is much stronger than for pulling the first proton off a neutral molecule. The gap is typically 4 to 6 orders of magnitude, here close to 4.

The student announces pH $=0.70$ instead of 3.68: nearly 3 units off, so a factor of a thousand on $[H_{3}O^{+}]$. The error is treating a WEAK acid as if it were strong AND fully diprotic: the student assumes both protons leave, and leave completely. In reality the first leaves only to the extent of 0.2%, and the second practically never. A reaction equation written with a single arrow is a chemical hypothesis, not a convenient shorthand.

Strong acid: $[H_{3}O^{+}]=0.100$ mol/L, $pH=1.00$.

Initial moles of acid: $(25.0)(0.100)=2.50$ mmol; moles of base added: 1.25 mmol. There remains 1.25 mmol of $H_{3}O^{+}$ in $25.0+12.5=37.5$ mL: $[H_{3}O^{+}]=\frac{1.25}{37.5}=0.0333$ mol/L, $pH=1.48$. At the halfway point of the titration the pH has barely moved: the curve is flat at the start, and all the action is elsewhere.

Equivalence occurs when the moles of base equal the moles of acid: 2.50 mmol, that is 25.0 mL. Only $Na^{+}$ and $Cl^{-}$ then remain in solution, two SPECTATOR ions (conjugates of a strong base and a strong acid: neither touches the protons of the water). Only the autoionization of water sets the pH: exactly 7.00 at 25 °C. This perfect neutrality is unique to the strong-strong titration: with a weak acid, the equivalence point would be basic.

Excess base: $(26.0-25.0)(0.100)=0.10$ mmol of $OH^{-}$ in 51.0 mL: $[OH^{-}]=1.96\times 10^{-3}$ mol/L, $pOH=2.71$, $pH=11.29$. One single millilitre past the equivalence point, and the pH has jumped from 7.0 to 11.3: the curve is nearly vertical around the equivalence point. It is this jump, more than 4 units for 1 mL, that makes the titration precise.

Yes, both are suitable: their transition ranges (6 to 8, and 8 to 10) both fall INSIDE the pH jump (roughly from 4 to 11 around the equivalence point). Since the jump spans several units in a single drop, any indicator that changes in this window changes colour at the right volume, to within one drop. The choice becomes critical only when the jump is short, as in the titration of a weak acid, where only phenolphthalein, on the right side of the jump, is suitable.

$pH=pK_{a}+\log\frac{[base]}{[acid]}=-\log(1.8\times 10^{-5})+\log\frac{0.50}{0.50}=4.74+0=4.74$. When the conjugate base and acid are equal, the logarithm vanishes: $pH=pK_{a}$. This is the centre of gravity of the buffer, where it is most effective.

The HCl added (a strong acid) is neutralized by the base of the buffer: $CH_{3}COO^{-}+H_{3}O^{+}\rightarrow CH_{3}COOH+H_{2}O$. The base goes from 0.50 to 0.45 mol/L, the acid from 0.50 to 0.55 mol/L. $pH=4.74+\log\frac{0.45}{0.55}=4.74-0.09=4.66$. The pH has barely moved, a drop of only 0.09.

In pure water, nothing absorbs the acid: $[H_{3}O^{+}]=0.050$ mol/L and $pH=-\log(0.050)=1.30$. The water goes from pH 7.0 to pH 1.3: a drop of 5.7 units, against 0.09 for the buffer. The same attack, met by a response hundreds of thousands of times gentler in ion concentration: that is the operational definition of a buffer.

The buffer holds BOTH halves of a pair IN RESERVE: its conjugate base swallows added acids, its acid neutralizes added bases; an addition only shifts the base/acid ratio slightly, and the logarithm damps that shift further. Blood works this way with the pair $H_{2}CO_{3}/HCO_{3}^{-}$: enzymes tolerate only a narrow pH range around 7.4, whereas physical effort continuously releases CO2 and metabolic acids. Without a buffer, the first sprint would be the last: the chemistry of solutions in this chapter is, quite literally, a matter of survival.

Weak acid alone: $[H_{3}O^{+}]=\sqrt{K_{a}C}=\sqrt{(1.8\times 10^{-5})(0.100)}=1.34\times 10^{-3}$ mol/L, pH $=2.87$.

12.50 mL is the HALF-EQUIVALENCE point: half of the acid has been converted to acetate, so $[CH_{3}COOH]=[CH_{3}COO^{-}]$. Henderson-Hasselbalch then gives pH $=pK_{a}+\log 1=pK_{a}=4.74$. It is the most convenient point of the curve because it delivers the $pK_{a}$ by simple READING, with no calculation at all, and without even knowing the concentrations: it is the standard experimental method for determining a $K_{a}$.

$n_{acid}=(0.02500)(0.100)=2.50\times 10^{-3}$ mol, so $V_{eq}=25.00$ mL. At the equivalence point ONLY acetate remains, in $50.00$ mL: $[CH_{3}COO^{-}]=\frac{2.50\times 10^{-3}}{0.05000}=0.0500$ mol/L. This weak base hydrolyzes with $K_{b}=\frac{K_{w}}{K_{a}}=\frac{1.0\times 10^{-14}}{1.8\times 10^{-5}}=5.56\times 10^{-10}$. $[OH^{-}]=\sqrt{(5.56\times 10^{-10})(0.0500)}=5.27\times 10^{-6}$, pOH $=5.28$, so pH $=8.72$.

Beyond the equivalence point, it is the excess strong base that sets everything: $n_{excess}=(0.03000-0.02500)(0.100)=5.00\times 10^{-4}$ mol in $55.00$ mL, that is $[OH^{-}]=9.09\times 10^{-3}$ mol/L. pOH $=2.04$, pH $=11.96$. Note that the acetate present no longer matters: its contribution is swamped by the free sodium hydroxide.

Because at the equivalence point the solution is not neutral salt water: it contains the conjugate base $CH_{3}COO^{-}$, which hydrolyzes and makes the medium BASIC. The equivalence pH of a weak acid-strong base titration is always above 7 (and the more so the weaker the acid). An indicator is therefore needed whose transition range brackets 8.72: PHENOLPHTHALEIN. Methyl red would change colour around pH 5, that is, well before the equivalence point, right in the buffer region where the pH rises slowly: the titration error would be gross.

$NaCl$: strong base ($NaOH$) + strong acid ($HCl$), neither ion reacts, solution NEUTRAL. $CH_{3}COONa$: strong base + WEAK acid, the anion hydrolyzes, solution BASIC. $NH_{4}Cl$: WEAK base + strong acid, the cation hydrolyzes, solution ACIDIC. $CH_{3}COONH_{4}$: weak base + weak acid, both react, so the constants must be compared.

$CH_{3}COO^{-}$ is a weak base with constant $K_{b}=\frac{K_{w}}{K_{a}}=\frac{1.0\times 10^{-14}}{1.8\times 10^{-5}}=5.56\times 10^{-10}$. $[OH^{-}]=\sqrt{(5.56\times 10^{-10})(0.10)}=7.45\times 10^{-6}$ mol/L, pOH $=5.13$, pH $=8.87$: basic, as predicted.

$NH_{4}^{+}$ is a weak acid with constant $K_{a}=\frac{K_{w}}{K_{b}}=\frac{1.0\times 10^{-14}}{1.8\times 10^{-5}}=5.56\times 10^{-10}$. $[H_{3}O^{+}]=\sqrt{(5.56\times 10^{-10})(0.10)}=7.45\times 10^{-6}$ mol/L, pH $=5.13$. The two pH values are SYMMETRICAL about 7: $8.87+5.13=14.00$. This is not a coincidence of the calculation but a consequence of the fact that $K_{a}(CH_{3}COOH)$ and $K_{b}(NH_{3})$ are both equal to $1.8\times 10^{-5}$, which makes the two ions exactly as weak as each other.

In $CH_{3}COONH_{4}$, the acetate pulls the pH up with $K_{b}=5.56\times 10^{-10}$ and the ammonium pulls it down with $K_{a}=5.56\times 10^{-10}$. The two tendencies are rigorously equal, so they cancel: the solution is NEUTRAL, pH $=7.00$. The general rule is the comparison: if $K_{a}(\text{cation})>K_{b}(\text{anion})$ the solution is acidic, if $K_{a}<K_{b}$ it is basic, and equality gives neutrality. A salt “of a weak acid and a weak base” is therefore not neutral by nature: it is neutral here by numerical equality.

Because $Cl^{-}$ is the conjugate base of $HCl$, a STRONG acid. A strong acid is by definition completely dissociated in water, which means its conjugate base has no affinity for the proton: it is infinitely weak and cannot take it back from the water. The stronger an acid, the weaker its conjugate base; at the limit of strong acids, it is chemically inert toward water. Since $CH_{3}COOH$ is weak, its acetate keeps, by contrast, a real affinity for $H^{+}$ and hydrolyzes.

A buffer is effective when the target pH is close to the $pK_{a}$ of the pair, say within one unit, because that is where the ratio of the two forms stays close to 1 and both reserves are plentiful. Here $pK_{a}=-\log(1.8\times 10^{-5})=4.74$, 0.24 unit from the target: ideal. Formic acid ($3.75$) would require a ratio of $10^{0.75}\approx 5.6$ and ammonium ($9.25$) a ratio of $10^{-4.75}$, meaning one of the two forms is almost absent: the buffer would have no reserve on one side and would collapse at the first addition.

Henderson-Hasselbalch: $4.50=4.74+\log\frac{[A^{-}]}{[HA]}$, so $\log\frac{[A^{-}]}{[HA]}=-0.245$ and $\frac{[A^{-}]}{[HA]}=10^{-0.245}=0.569$. A little more acid than conjugate base is therefore needed, which is consistent with a pH below the $pK_{a}$.

With $[A^{-}]+[HA]=0.200$ and $[A^{-}]=0.569[HA]$: $[HA]=\frac{0.200}{1.569}=0.127$ mol/L and $[A^{-}]=0.0725$ mol/L. Mass of sodium acetate: $(0.0725)(82.03)=5.95$ g, to be completed with 0.127 mol of acetic acid.

The $HCl$ added converts acetate into acetic acid, mole for mole: $[A^{-}]=0.0725-0.050=0.0225$ mol/L and $[HA]=0.127+0.050=0.177$ mol/L. pH $=4.74+\log\frac{0.0225}{0.177}=4.74-0.896=3.85$. The drop is only 0.65 unit, where the same addition to a litre of pure water would send the pH down to 1.3.