WHAT. An amino acid has two reactive groups on the same central carbon: a carboxyl group written −COOH and an amino group written −NH2. Everything about charge comes from these two.
WHY start here. Before we can talk about net charge, we must know exactly which pieces can gain or lose a proton (a proton is just a hydrogen nucleus, symbol H+ — a hydrogen atom that has dropped its one electron, so it carries a single +).
PICTURE. In the figure the central carbon (the α-carbon) holds four things: an −H, a side chain labelled R, the give-away hand −COOH (in coral), and the grabby hand −NH2 (in lavender).
WHAT. Left alone in neutral water, the carboxyl hands its proton to the amino group on the same molecule. We get −COO− and −NH3+ at the same time.
WHY it happens.−COOH is a moderately strong acid; −NH2 is a good base. A strong acid next to a good base loses its proton to that base — even when both are on one molecule. The lower-energy product wins.
PICTURE. The coral hand has passed a small ball (the proton) across to the lavender hand. Now one end is − and one end is +, but the whole molecule sums to zero.
Read the equation left to right: the +1 on the amino nitrogen and the −1 on the carboxyl oxygen are drawn right under the groups that carry them; add them: (+1)+(−1)=0.
WHAT. We now change the pH of the water. pH measures how many free protons H+ float around: low pH = lots of protons; high pH = very few protons (see Acids and bases — pKa and pH).
WHY this knob. Charge on the molecule is entirely decided by which groups are holding a proton. Flooding with protons pushes protons onto the groups; starving of protons pulls protons off. So pH is the single control that moves the molecule through its charge states.
PICTURE. A horizontal pH axis. On the left (low pH) the molecule is protonated everywhere → net +1 (cation). In the middle it is the zwitterion → net 0. On the right (high pH) it has lost the extra proton → net −1 (anion).
So as pH rises, net charge marches +1→0→−1. Somewhere in the middle it passes exactly through 0. That crossing pH is what we are hunting.
WHAT. There are two proton-releasing steps, so two pKa values. Number them by which proton leaves.
Step ① pKa1≈2: the −COOH loses its proton. This turns cation (+1) into zwitterion (0).
Step ② pKa2≈9: the −NH3+ loses its proton. This turns zwitterion (0) into anion (−1).
WHY pKa and not something else. The pKa of a group is the pH at which that group is exactly half protonated, half deprotonated — the tipping point of that one reaction. It answers precisely the question we need: "at what pH does this proton come off?" A small pKa means the proton comes off easily (strong acid); a large pKa means it clings on.
PICTURE. The reaction ladder. Two rungs, each labelled with its pKa, showing the charge before and after each proton departs.
\ \xrightarrow{\,pK_{a1}\,}\
\underbrace{\text{H}_3\overset{+}{\text{N}}-\text{CHR}-\text{COO}^-}_{\text{zwitterion, net } 0}
\ \xrightarrow{\,pK_{a2}\,}\
\underbrace{\text{H}_2\text{N}-\text{CHR}-\text{COO}^-}_{\text{net } -1}$$
Term by term: the **left** species is the acid of step ①; the **middle** species is both the product of step ① *and* the acid of step ②; the **right** species is the product of step ②. The zwitterion is **flanked** by cation and anion — hold that word "flanked," it is the whole idea.
---
## Step 5 — The exact-balance condition (WHERE net charge = 0)
**WHAT.** At the isoelectric point the average net charge is zero. Because the zwitterion is already $0$, the only way the *average* stays $0$ is if the leftover **cation ($+1$)** and **anion ($-1$)** are present in **equal amounts** — they cancel pair-for-pair.
**WHY this is the condition.** Average charge $=$ (fraction cation)$\times(+1)$ + (fraction zwit)$\times 0$ + (fraction anion)$\times(-1)$. Set that to $0$: the two non-zero terms must be equal in size, so $[\text{cation}]=[\text{anion}]$.
**PICTURE.** A balance scale: a small pile of cations on one pan, an equal small pile of anions on the other, the big zwitterion pile sitting on the pivot contributing nothing. The scale is level — that level state *is* the pI.
![[deepdives/dd-chemistry-4.5.02-d2-s05.png]]
> [!formula] The balance condition
> $$[\text{cation}] = [\text{anion}] \quad\Longleftrightarrow\quad \text{net charge} = 0 \quad\Longleftrightarrow\quad \text{pH} = \text{pI}$$
---
## Step 6 — Write each equilibrium with Henderson–Hasselbalch
**WHAT.** We use the [[Henderson–Hasselbalch equation]] once per rung. It links pH, pKa, and the ratio of the base-form to the acid-form of that rung:
$$\text{pH} = pK_a + \log\frac{[\text{base form}]}{[\text{acid form}]}$$
**WHY this equation.** It is the direct translation of "pKa = the tipping pH." When the two forms are equal, the $\log$ of $1$ is $0$ and $\text{pH}=pK_a$. It is the algebraic bridge from *pictures of piles* to a *number for pH*.
**PICTURE.** Two labelled equilibrium bars: rung ① relates cation ↔ zwitterion via $pK_{a1}$; rung ② relates zwitterion ↔ anion via $pK_{a2}$.
![[deepdives/dd-chemistry-4.5.02-d2-s06.png]]
Rung ① — cation is the acid form, zwitterion is the base form:
$$\text{pH} = pK_{a1} + \log\frac{[\text{zwit}]}{[\text{cation}]}$$
Rung ② — zwitterion is the acid form, anion is the base form:
$$\text{pH} = pK_{a2} + \log\frac{[\text{anion}]}{[\text{zwit}]}$$
In each: the $pK_a$ pins the tipping pH; the $\log$ term shifts pH up or down depending on which form dominates.
---
## Step 7 — Add the two equations and collapse the log
**WHAT.** Add rung ① and rung ②. On the left, $\text{pH}+\text{pH}=2\,\text{pH}$.
**WHY add.** Adding lets the two $\log$ terms merge into one (log of a product = sum of logs), and the zwitterion — which we don't want to track — **cancels itself out**.
$$2\,\text{pH} = pK_{a1} + pK_{a2} + \log\!\left(\frac{[\text{zwit}]}{[\text{cation}]}\cdot\frac{[\text{anion}]}{[\text{zwit}]}\right)$$
The $[\text{zwit}]$ on top and bottom cancel:
$$2\,\text{pH} = pK_{a1} + pK_{a2} + \log\frac{[\text{anion}]}{[\text{cation}]}$$
**Now use Step 5's balance:** $[\text{anion}]=[\text{cation}]$, so the ratio is $1$ and $\log 1 = 0$:
$$2\,\text{pH} = pK_{a1} + pK_{a2}$$
**PICTURE.** The two stacked equations physically summing, the $[\text{zwit}]$ terms crossing out, and the $\log 1 \to 0$ term evaporating.
![[deepdives/dd-chemistry-4.5.02-d2-s07.png]]
> [!formula] The result, earned
> $$\boxed{\ \text{pI} = \dfrac{pK_{a1}+pK_{a2}}{2}\ }$$
> The pI is the **midpoint** of the two pKa's that flank the zwitterion — exactly the two reactions that create the cancelling $+$ and $-$.
---
## Step 8 — Edge & degenerate cases (never leave a gap)
**Case A — an *acidic* side chain (three pKa's, e.g. aspartic acid).** The side chain is a *second* $-\text{COOH}$. Now two protons must both come off to reach net-zero, so the neutral form is flanked by the **two lowest** pKa's.
$$\text{pI}_{\text{Asp}} = \frac{2.0 + 3.9}{2} = 2.95$$
The acidic R drags pI **down**.
**Case B — a *basic* side chain (e.g. lysine).** The extra basic nitrogen means the neutral form is flanked by the **two highest** pKa's.
$$\text{pI}_{\text{Lys}} = \frac{9.0 + 10.5}{2} = 9.75$$
The basic R pushes pI **up**.
**Case C — glycine, R = H (no ionisable side chain).** Only the two backbone pKa's exist; the general formula reduces straight back to Step 7:
$$\text{pI}_{\text{Gly}} = \frac{2.34 + 9.60}{2} = 5.97$$
**Degenerate limit — pH far from pI.** If pH $\gg$ pI the molecule is essentially all anion ($-1$); if pH $\ll$ pI it is essentially all cation ($+1$). The balance only holds at the single midpoint. This is exactly why in [[Electrophoresis and separation techniques]] a molecule sits still **only** at its pI and drifts otherwise — and why [[Buffers — why pI matters for solubility]] matters: at pI solubility is lowest.
**PICTURE.** Three mini charge-vs-pH curves side by side — acidic (pI shifted left), neutral (pI central), basic (pI shifted right) — each crossing zero at its own pI.
![[deepdives/dd-chemistry-4.5.02-d2-s08.png]]
> [!mistake] Don't average all three
> With three pKa's, average only the **two flanking the neutral form**, never all three. The third proton is a spectator at the pI.
---
## The one-picture summary
Everything on one canvas: the molecule's charge sliding from $+1$ (left, acid) through the zwitterion ($0$, middle) to $-1$ (right, base) as pH rises; the two pKa's marking the tipping points; and the pI sitting exactly halfway between them where the charge curve crosses zero.
![[deepdives/dd-chemistry-4.5.02-d2-s09.png]]
> [!recall]- Feynman retelling — the whole walkthrough in plain words
> Picture a little person with a give-away hand ($-\text{COOH}$) and a grabby hand ($-\text{NH}_2$). Left alone, the give-away hand passes a ball (a proton) to the grabby hand: now one hand is minus, one is plus, but the person is balanced — the zwitterion.
>
> Now play with the water. Throw *lots* of balls at it (acid, low pH): the minus hand grabs one back, the person turns plus. Snatch balls away (base, high pH): the plus hand loses one, the person turns minus. So there is a special pH in the middle where the pluses and minuses in the crowd exactly cancel — that's the isoelectric point.
>
> Each hand has a "tipping pH" (its pKa) where it's half-holding, half-empty. Writing the balance rule for both hands and adding them, the zwitterion cancels itself, the "equal-piles" fact kills the last log term, and out pops: sit at the **average of the two tipping pH's that hug the zwitterion.** That is the pI. Acidic side chains add a low tipping point (pI drops); basic side chains add a high one (pI rises).
> [!recall]- Quick self-test
> At the pI, are cation and anion present? ::: Yes — in **equal** amounts, so they cancel to net zero (the zwitterion dominates but the traces balance).
> Why average only two of three pKa's for lysine? ::: Only the two pKa's **flanking the neutral form** create the cancelling charges; the third is a spectator at that pH — so average the two **highest**.
> pI of glycine from $pK_{a1}=2.34,\ pK_{a2}=9.60$? ::: $\frac{2.34+9.60}{2}=5.97$.
> At pH below pI, is the molecule + or –? ::: **Positive** (cation) — excess $\text{H}^+$ protonates the groups; it migrates to the cathode.
