= Glycine speciation model
In an ideal aqueous <chemical solution>, set $h=[\mathrm H_3\mathrm O^+]/c^\circ=10^{-\mathrm{pH}}$, with $c^\circ=1\,\mathrm{mol\,L^{-1}}$, and approximate <activity coefficients> by one. Put $K_1=10^{-2.35}$ and $K_2=10^{-9.78}$ for the stipulated <dimensionless> <acid dissociation constants>. The <Henderson-Hasselbalch equation> gives the adjacent-state <ratios>
$$
\frac{[G^0]}{[G^+]}=\frac{K_1}{h},\qquad\frac{[G^-]}{[G^0]}=\frac{K_2}{h}.
$$
With <analytical concentration> $C_T=[G^+]+[G^0]+[G^-]$, normalization gives the three fractions and mean <electric charge> in units of the <elementary charge>:
$$
D=h^2+K_1h+K_1K_2,\qquad
\alpha_+=\frac{h^2}{D},\quad\alpha_0=\frac{K_1h}{D},\quad\alpha_-=\frac{K_1K_2}{D},\qquad
\bar z=\alpha_+-\alpha_-.
$$
All three fractions are nonnegative and sum to one. At <pH> $7$, the model gives about $99.832\%$ <zwitterion>, $0.00224\%$ <cation> and $0.1657\%$ <anion>; $\bar z\approx-0.00163$. Predominantly <zwitterionic> therefore need not mean exactly at the <isoelectric point>.
Setting $\bar z=0$ requires $h^2=K_1K_2$. Taking the <logarithm> yields
$$
\mathrm{pI}=\frac{\mathrm pK_{a1}+\mathrm pK_{a2}}2=\frac{2.35+9.78}2=6.065.
$$
At this model <isoelectric point>, $\alpha_+=\alpha_-\approx0.0001927$ and $\alpha_0\approx0.9996146$: about $0.0193\%$ of each charged form remains. At either <pKa>, the relevant adjacent fractions are equal, but the third fraction is not mathematically zero. This three-state model neglects the minor uncharged form $\mathrm{H_2N{-}CH_2{-}COOH}$; macroscopic <pKa> values alone do not determine its fraction separately from the <zwitterion>.
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