Take the pKa values of phosphoric acid as 2.15, 7.20 and 12.35. A 100 mL portion of 0.200 M Na3PO4 is mixed with 100 mL of 0.250 M HCl, and to the…

Work in millimoles. Initially there are $20.0\ \mathrm{mmol\ PO_4^{3-}}$ and $25.0\ \mathrm{mmol\ H^+}$. The acid first protonates $\mathrm{PO_4^{3-}}$ to $\mathrm{HPO_4^{2-}}$ (all $20.0$ mmol), and the remaining $5.0\ \mathrm{mmol\ H^+}$ converts $5.0\ \mathrm{mmol}$ of $\mathrm{HPO_4^{2-}}$ to $\mathrm{H_2PO_4^-}$, leaving $15.0\ \mathrm{mmol\ HPO_4^{2-}}$ and $5.0\ \mathrm{mmol\ H_2PO_4^-}$ (pH $=7.20+\log3=7.68$ at this stage). Now $20.0\ \mathrm{mL}\times0.500\ \mathrm{M}=10.0\ \mathrm{mmol\ OH^-}$ is added. The first $5.0$ mmol turns all of the $\mathrm{H_2PO_4^-}$ into $\mathrm{HPO_4^{2-}}$ (giving $20.0$ mmol), and the second $5.0$ mmol converts $5.0\ \mathrm{mmol\ HPO_4^{2-}}$ to $\mathrm{PO_4^{3-}}$. The mixture is $15.0\ \mathrm{mmol\ HPO_4^{2-}}$ and $5.0\ \mathrm{mmol\ PO_4^{3-}}$, a buffer governed by $\mathrm{pK_{a3}}$: $\mathrm{pH}=12.35+\log(5/15)=12.35-0.477=11.87$. Volume cancels in the ratio. (A full equilibrium treatment, which includes the appreciable hydrolysis of $\mathrm{PO_4^{3-}}$ at this dilution, gives about $11.72$; the question fixes the Henderson–Hasselbalch approximation.)