Chemistry - Ionic Equilibrium Question with Solution | TestHub

At the second half-equivalence point:

${\mathrm{Na}}_2{\mathrm{HPO}}_4+\mathrm{HCl}\to \mathrm{NaCl}+{\mathrm{NaH}}_2{\mathrm{PO}}_4$ Initial moles: 100 50 - - 50 × 50 50

$$\mathrm{pH}={\mathrm{pK}}_2+\log \left(\frac{[{\mathrm{Na}}_2{\mathrm{HPO}}_4]}{[{\mathrm{NaH}}_2{\mathrm{PO}}_4]}\right)$$

$$\mathrm{pH}={\mathrm{pK}}_2+\log \left(\frac{50}{50}\right)={\mathrm{pK}}_2+\log (1)={\mathrm{pK}}_2$$

Given ${\mathrm{pK}}_2=8$,

$$\mathrm{pH}=8$$