$$\begin{gathered} {\mathrm{H}}_2\mathrm{O}+{\mathrm{H}}_3{\mathrm{PO}}_4\rightleftharpoons {\mathrm{H}}_3{\mathrm{O}}^{+}+{\mathrm{H}}_2{\mathrm{PO}}_4^{-},{\mathrm{pK}}_1=2.15 \\ {\mathrm{H}}_2\mathrm{O}+{\mathrm{H}}_2{\mathrm{PO}}_4^{-}\rightleftharpoons {\mathrm{H}}_3{\mathrm{O}}^{+}+{\mathrm{HPO}}_4^{2-},{\mathrm{pK}}_2=7.20 \end{gathered}$$

Hence, pH of 0.01 M NaH₂PO₄ is:

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