H 2 PO 4 − {\displaystyle {\textrm {H}}_{2}{\textrm {PO}}_{4}^{-}} : dihydrogenophosphate
HPO 4 2 − {\displaystyle {\textrm {HPO}}_{4}^{2-}} : hydrogenophosphate
{ pH = 7,2 + log [ H 2 PO 4 − ] [ HPO 4 2 − ] [ H 2 PO 4 − ] + [ HPO 4 2 − ] = 0.1 ⇔ {\displaystyle \left\{{\begin{matrix}{\textrm {pH}}={\textrm {7,2}}+\log {\frac {\left[{\textrm {H}}_{2}{\textrm {PO}}_{4}^{-}\right]}{\left[{\textrm {HPO}}_{4}^{2-}\right]}}\\\left[{\textrm {H}}_{2}{\textrm {PO}}_{4}^{-}\right]+\left[{\textrm {HPO}}_{4}^{2-}\right]=0.1\end{matrix}}\right.\Leftrightarrow }
pH final = 7,2 + log [ HPO 4 2 − ] − [ HCl ] [ H 2 PO 4 − ] + [ HCl ] {\displaystyle {\textrm {pH}}_{\textrm {final}}={\textrm {7,2}}+\log {\frac {\left[{\textrm {HPO}}_{4}^{2-}\right]-[{\textrm {HCl}}]}{\left[{\textrm {H}}_{2}{\textrm {PO}}_{4}^{-}\right]+[{\textrm {HCl}}]}}}
[ HPO 4 2 − ] − [ HCl ] [ H 2 PO 4 − ] + [ HCl ] {\displaystyle {\frac {\left[{\textrm {HPO}}_{4}^{2-}\right]-[{\textrm {HCl}}]}{\left[{\textrm {H}}_{2}{\textrm {PO}}_{4}^{-}\right]+[{\textrm {HCl}}]}}}
c = n V ⇔ V = n c {\displaystyle c={\frac {n}{V}}\Leftrightarrow V={\frac {n}{c}}}
Δ pH Δ T = 6.8 − 6.9 37 − 0 = − 0.0027 K − 1 {\displaystyle {\frac {\Delta {\textrm {pH}}}{\Delta {\textrm {T}}}}={\frac {6.8-6.9}{37-0}}=-0.0027\ {\textrm {K}}^{-1}}