Answer:
[tex]\frac{[HPO_4^{2-}]}{[H_2PO_4^{-}]}=1.26[/tex]
Explanation:
Considering the Henderson- Hasselbalch equation for the calculation of the pH of the buffer solution as:
pH=pKa+log[base]/[acid]
For the equilibrium buffer of dihydrogen phosphate, pKa = 7.10
pH = 7.20
[tex]pH=pKa+\log\frac{[base]}{[acid]}[/tex]
[tex]7.20=7.10+\log\frac{[HPO_4^{2-}]}{[H_2PO_4^{-}]}[/tex]
[tex]\log\frac{[HPO_4^{2-}]}{[H_2PO_4^{-}]}=0.10[/tex]
[tex]\frac{[HPO_4^{2-}]}{[H_2PO_4^{-}]}=1.26[/tex]
