Answer:
The pH of the mixed buffer will be
[tex]\frac{[Base]}{[Acid]}[/tex] = 1.737
Explanation:
Buffer is prepared by mixing solutions of KH₂PO₄ and K₂HPO₄
H₂PO₄⁻ ⇄ HPO₄²⁻ + H⁺
Using Henderson equation
⇒ pH = pKa₂ + log[tex]\frac{[Base]}{[Acid]}[/tex]
⇒ 7.45 = -log(6.2 x 10⁻⁸) + log[tex]\frac{[Base]}{[Acid]}[/tex]
⇒ log[tex]\frac{[Base]}{[Acid]}[/tex] = 7.45 - 7.2
⇒ [tex]\frac{[Base]}{[Acid]}[/tex] = [tex]10^{0.24}[/tex]
⇒ [tex]\frac{[Base]}{[Acid]}[/tex] = 1.737
So, pH of Buffer solution is equal to 1.737
