Question

A buffer at pH 7.45 is prepared by mixing solutions of KH2PO4 and K2HPO4. Which of the following ratios of [base]/[acid] is required? For phosphoric acid, Ka1 = 7.5 × 10-3 Ka2 = 6.2 × 10-8 Ka3 = 4.2 × 10-13

Answer 1

The pH of the mixed buffer will be

`([Base])/([Acid])` = 1.737

Step-by-step explanation:

Buffer is prepared by mixing solutions of KH₂PO₄ and K₂HPO₄

H₂PO₄⁻ ⇄ HPO₄²⁻ + H⁺

Using Henderson equation

⇒ pH = pKa₂ + log
`([Base])/([Acid])`

⇒ 7.45 = -log(6.2 x 10⁻⁸) + log
`([Base])/([Acid])`

⇒ log
`([Base])/([Acid])` = 7.45 - 7.2

⇒
`([Base])/([Acid])` =
`10^(0.24)`

⇒
`([Base])/([Acid])` = 1.737

So, pH of Buffer solution is equal to 1.737