Question

buffer consists of 0.50 M NaH2PO4 and 0.40 M Na2HPO4. Phosphoric acid is a triprotic acid (K a 1 = 7.2 × 10 − 3, K a 2 = 6.3 × 10 − 8, and K a 3 = 4.2 × 10 − 13). (a) Which Ka value is most important to this buffer? (b) What is the buffer pH?

Answer 1

Answer:

The Ka value used for the buffer system is :- Ka2 = $6.3\times 10^{-8}$ pKa2 = 7.2

pH = 7.1

Explanation:

The buffer system given in the question is :-

$NaH_2PO_4$ and $Na_2HPO_4$

The reaction taking place is:-

$H_2PO_4^-\rightleftharpoons HPO_4^{2-}$

The Ka value used for the buffer system is :- Ka2 = $6.3\times 10^{-8}$ pKa2 = 7.2

The pH can be calculated as:-

$pH=pKa+\log\frac{Na_2HPO_4}{NaH_2PO_4}$

pH = 7.2 + log 0.40/0.50 = 7.1

Answer 2

The most important Ka for the buffer consisting of 0.50 M NaH₂PO₄ and 0.40 M Na₂HPO₄ is  Ka₂ (6.3 × 10⁻⁸) and the pH of the buffer is 7.1.

A buffer consists of 0.50 M NaH₂PO₄ and 0.40 M Na₂HPO₄. The net ionic equation that represents the relation between both species is:

H₂PO₄⁻ =  HPO₄²⁻ + H⁺      Ka₂ = 6.3 × 10⁻⁸

We can calculate the pH of this buffer using Henderson-Hasselbach's equation.

$pH = pKa_2 + log \frac{[HPO_4^{2-} ]}{[H_2PO_4^{-} ]} \\\\pH = -log 6.3 \times 10^{-8} + log \frac{0.40}{0.50} = 7.1$

The most important Ka for the buffer consisting of 0.50 M NaH₂PO₄ and 0.40 M Na₂HPO₄ is  Ka₂ (6.3 × 10⁻⁸) and the pH of the buffer is 7.1.

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