How do we calculate the $p H$ $pH$ of a buffer that is composed of $H P O_{4}^{2 -}$ $HPO_4^(2-)$ and $H_{2} P O_{4}^{-}$ $H_2PO_4^(-)$ ?

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How do we calculate the $p H$ $pH$ of a buffer that is composed of $H P O_{4}^{2 -}$ $HPO_4^(2-)$ and $H_{2} P O_{4}^{-}$ $H_2PO_4^(-)$ ?

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Answer 1 · 2017-04-13T07:58:43

The buffer equation tells us that $p H = p K_{a} + {log}_{10} (\frac{[H P O_{4}^{2 -}]}{[H_{2} P O_{4}^{-}]})$ $pH=pK_a+log_10([[HPO_4^(2-)]]/[[H_2PO_4^(-)]])$ , but here $p K_{a} = 7.20$ $pK_a=7.20$ for $H_{2} P O_{4}^{-}$ $H_2PO_4^(-)$ .

Explanation:

A buffer is formed by mixing appreciable quantities of a weak acid and its conjugate base. Such a mixture keeps the $p H$ $pH$ of the solution tolerably close to the $p K_{a}$ $pK_a$ of the weak acid.

Now this site gives $p K_{a} = 7.20$ $pK_a=7.20$ for the following reaction:

$H_{2} P O_{4}^{-} + H_{2} O (l) ⇌ H P O_{4}^{2 -}$ $H_2PO_4^(-) + H_2O(l) rightleftharpoonsHPO_4^(2-)$ ,

And thus solutions which contain tolerably equal concentrations of dihydrogen phosphate, and biphosphate, should have $p H$ $pH$ reasonably close to $7.20$ $7.20$ .

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How do we calculate the $p H$ $pH$ of a buffer that is composed of $H P O_{4}^{2 -}$ $HPO_4^(2-)$ and $H_{2} P O_{4}^{-}$ $H_2PO_4^(-)$ ?

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How do we calculate the $p H$ $pH$ of a buffer that is composed of $H P O_{4}^{2 -}$ $HPO_4^(2-)$ and $H_{2} P O_{4}^{-}$ $H_2PO_4^(-)$ ?