# Chapter 17 - Questions and Problems - Page 825: 17.19

$Na_2A$ and $NaHA$ are the best choice for $pH = 5.80$.

#### Work Step by Step

1. Calculate the $pK_{a1}$ Value $pK_{a1} = -log(K_{a1})$ $pK_{a1} = -log( 1.1 \times 10^{- 3})$ $pK_{a1} = 2.959$ 2. Calculate the $pK_{a2}$ Value $pK_{a2} = -log(K_{a2})$ $pK_{a2} = -log( 2.5 \times 10^{- 6})$ $pK_{a2} = 5.602$ Therefore, the compounds in the second ionization are more appropriate, because the $pK_{a2}$ is closer to 5.80; Second ionization of a diprotic acid: $HA^-(aq) + H_2O(l) \lt -- \gt A^{2-}(aq) + H_3O^+(aq)$ Since we are using sodium salts: $NaHA$ and $Na_2A$

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