Question 16.19: Predict whether the salt Na2HPO4 forms an acidic solution or...
Predict whether the salt Na_{2}HPO_{4} forms an acidic solution or a basic solution when dissolved in water.
Analyze We are asked to predict whether a solution of Na_{2}HPO_{4} is acidic or basic. This substance is an ionic compound composed of Na^{+} and HPO^{2-}_{4} ions.
Plan We need to evaluate each ion, predicting whether it is acidic or basic. Because Na^{+} is a cation of Group 1, it has no influence on pH. Thus, our analysis of whether the solution is acidic or basic must focus on the behavior of the HPO^{2-}_{4} ion. We need to consider that HPO^{2-}_{4} can act as either an acid or a base:
As acid HPO^{2-}_{4} (aq) ⇌ H^{+} (aq) + PO^{3-}_{4} (aq) [16.46]
As base HPO^{2-}_{4} (aq) + H_{2}O ⇌ H_{2}PO^{-}_{4} (aq) + OH^{-} (aq) [16.47]
Of these two reactions, the one with the larger equilibrium constant determines whether the solution is acidic or basic.
Solve The value of K_{a} for Equation 16.46 is K_{a3} for H_{3}PO_{4}: 4.2 × 10^{-13} (Table 16.3). For Equation 16.47, we must calculate K_{b} for the base HPO^{2-}_{4} from the value of K_{a} for its conjugate acid, H_{2}PO^{-}_{4}, and the relationship K_{a} × K_{b} = K_{w} (Equation 16.40). The relevant value of K_{a} for H_{2}PO^{-}_{4} is K_{a2} for H_{3}PO_{4}: 6.2 × 10^{-8} (from Table 16.3). We therefore have
| TABLE 16.3 Acid-Dissociation Constants of Some Common Polyprotic Acids |
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| Polyprotic Acids | Formula | K_{a1} | K_{a2} | K_{a3} |
| Ascorbic | H_{2}C_{6}H_{6}O_{6} | 8.0 × 10^{-5} | 1.6 × 10^{-12} | |
| Carbonic | H_{2}CO_{3} | 4.3 × 10^{-7} | 5.6 × 10^{-11} | |
| Citric | H_{3}C_{6}H_{5}O_{7} | 7.4 × 10^{-4} | 1.7 × 10^{-5} | 4.0 × 10^{-7} |
| Oxalic | HOOC—COOH | 5.9 × 10^{-2} | 6.4 × 10^{-5} | |
| Phosphoric | H_{3}PO_{4} | 7.5 × 10^{-3} | 6.2 × 10^{-8} | 4.2 × 10^{-13} |
| Sulfurous | H_{2}SO_{3} | 1.7 × 10^{-2} | 6.4 × 10^{-8} | |
| Sulfuric | H_{2}SO_{4} | Large | 1.2 × 10^{-2} | |
| Tartaric | C_{2}H_{2}O_{2}(COOH)_{2} | 1.0 × 10^{-3} | 4.6 × 10^{-5} | |
K_{a} × K_{b} = K_{w} (for a conjugate acid-base pair) [16.40]
K_{b}(HPO^{2-}_{4} ) × K_{a}(H_{2}PO^{-}_{4}) = K_{w} = 1.0 × 10^{-14}K_{b}(HPO^{2-}_{4} ) =\frac{1.0 × 10^{-14}}{6.2 × 10^{-8}} = 1.6 × 10^{-7}.
This K_{b} value is more than 10^{5} times larger than K_{a} for HPO^{2-}_{4}; thus, the reaction in Equation 16.47 predominates over that in Equation 16.46, and the solution is basic.