Question

5.单选题 (10分) 已知h₃po₄的pka₁ = 2.16, pka₂ = 7.21, pka₃ = 12.32。0.20 mol·l⁻¹ h₃po₄溶液与0.20 mol·l⁻¹的na₃po₄溶液等体积混合后溶液的ph为 a 4.69 b 7.21 c 6.79 d 9.77

Explanation:

Step1: Determine the reaction

The reaction between \(H_3PO_4\) and \(Na_3PO_4\) is \(H_3PO_4+Na_3PO_4 = 2NaH_2PO_4\). But since they are in equal volume and concentration, the actual reaction is \(H_3PO_4+Na_3PO_4=2Na_2HPO_4\). The resulting solution is a buffer solution. The Henderson - Hasselbalch equation for a buffer is \(pH = pK_{a2}+\log\frac{[HPO_4^{2 -}]}{[H_2PO_4^-]}\). In this case, after the reaction, the concentrations of \(H_2PO_4^-\) and \(HPO_4^{2 -}\) are equal (because of the stoichiometry of the reaction \(H_3PO_4+Na_3PO_4 = 2Na_2HPO_4\), assume volume doubles, original \(n(H_3PO_4)=n(Na_3PO_4)\), after reaction \(n(H_2PO_4^-)=n(HPO_4^{2 -})\), and \(C=\frac{n}{V}\), \(V\) is the same for both species in the final solution).

Step2: Calculate pH

When \(\frac{[HPO_4^{2 -}]}{[H_2PO_4^-]} = 1\), \(\log\frac{[HPO_4^{2 -}]}{[H_2PO_4^-]}=0\). According to the Henderson - Hasselbalch equation \(pH=pK_{a2}\). Given \(pK_{a2}=7.21\)

Answer:

B. 7.21