Calculate the pH of 0.10 M [latex]Na_2HPO_4[/latex].

We treat [latex]{HPO_4}^{2–}[/latex] as the intermediate form of a diprotic weak acid [latex]{H_2PO_4}^–(aq) \rightleftharpoons {HPO_4}^{2–}(aq) \rightleftharpoons PO_4 ^{3–}(aq)[/latex] where the equilibrium constants are [latex]K_{a2} = 6.32 × 10^{–8} and K_{a3} = 4.5 × 10^{–13} .[/latex] Since the value of [latex]K_{a3} [/latex] is so small, we use equation 6.41 instead of equation 6.42. [latex][H_3O^+]=\sqrt{\frac{K_{a1}K_{a2}C_{HL}+K_{a1}K_{\mathrm{w}}}{C_{HL}+K_{a1}}}[/latex] 6.41 [latex][H_3O^+] = \sqrt{K_{a1}K_{a2}}[/latex] 6.42 [latex][H_3O^+] = \sqrt{\frac{(6.32 \times 10^{-8}) (4.5\times 10^{-13}) (0.10)+ (6.32\times 10^{-8}) (1.00 \times 10^{-14})}{0.10 + 6.32\times 10^{-8}} } [/latex] [latex]=1.86 × 10^{-10} [/latex] or a pH of 9.73.

Question 6.12: Calculate the pH of 0.10 M Na2HPO4.

Modern Analytical Chemistry

Calculate the pH of 0.10 M $Na_2HPO_4$ .

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We treat ${HPO_4}^{2–}$ as the intermediate form of a diprotic weak acid

${H_2PO_4}^–(aq) \rightleftharpoons {HPO_4}^{2–}(aq) \rightleftharpoons PO_4 ^{3–}(aq)$

where the equilibrium constants are $K_{a2} = 6.32 × 10^{–8} and K_{a3} = 4.5 × 10^{–13} .$ Since the value of $K_{a3}$ is so small, we use equation 6.41 instead of equation 6.42.

$[H_3O^+]=\sqrt{\frac{K_{a1}K_{a2}C_{HL}+K_{a1}K_{\mathrm{w}}}{C_{HL}+K_{a1}}}$ 6.41

$[H_3O^+] = \sqrt{K_{a1}K_{a2}}$ 6.42