If enough of a monoprotic acid is dissolved in water to produce a 0.0200 M solution with a pH of 6.70, what is the equilibrium constant, $K_a$, for the acid? Number $K_a = 2 imes 10^{-12}$ Incorrect. You have not accounted for the autoprotolysis of water when calculating the $K_a$ value. For weak acids that are so dilute or so weak that the pH of the solution lies between 6 and 7, the autoprotolysis of water must be taken into account when determining the $K_a$ value. Start by writing four equations: 1) the $K_w$ expression 2) the $K_a$ expression 3) the charge balance of the solution 4) the material (mass) balance of the solution Use these four equations to develop an expression for $K_a$ in terms of $K_w$, $[H_3O^+]$, and the initial concentration of the acid, $[HA]_{initial}$. Recall that $[H_3O^+] = 10^{-pH}$
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Step 1: Write the equation for the autoprotolysis of water: \[2H_2O \rightleftharpoons H_3O^+ + OH^-\] Show more…
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You have not accounted for the autoprotolysis of water when calculating the Ka value. For weak acids that are so dilute or so weak that the pH of the solution lies between 6 and 7, the autoprotolysis of water must be taken into account when determining the Ka value. Start by writing four equations: 1. The Kw expression 2. The Ka expression 3. The charge balance of the solution 4. The material (mass) balance of the solution Use these four equations to develop an expression for Ka in terms of Kw, [H3O+], and the initial concentration of the acid, [HA]initial. Recall that [H3O+] = 10^-pH
David C.
A The hydrogen phthalate ion, $\mathrm{C}_{8} \mathrm{H}_{5} \mathrm{O}_{4}^{-}$, is a weak acid with $K_{\mathrm{a}}=3.91 \times 10^{-6}$ $\mathrm{C}_{8} \mathrm{H}_{5} \mathrm{O}_{4}^{-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightleftarrows \mathrm{C}_{8} \mathrm{H}_{4} \mathrm{O}_{4}^{2-}(\mathrm{aq})+\mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})$ What is the pH of a $0.050 \mathrm{M}$ solution of potassium hydrogen phthalate, $\mathrm{KC}_{8} \mathrm{H}_{5} \mathrm{O}_{4} ?$ Note: To find the $\mathrm{pH}$ for a solution of the anion, we must take into account that the ion is amphiprotic. It can be shown that, for most cases of amphiprotic ions, the $\mathrm{H}_{3} \mathrm{O}^{+}$ concentration is (EQUATION CAN'T COPY) where $K_{1}$ here is for phthalic acid, $\mathrm{C}_{8} \mathrm{H}_{6} \mathrm{O}_{4}(=1.12 \times$ $\left.10^{-3}\right),$ and $K_{2}$ is for the hydrogen phthalate anion $\left(=3.91 \times 10^{-6}\right)$
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