QUESTION IMAGE
Question
- use tables in chapter 16 to determine which side of the following equilibrium equations is favored at equilibrium.
a. ch₃cooh(aq) + co₃²⁻(aq) ⇌ ch₃coo⁻(aq) + hco₃⁻(aq)
b. hso₄⁻(aq) + no₂⁻(aq) ⇌ so₄²⁻(aq) + hno₂(aq)
c. nh₄⁺(aq) + ch₃o⁻(aq) ⇌ nh₃(aq) + ch₃oh(aq)
Step1: Recall acid - base equilibrium principle
The side of the equilibrium that is favored is the side with the weaker acid and weaker base. We will use acid - dissociation constants ($K_a$) values from the tables in Chapter 16 to compare the acids on both sides of the equation.
Step2: Analyze equation a
For the reaction $\mathrm{CH_3COOH(aq)+CO_3^{2 - }(aq)
ightleftharpoons CH_3COO^-(aq)+HCO_3^-(aq)}$, the acid on the left - hand side is $\mathrm{CH_3COOH}$ and on the right - hand side is $\mathrm{HCO_3^-}$. Looking up the $K_a$ values, $K_a(\mathrm{CH_3COOH}) = 1.8\times10^{-5}$ and $K_a(\mathrm{HCO_3^-})=4.8\times 10^{-11}$. Since $K_a(\mathrm{HCO_3^-})<K_a(\mathrm{CH_3COOH})$, $\mathrm{HCO_3^-}$ is the weaker acid. The right - hand side is favored at equilibrium.
Step3: Analyze equation b
For the reaction $\mathrm{HSO_4^-(aq)+NO_2^-(aq)
ightleftharpoons SO_4^{2 - }(aq)+HNO_2(aq)}$, the acid on the left - hand side is $\mathrm{HSO_4^-}$ and on the right - hand side is $\mathrm{HNO_2}$. $K_a(\mathrm{HSO_4^-})=1.2\times10^{-2}$ and $K_a(\mathrm{HNO_2}) = 4.5\times10^{-4}$. Since $K_a(\mathrm{HNO_2})<K_a(\mathrm{HSO_4^-})$, $\mathrm{HNO_2}$ is the weaker acid. The right - hand side is favored at equilibrium.
Step4: Analyze equation c
For the reaction $\mathrm{NH_4^+(aq)+CH_3O^-(aq)
ightleftharpoons NH_3(aq)+CH_3OH(aq)}$, the acid on the left - hand side is $\mathrm{NH_4^+}$ and on the right - hand side is $\mathrm{CH_3OH}$. $K_a(\mathrm{NH_4^+})=5.6\times10^{-10}$ and $K_a(\mathrm{CH_3OH})\approx1\times10^{-16}$. Since $K_a(\mathrm{CH_3OH})<K_a(\mathrm{NH_4^+})$, $\mathrm{CH_3OH}$ is the weaker acid. The right - hand side is favored at equilibrium.
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