Writean equation for the reaction of chloroaceticacid \(\left(K_{\mathrm{a}}=1.5 \times 10^{-3}\right)\) with trimethylamine \(\left(K_{\mathrm{b}}=5.9 \times 10^{-5}\right) .\) Calculate the equilibrium constant for the reaction. If \(0.10 M\) solutions of these two species are mixed, what will be their concentrations at equilibrium?

The reaction between chloroacetic acid (HClCH2COOH) and trimethylamine (N(CH3)3) involves the transfer of a proton from the acid to the base, producing the chloroacetate ion (ClCH2COO-) and the trimethylammonium ion (N(CH3)3H+). The balanced chemical equation is as follows: $$\text{HClCH}_{2}\text{COOH} + \text{N}(\text{CH}_{3})_{3} \rightleftharpoons \text{ClCH}_{2}\text{COO}^- + \text{N}(\text{CH}_{3})_{3}\text{H}^+$$

We are given the \(K_{a}\) value for chloroacetic acid (1.5 × 10⁻³) and the \(K_{b}\) value for trimethylamine (5.9 × 10⁻⁵). We can calculate the equilibrium constant for the reaction (K) using the expression: $$K = \frac{K_w}{K_{a} \times K_{b}}$$ where \(K_w\) is the ionic product of water (\(1.0 \times 10^{-14}\) at 25 °C). Plugging in the given values, we have: $$K = \frac{1.0 \times 10^{-14}}{(1.5 \times 10^{-3})(5.9 \times 10^{-5})}$$ $$K = 1.13 \times 10^{3}$$

We will now set up an ICE (Initial, Change, Equilibrium) table for the reaction, using the initial concentrations of 0.10 M for both the acid and the base: | | HClCH2COOH | N(CH3)3 | ClCH2COO- | N(CH3)3H+ | |------------------|------------|-----------|-----------|-----------| | Initial (M) | 0.10 | 0.10 | 0 | 0 | | Change (M) | -x | -x | +x | +x | | Equilibrium (M) | 0.10 - x | 0.10 - x | x | x | Next, we will use the expressions for K provided above as well as the equilibrium concentrations in the table to solve for x. $$K = \frac{[\text{ClCH}_{2}\text{COO}^-][\text{N}(\text{CH}_{3})_{3}\text{H}^+]}{[\text{HClCH}_{2}\text{COOH}][\text{N}(\text{CH}_{3})_{3}]}$$ $$1.13 \times 10^{3} = \frac{x^2}{(0.10 - x)^2}$$ Here, we assume that x is much smaller than 0.10, so we approximate \((0.10 - x) \approx 0.10\). The equation then becomes: $$1.13 \times 10^{3} = \frac{x^2}{(0.10)^2}$$ Solving for x, we get: $$x^2 = 1.13 \times 10^{-1}$$ $$x = \sqrt{1.13 \times 10^{-1}}$$ $$x \approx 0.0336$$ Since we assumed x is much smaller than 0.10, this approximation is reasonable. Now that we have the value of x, we can calculate the equilibrium concentrations: $$[\text{HClCH}_{2}\text{COOH}] = 0.10 - x = 0.10 - 0.0336 \approx 0.0664 \, M$$ $$[\text{N}(\text{CH}_{3})_{3}] = 0.10 - x = 0.10 - 0.0336 \approx 0.0664 \, M$$ $$[\text{ClCH}_{2}\text{COO}^-] = x = 0.0336 \, M$$ $$[\text{N}(\text{CH}_{3})_{3}\text{H}^+] = x = 0.0336 \, M$$ The equilibrium concentrations of chloroacetic acid, trimethylamine, chloroacetate ion, and trimethylammonium ion are approximately 0.0664 M, 0.0664 M, 0.0336 M, and 0.0336 M, respectively.