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Chapter 22: Problem 28

When the conjugate acid of aniline, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}{ }^{+}\), reacts with the acetate ion, the following reaction takes place: \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}{ }^{+}(a q)+\mathrm{CH}_{3} \mathrm{COO}^{-}(a q) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}(a q)+\mathrm{CH}_{3} \mathrm{COOH}(a q)\) If \(K_{\mathrm{a}}\) for \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}{ }^{+}\) is \(1.35 \times 10^{-5}\) and \(K_{\mathrm{a}}\) for \(\mathrm{CH}_{3} \mathrm{COOH}\) is \(1.86 \times 10^{-5}\) what is \(K\) for the reaction?

Short Answer

Expert verified
Answer: The equilibrium constant (K) for the reaction between aniline and acetic acid is approximately 1.38.

Step by step solution

01

Identify the Ka values provided

We are given that: \(K_{a}\)(C6H5NH3+) = \(1.35 \times 10^{-5}\), this is for the conjugate acid of aniline \(K_{a}\)(CH3COOH) = \(1.86 \times 10^{-5}\), this is for the acetic acid
02

Apply the relationship to find the equilibrium constant of the reaction

We know that \(K_{reaction} = \frac{K_{a \space acid}}{K_{a \space conjugate \space acid}}\). In this case, our reaction involves the acid CH3COOH and the conjugate acid C6H5NH3+. So, we can plug in the given Ka values to find the equilibrium constant: \(K = \frac{K_{a \space (CH3COOH)}}{K_{a \space (C6H5NH3+)}}\) \(K = \frac{1.86 \times 10^{-5}}{1.35 \times 10^{-5}}\)
03

Calculate the equilibrium constant K

Now divide the given values to find the equilibrium constant K: \(K \approx \frac{1.86}{1.35}\) \(K \approx 1.38\) Thus, the equilibrium constant K for the reaction is approximately 1.38.

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