Question

Identify the strongest acid and the strongest base among the four species in the equilibrium equation below, using only the \(K\) value given.

$$ \mathrm{H}_{2} \mathrm{~S}+\mathrm{OCl}^{-} \leftrightarrow \mathrm{HOCl}+\mathrm{HS}^{-} \quad K=10^{+0.6} $$

2. The transfer of a proton from butyric acid (stomach acid) to acetate can be described by the following reaction:

$$ \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH}+\mathrm{CH}_{3} \mathrm{COO}^{-} \leftrightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COO}^{-}+\mathrm{CH}_{3} \mathrm{COOH} $$

Butyric acid \(+\) Acetate \(\leftrightarrow\) Butyrate \(+\) Acetic acid

a. The equilibrium constant for the above reaction is \(K=0.87,\) and \(K_{a}\) for acetic acid is \(1.74 \times 10^{-5}\). Compute the acidity constant for butyric acid.

b. Write the reaction and determine the value of \(K_{b}\) (the basicity constant) for butyrate.

c. A solution is made by adding some acetic acid and some sodium butyrate to water. List all the species in the equilibrium solution that can act as acids, and rank them from most acidic to least acidic. Do the same for all species that can act as bases.

Answer 1

(a)

\(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH}+\mathrm{CH}_{3} \mathrm{COO} \rightleftharpoons \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COO}+\mathrm{CH}_{3} \mathrm{COOH}\)

\(\mathrm{K}=\frac{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COO}\right]\left[\mathrm{CH}_{3} \mathrm{COOH}\right]}{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH}\right]\left[\mathrm{CH}_{3} \mathrm{COO}\right]}=\frac{\left(\frac{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]}{\left[\mathrm{CH}_{3} \mathrm{COO}\right]\left[\mathrm{H}^{-}\right]}\right)}{\left(\frac{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH}\right]}{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COO}\right]\left[\mathrm{H}^{-}\right]}\right)}=\frac{\left(\frac{1}{\mathrm{~K}_{\mathrm{a}}\left(\mathrm{CH}_{3} \mathrm{COOH}\right)}\right)}{\left(\frac{1}{\mathrm{~K}_{\mathrm{a}}\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH}\right)}\right)}\)

\(\mathrm{K}=\frac{\mathrm{K}_{\mathrm{a}}\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH}\right)}{\mathrm{K}_{\mathrm{a}}\left(\mathrm{CH}_{3} \mathrm{COOH}\right)}\)

\(\mathrm{K}_{\mathrm{a}}\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH}\right)=\mathrm{K} \times \mathrm{K}_{\mathrm{a}}\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\)

\(\mathrm{K}_{\mathrm{a}}\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH}\right)=0.87 \times 1.74 \times 10^{-5}\)

\(\mathrm{K}_{\mathrm{a}}\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH}\right)=1.51 \times 10\)

(b)

\(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COO}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH}+\mathrm{OH}\)

\(\mathrm{K}_{\mathrm{b}}\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH}\right)=\frac{\mathrm{K}_{w}}{\mathrm{~K}_{\mathrm{a}}\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH}\right)}=\frac{1.0 \times 10^{-14}}{1.51 \times 10^{-5}}=6.62 \times 10^{-10}\)

(c)

\(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COONa}+\mathrm{CH}_{3} \mathrm{COOH} \rightleftharpoons \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH}+\mathrm{CH}_{3} \mathrm{COONa}\)

Order of acids from most acidic to least acidic:

\(\mathrm{H}_{3} \mathrm{O}^{-}\)

\(\mathrm{CH}_{3} \mathrm{COOH}\)

\(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH}\)

Order of bases from most basic to least basic:

\(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COO}\)

\(\mathrm{CH}_{3} \mathrm{COO}\)

OH