Butyric acid is responsible for the foul smell of rancid butter. The \(\mathrm{pK}_{a}\) of butyric acid is 4.84 (a) Calculate the pK \(_{b}\) for the butyrate ion. (b) Calculate the pH of a 0.050 \(M\) solution of butyric acid. (c) Calculate the pH of a 0.050\(M\) solution of sodium butyrate.

To find the pKb for the butyrate ion, we can use the relationship between the dissociation constants of an acid and its conjugate base: \(K_a \times K_b = K_w\) \(K_w\) is the ion-product constant of water, and its value is \(1.0 \times 10^{-14}\). First, we need to find the value of Ka for butyric acid using the given pKa: \(K_a = 10^{-pKa}\) \(K_a = 10^{-4.84}\) Now, we can find the value of Kb for the butyrate ion: \(K_b = \frac{K_w}{K_a}\) \(K_b = \frac{1.0 \times 10^{-14}}{10^{-4.84}}\) Finally, we can find the pKb for the butyrate ion: \(pK_b = -\log{K_b}\)

To calculate the pH of the butyric acid solution, we can use the equilibrium expression for the dissociation of butyric acid: \(K_a = \frac{[H^+][A^-]}{[HA]}\) Since the initial concentration of H+ and A- is 0 and the concentration of HA is 0.050 M, we can use an ICE table to find the equilibrium concentrations. Note that x is the dissociation value: \(HA \longleftrightarrow H^+ + A^-\) Initial: \(0.050 \qquad 0 \qquad 0\) Change: \(-x \qquad +x \qquad +x\) Equilibrium: \(0.050 - x \qquad x \qquad x\) Now, we can substitute the values in the Ka expression: \(K_a = \frac{x^2}{0.050 - x}\) We can use the quadratic formula, or we can assume that x is very small compared to 0.050 to simplify our calculation: \(K_a = \frac{x^2}{0.050}\) Solve for x, which represents the [H+]: Now, we can use the pH formula: \(pH = -\log{[H^+]}\)

Sodium butyrate is the salt of a weak acid (butyric acid) and a strong base (NaOH). When sodium butyrate dissolves in water, it produces the conjugate base of butyric acid, the butyrate ion, which can accept a proton (H+) from water: \(A^- (aq) + H_2O (l) \longleftrightarrow HA (aq) + OH^- (aq)\) The equilibrium constant for the reaction is Kb. Since the concentration of sodium butyrate is 0.050 M, we can assume that the initial concentration of A- is also 0.050 M (considering 100% dissociation). We can use the ICE table method and the Kb expression to calculate the [OH-] concentration: \(A^- + H_2O \longleftrightarrow HA + OH^-\) Initial: \(0.050 \qquad \qquad\qquad 0\) Change: \(-x \qquad\qquad\qquad\qquad+x\) Equilibrium: \(0.050 - x \qquad\qquad\qquad x\) \(K_b = \frac{[HA][OH^-]}{[A^-]}\) Now, we can substitute the values in the Kb expression: \(K_b = \frac{x^2}{0.050 - x}\) We can use the quadratic formula, or we can assume that x is very small compared to 0.050 to simplify our calculation: \(K_b = \frac{x^2}{0.050}\) Solve for x, which represents the [OH-]: Now, we can use the pOH formula and then find the pH: \(pOH = -\log{[OH^-]}\) \(pH = 14 - pOH\)