Question

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

Short answer

The pKb of the butyrate ion is 9.16. The pH of a 0.050 M solution of butyric acid is 2.42, and the pH of a 0.050 M solution of sodium butyrate is 8.42.

Step-by-step solution

Part (a): Calculate the pKb for the butyrate ion

To find the pKb for the butyrate ion, we can use the relationship between pKa and pKb: pKb = 14 - pKa Given the pKa of butyric acid is 4.84: pKb = 14 - 4.84 = 9.16 So, the pKb for the butyrate ion is 9.16.

Part (b): Calculate the pH of a 0.050 M solution of butyric acid

To find the pH of a 0.050 M solution of butyric acid, we can use the relation between pH and the pKa of the acid: $pH=pKa+{\log }_{10}\frac{[A^{-}]}{[HA]}$ Since the solution is 0.050 M, we can denote the concentration of butyric acid ([HA]) as 0.050-x and the concentration of the butyrate ion ([A^-]) as x. Now, we use the approximation that the concentration of butyric acid remains nearly equal to the initial concentration of butyric acid, so [HA] ≈ 0.050: $pH=4.84+{\log }_{10}\frac{x}{0.050}$ By substituting the Ka value, we can find x: $Ka=\frac{x\cdot [H^{+}]}{(0.050-x)}={10}^{-4.84}$ Since it is a weak acid, x should be much smaller than the initial concentration, so we can approximate $Ka=\frac{x\cdot [H^{+}]}{0.050}={10}^{-4.84}$ Dividing both sides by the acid concentration and taking the square root, we get: $x=\sqrt{Ka\cdot 0.050}=\sqrt{1.445\cdot {10}^{-5}}$ $x=3.801\cdot {10}^{-3}$ The concentration of [H+] is equal to x, so pH = -log([H+]): pH = -log(3.801 x 10^{-3}) = 2.42 So, the pH of a 0.050 M solution of butyric acid is 2.42.

Part (c): Calculate the pH of a 0.050 M solution of sodium butyrate

To find the pH of a 0.050 M solution of sodium butyrate, first, we need to understand that sodium butyrate dissociates completely in water, providing the butyrate ion, which is a weak base. To calculate the pH, we can use the pKb and Kb value that we found in part (a). $pOH=pKb+{\log }_{10}\frac{[B{H}^{+}]}{[B]}$ Since the solution is 0.050 M, we can denote the concentration of the butyrate ion ([B], which is equal to the initial concentration in this case) as 0.050. The concentration of the conjugate acid ([BH+]) can be denoted as x. Now, we use the approximation that the concentration of butyrate ion remains nearly equal to the initial concentration of butyrate, so [B] ≈ 0.050: $pOH=9.16+{\log }_{10}\frac{x}{0.050}$ By substituting the Kb value, we can find x: $Kb=\frac{x\cdot [O{H}^{-}]}{(0.050-x)}={10}^{-9.16}$ Since it is a weak base, x should be much smaller than the initial concentration, so we can approximate $Kb=\frac{x\cdot [O{H}^{-}]}{0.050}={10}^{-9.16}$ Dividing both sides by the base concentration and taking the square root, we get: $x=\sqrt{Kb\cdot 0.050}=\sqrt{6.944\cdot {10}^{-11}}$ $x=2.635\cdot {10}^{-6}$ The concentration of [OH-] is equal to x, so pOH = -log([OH-]): pOH = -log(2.635 x 10^{-6}) = 5.58 To find the pH, we can use the relation: pH = 14 - pOH = 14 - 5.58 = 8.42 So, the pH of a 0.050 M solution of sodium butyrate is 8.42.

Key concepts

pKa and pKb Relationship

Understanding the relationship between the acidity constant ($p{K}_a$ and the basicity constant ($p{K}_b$ is essential when dealing with acid-base chemistry. The constants are related through the equation:
$$p{K}_a+p{K}_b=14$$
This equation stems from the water ion product, which states that the concentration of hydrogen ions ($[H^{+}]$ and hydroxide ions ($[O{H}^{-}]$ in pure water at 25°C equals to $1\times {10}^{-14}$. This also reflects the idea that the stronger an acid is, the weaker its conjugate base will be, and vice versa. For butyric acid with a $p{K}_a$ of 4.84, we calculate the $p{K}_b$ of its conjugate base, the butyrate ion, to be 9.16.
This relationship is crucial as it allows us to predict the behavior of an acid or base in water, and subsequently, the pH of the solution.

pH Calculation of Weak Acid

To calculate the pH of a weak acid, such as butyric acid, we must first acknowledge that weak acids do not dissociate completely in water. This incomplete dissociation leads to an equilibrium between the acid ($HA$ and its conjugate base ($A^{-}$ which can be represented by the acid dissociation constant ($K_a$ or its negative logarithm, the $p{K}_a$ value.
The pH can be estimated using the Henderson-Hasselbalch equation:
$$pH=pKa+{\text{log}}_{10}\frac{[A^{-}]}{[HA]}$$
However, calculating the exact pH requires knowledge of the ionization rate of the acid in solution, which involves some initial assumptions and approximations, as detailed in the exercise solution. The pH for a 0.050 M solution of butyric acid was calculated using an approximation based on the weak acid assumption, resulting in a pH of 2.42.

pH Calculation of Salt Solution

When it comes to salt solutions such as sodium butyrate, we're dealing with the salt of a weak acid and a strong base. These salts will often produce basic solutions upon dissolving in water, due to the hydrolysis of the anion (the conjugate base of the weak acid). In this case, sodium butyrate dissociates entirely in water, and the resultant butyrate ion ($B$ slightly increases the pH by accepting a hydrogen ion from water.
For sodium butyrate solutions, we can calculate the pH by first determining the $pOH$, using the known $p{K}_b$ and the approximation for the hydroxide ion concentration, derived from the base dissociation constant ($K_b$ similar to the process with $K_a$ for acids. After finding the $pOH$, we convert it to pH via the relationship:
$$pH=14-pOH$$
This gives us the pH of the sodium butyrate solution, which for a 0.050 M concentration is 8.42, indicating a basic solution.