Question

Calculate the \(\mathrm{pH}\) of each of the following solutions. a. \(0.100 M\) propanoic acid \(\left(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}, K_{\mathrm{a}}=1.3 \times 10^{-5}\right)\) b. \(0.100 M\) sodium propanoate \(\left(\mathrm{NaC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\right)\) c. pure \(\mathrm{H}_{2} \mathrm{O}\) d. a mixture containing \(0.100 \mathrm{M} \mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\) and \(0.100 \mathrm{M}\) \(\mathrm{NaC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\)

Short answer

The pH values of the following solutions are: a. \(0.100 M\) propanoic acid: \(pH \approx 2.94\) b. \(0.100 M\) sodium propanoate: \(pH = 13\) c. pure water: \(pH = 7\) d. mixture of \(0.100 \mathrm{M} \mathrm{HC}_{3} \mathrm{H}_{5}\mathrm{O}_{2}\) and \(0.100 \mathrm{M} \mathrm{NaC}_{3} \mathrm{H}_{5}\mathrm{O}_{2}\): \(pH = 4.89\)

Step-by-step solution

Write the ionization reaction for propanoic acid.

The ionization reaction of propanoic acid can be represented as: \[\mathrm{HC}_3\mathrm{H}_5\mathrm{O}_2 (\mathrm{aq}) \rightleftharpoons \mathrm{H^+} (\mathrm{aq}) + \mathrm{C}_3\mathrm{H}_5\mathrm{O}_2^- (\mathrm{aq})\]

Set up an equilibrium expression.

Using the equilibrium constant, \(K_\mathrm{a}\), we can set up the expression as: \[K_\mathrm{a} = \frac{[\mathrm{H^+}] [\mathrm{C}_3\mathrm{H}_5\mathrm{O}_2^-]}{[\mathrm{HC}_3\mathrm{H}_5\mathrm{O}_2]}\]

Calculate the concentration of \(\mathrm{H^+}\) ions.

Let \(x\) be the concentration of \(\mathrm{H^+}\) ions in the solution. At equilibrium, the concentrations will be: \[ [\mathrm{H^+}] = x \\ {[\mathrm{C}_3\mathrm{H}_5\mathrm{O}_2^-]} = x \\ [\mathrm{HC}_3\mathrm{H}_5\mathrm{O}_2] = 0.100 - x \] Now, substitute these concentrations into the expression for \(K_\mathrm{a}\): \[1.3 \times 10^{-5} = \frac{x^2}{0.100 - x}\] Since \(K_\mathrm{a}\) is very small (\(1.3 \times 10^{-5}\)), the amount of ionization is very small, so \(x\) is much smaller than \(0.100\), and we can approximate \(0.100 - x \approx 0.100\). Now, we can solve for x: \(x^2 = 1.3 \times 10^{-5} \times 0.100 \\ x = \sqrt{1.3 \times 10^{-6}} \\ x \approx 1.14 \times 10^{-3}\) Therefore, the concentration of \(\mathrm{H^+}\) ions in the solution is approximately \(1.14 \times 10^{-3} M\).

Calculate the pH of the solution.

Now that we have the concentration of \(\mathrm{H^+}\) ions, we can calculate the pH using the formula: \[pH = -\log [\mathrm{H^+}]\] Substitute the value of \([\mathrm{H^+}]\) into the formula: \(pH = -\log(1.14 \times 10^{-3}) \\ pH \approx 2.94\) The pH of the \(0.100 M\) propanoic acid solution is approximately 2.94. #b. pH of \(0.100 M\) sodium propanoate.#

Write the reaction for sodium propanoate in water.

The reaction of sodium propanoate in water can be represented as: \[\mathrm{NaC}_3\mathrm{H}_5\mathrm{O}_2 (\mathrm{aq}) \rightleftharpoons \mathrm{OH^-} (\mathrm{aq}) + \mathrm{C}_3\mathrm{H}_5\mathrm{O}_2^- (\mathrm{aq})\]

Calculate the concentration of \(\mathrm{OH^-}\) ions in the solution.

Since sodium propanoate is a strong electrolyte, it will completely dissociate in water. Therefore, the concentration of \(\mathrm{OH^-}\) ions in the solution will be the same as the concentration of sodium propanoate, which is \(0.100 M\).

Calculate the pOH of the solution.

Now that we have the concentration of \(\mathrm{OH^-}\) ions in the solution, we can calculate the pOH of the solution using the formula: \[pOH = -\log [\mathrm{OH^-}]\] \[pOH = -\log (0.100) \\ pOH = 1\] The pOH of the \(0.100 M\) sodium propanoate solution is 1.

Calculate the pH of the solution.

Now, we can determine the pH by using the relation between pH and pOH: \[pH = 14 - pOH\] \[pH = 14 - 1 \\ pH = 13\] The pH of the \(0.100 M\) sodium propanoate solution is 13. #c. pH of pure water. Since pure water is neutral, it has equal concentrations of \(\mathrm{H^+}\) and \(\mathrm{OH^-}\) ions, both at a concentration of \(10^{-7} M\). The pH of a neutral aqueous solution like pure water is 7. #d. pH of a mixture containing \(0.100 \mathrm{M} \mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\) and \(0.100 \mathrm{M} \mathrm{NaC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\).

Write the ionization reaction for the mixture.

The ionization reaction for the mixture can be represented as: \[\mathrm{HC}_3\mathrm{H}_5\mathrm{O}_2 (\mathrm{aq}) + \mathrm{OH^-} (\mathrm{aq}) \rightleftharpoons \mathrm{H_2O} (\mathrm{l}) + \mathrm{C}_3\mathrm{H}_5\mathrm{O}_2^- (\mathrm{aq})\]

Use the Henderson-Hasselbalch Equation.

To calculate the pH of a buffer solution, we can use the Henderson-Hasselbalch equation: \[pH = pK_\mathrm{a} + \log \frac{[\mathrm{C}_3\mathrm{H}_5\mathrm{O}_2^-]}{[\mathrm{HC}_3\mathrm{H}_5\mathrm{O}_2]}\] \(pK_\mathrm{a} = -\log K_\mathrm{a} \\ pK_\mathrm{a} = -\log (1.3 \times 10^{-5}) \\ pK_\mathrm{a} \approx 4.89\) Since the concentrations of both the acid and the conjugate base are equal (\(0.100 M\)), the ratio in the equation becomes: \(\frac{[\mathrm{C}_3\mathrm{H}_5\mathrm{O}_2^-]}{[\mathrm{HC}_3\mathrm{H}_5\mathrm{O}_2]} = \frac{0.100}{0.100} = 1\) Now, we can use the values in the Henderson-Hasselbalch equation: \[pH = 4.89 + \log 1 \\ pH = 4.89\] The pH of the mixture containing \(0.100 \mathrm{M} \mathrm{HC}_{3} \mathrm{H}_{5}\mathrm{O}_{2}\) and \(0.100 \mathrm{M} \mathrm{NaC}_{3} \mathrm{H}_{5}\mathrm{O}_{2}\) is 4.89.

Key concepts

Propanoic Acid

Propanoic acid, also known as propionic acid, is an organic compound with the formula \( \mathrm{HC}_3\mathrm{H}_5\mathrm{O}_2 \). It belongs to the carboxylic acids family, characterized by a carboxyl group \(-\text{COOH}\). This acid is weak, which means it does not completely dissociate in water. In an aqueous solution, propanoic acid partially ionizes into hydrogen ions (\( \mathrm{H^+} \)) and propanoate ions (\( \mathrm{C}_3\mathrm{H}_5\mathrm{O}_2^- \)). The equilibrium of this ionization is important for determining the pH of the solution. - Propanoic acid is used in the food industry as a preservative.- It naturally occurs in biological systems and is part of metabolism in animals and humans. The acid's strength is quantified as an equilibrium constant known as \( K_a \). This value is quite small for propanoic acid, indicating only slight ionization, which affects the hydrogen ion concentration and the pH.

Sodium Propanoate

Sodium propanoate, or sodium propionate, is the sodium salt of propanoic acid, with the chemical formula \( \mathrm{NaC}_3\mathrm{H}_5\mathrm{O}_2 \). When dissolved in water, it disassociates completely into sodium ions (\( \mathrm{Na^+} \)) and propanoate ions (\( \mathrm{C}_3\mathrm{H}_5\mathrm{O}_2^- \)). This complete dissociation makes sodium propanoate a good base in solution. The propanoate ion acts as a conjugate base of propanoic acid and can accept protons from water, forming \( \mathrm{OH^-} \) ions. As a result, the solution becomes basic.- Commonly used as a food preservative due to its ability to inhibit mold and bacterial growth.- It helps maintain stability in buffer systems.Understanding its dissociation is critical when calculating the pH of a sodium propanoate solution, effectively determining how basic the solution becomes.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a mathematical formula that relates the pH of a solution to the pKa of the acid and the ratio of the concentrations of the conjugate base to the acid. It is a crucial tool for calculating the pH of buffers. This equation is written as:\[ pH = pK_a + \log \left( \frac{[\mathrm{A^-}]}{[\mathrm{HA}]} \right) \]Where:- \( [\mathrm{A^-}] \) is the concentration of the conjugate base.- \( [\mathrm{HA}] \) is the concentration of the acidic form.- \( pK_a = -\log K_a \), representing the negative logarithm of the acid dissociation constant.- Useful in designing buffer solutions with desired pH levels.- Applied in many biological and chemical systems to maintain pH stability.The equation assumes the conditions of a buffer solution: the presence of both an acid and its conjugate base.

Buffer Solution

A buffer solution is a special type of solution that resists changes in pH upon the addition of small amounts of acid or base. It usually consists of a weak acid and its conjugate base, or a weak base and its conjugate acid. In the case of propanoic acid and sodium propanoate, the buffer solution stabilizes the pH around the pKa of the acid. When additional \( \mathrm{H^+} \) ions are introduced, the conjugate base (propanoate ion) can neutralize them, minimizing any pH change.Key characteristics:- Maintains pH in a narrow range.- Crucial for many biological systems, such as blood, where pH must remain constant.- Useful in chemical applications where a stable pH is necessary.This mixture, such as those containing equal concentrations of propanoic acid and sodium propanoate, efficiently maintains the pH even when challenged by outside influences.