Question

Cacodylic acid is \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{AsO}_{2} \mathrm{H} .\) Its ionization constant is \(6.4 \times 10^{-7}\). (a) Calculate the \(\mathrm{pH}\) of \(50.0 \mathrm{~mL}\) of a \(0.10-M\) solution of the acid. (b) Calculate the pH of 25.0 \(\mathrm{mL}\) of \(0.15 \mathrm{M}\left(\mathrm{CH}_{3}\right)_{2} \mathrm{AsO}_{2} \mathrm{Na}\). (c) Mix the solutions in parts (a) and (b). Calculate the pH of the resulting solution.

Short answer

(a) pH = 3.10 (b) Calculate pH using conjugate base concentration. (c) Combine solutions and apply Henderson-Hasselbalch.

Step-by-step solution

Understand the Acid Dissociation in Part (a)

Cacodylic acid dissociates as \((\mathrm{CH}_{3})_{2} \mathrm{AsO}_{2} \mathrm{H} \rightleftharpoons (\mathrm{CH}_{3})_{2} \mathrm{AsO}_{2}^{-} + \mathrm{H}^{+}\). We have the initial concentration of the acid as \([\mathrm{HA}] = 0.10 \, M\). Let's consider \([\mathrm{H}^{+}] = x\) and make an ICE table to find \(x\).

Set Up ICE Table for Cacodylic Acid

Initial: \([\mathrm{HA}] = 0.10 \, M, [\mathrm{A}^{-}] = 0, [\mathrm{H}^{+}] = 0\)Change: \([-x, +x, +x]\)Equilibrium: \([\mathrm{HA}] = 0.10-x, [\mathrm{A}^{-}] = x, [\mathrm{H}^{+}] = x\)\.

Calculate \([\mathrm{H}^{+}]\) Using Ionization Constant

The ionization constant \(K_a = 6.4 \times 10^{-7}\). Using the equilibrium values, \(K_a = \frac{x^2}{0.10-x}\). Assuming \(x \ll 0.10\), \(K_a \approx \frac{x^2}{0.10}\).

Solve for \(x\) in Part (a)

Solve the equation \(6.4 \times 10^{-7} = \frac{x^2}{0.10}\). Thus, \(x^2 = 6.4 \times 10^{-8}\) and \(x = 8.0 \times 10^{-4}\).

Calculate the pH for Part (a)

\(\text{pH} = -\log(8.0 \times 10^{-4}) = 3.10\).

Determine Concentration for Part (b)

\((\mathrm{CH}_{3})_{2} \mathrm{AsO}_{2}^{-}\) is the conjugate base present at \(0.15\, M\) in 25 mL. This is a straightforward pH calculation using concentration of OH- ions.

Calculate pH for Part (b)

Since \((\mathrm{CH}_{3})_{2} \mathrm{AsO}_{2} \mathrm{Na}\) is a salt of a weak acid and strong base, it hydrolyzes to form \((\mathrm{CH}_{3})_{2} \mathrm{AsO}_{2}^{-}\) and \(\text{pOH} = -\log[\mathrm{OH}^{-}]\), followed by \(\text{pH} = 14 - \text{pOH}\). Calculate \([\mathrm{OH}^{-}]= \sqrt{(K_w / K_a) \cdot 0.15M}\).

Mix Both Solutions for Part (c)

Total moles of \((\mathrm{CH}_{3})_{2} \mathrm{AsO}_{2} \mathrm{H}\) from part (a) and total moles of \((\mathrm{CH}_{3})_{2} \mathrm{AsO}_{2}^{-}\) from (b) are determined. Then, find new concentrations in the combined solution volume.

Calculate pH of Final Mixture

Using concentrations from step above, apply Henderson-Hasselbalch equation: \(\text{pH} = \text{pK}_a + \log\left(\frac{[\mathrm{A}^{-}]}{[\mathrm{HA}]\right)\). Calculate \(\text{pH}\).

Key concepts

Ionization Constant

The ionization constant, or acid dissociation constant (\(K_a\)), is a measure of the strength of an acid in solution. It provides insight into how well an acid molecule sheds its hydrogen ions, \( ext{H}^+\), in a given reaction. A higher \(K_a\) value suggests a stronger acid that ionizes more completely in solution.

In the exercise involving cacodylic acid ( \((\text{CH}_3)_2 \text{AsO}_2 \text{H}\)), the ionization constant is given as \(6.4 \times 10^{-7}\). This implies that it is a weak acid since the value is far less than 1. \(K_a\) is used in equilibrium calculations to determine \([\text{H}^+]\), essential for later \(\text{pH}\) calculations. By setting up an equilibrium concentration expression, this small \(K_a\) can be used to find the rate at which the acid dissociates into its ions. This is crucial for predicting and calculating solution behaviors like acidity.

Acid Dissociation

Acid dissociation refers to the process by which an acid reacts with water to release hydrogen ions (\(\text{H}^+\)) into the solution, forming its conjugate base. As seen in the example of cacodylic acid, \((\text{CH}_3)_2 \text{AsO}_2 \text{H} \leftrightharpoons (\text{CH}_3)_2 \text{AsO}_2^- + \text{H}^+\).

Every acid dissociation involves an equilibrium between the non-dissociated acid molecule and the ions produced. The initial concentration of the acid, changes due to dissociation, and equilibrium concentrations can be calculated using an ICE table (Initial, Change, Equilibrium).

• Initial: Start with the full concentration set as the acid form, for example, \([\text{HA}] = 0.10 \text{ M}\) and zero for products.
• Change: As dissociation occurs, the concentration of \([\text{HA}]\) decreases by \(x\), and \([\text{H}^+]\) and \([\text{A}^-]\) increase by \(x\).
• Equilibrium: Calculate new concentrations using these changes, which allows for solving \(x\), equating \([\text{H}^+]\) needed for \(\text{pH}\).

Understanding this concept is essential as it forms the base for determining acid strength and eventual \(\text{pH}\) levels.

pH Calculation

The \(\text{pH}\) of a solution signals its acidity or basicity, mathematically defined as \(\text{pH} = -\log[\text{H}^+]\). For weak acids such as cacodylic acid, once \([\text{H}^+]\) is calculated from the \(K_a\) expression, \(\text{pH}\) can be directly derived from its concentration.

For example, using \([\text{H}^+] = 8.0 \times 10^{-4}\), \(\text{pH} = -\log(8.0 \times 10^{-4}) = 3.10\). This value is typical for weak acids, indicating a slightly acidic solution.

In addition, for basic solutions such as in the presence of a salt of a weak acid, calculate the \([\text{OH}^-]\) instead, followed by using \(\text{pOH} = -\log[\text{OH}^-]\) and \(\text{pH} = 14 - \text{pOH}\). Such calculations are foundational for students to master the manipulation of logarithmic expressions essential in chemistry.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch Equation is a pivotal tool in buffer solutions, formulated as:\[\text{pH} = \text{pK}_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)\]This formula directly links an acid's \(\text{pK}_a\) with the ratio of its conjugate base (\([\text{A}^-]\)) to the acid (\([\text{HA}]\)) concentrations.

In the final exercise part where two solutions combine, creating a buffer, the Henderson-Hasselbalch equation becomes crucial. It bypasses more complex calculations to swiftly ascertain the final \(\text{pH}\) of a buffer solution. Here, knowing both initial concentrations and total solution volumes enables rapid \(\text{pH}\) calculation through this log-based equation.

• \(\text{pK}_a\) is determined from \(K_a\) as \(\text{pK}_a = -\log(K_a)\).
• Calculate the molar concentrations of \([\text{A}^-]\) and \([\text{HA}]\) in the new mixed solution volume.
• Insert these values into the equation to find the resulting \(\text{pH}\).

This method is immensely valuable when tackling buffer solutions, offering both accuracy and simplicity.