Problem 60 - Chemistry The Science in Context

Question

The awful odor of dead fish is due mostly to trimethylamine, \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N},\) one of three compounds related to ammonia in which methyl groups replace \(1,2,\) or all 3 of the H atoms in ammonia. a. The \(K_{b}\) of trimethylamine \(\left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}\right]\) is \(6.5 \times 10^{-5}\) at \(25^{\circ} \mathrm{C} .\) Calculate the \(\mathrm{pH}\) of a \(3.00 \times 10^{-4} M\) solution of trimethylamine. b. The \(K_{b}\) of methylamine \(\left[\left(\mathrm{CH}_{3}\right) \mathrm{NH}_{2}\right]\) is \(4.4 \times 10^{-4}\) at \(25^{\circ} \mathrm{C} .\) Calculate the \(\mathrm{pH}\) of a \(2.88 \times 10^{-3} \mathrm{M}\) solution of methylamine. *c. The \(K_{\mathrm{b}}\) of dimethylamine \(\left[\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}\right]\) is \(5.9 \times 10^{-4}\) at \(25^{\circ} \mathrm{C} .\) What concentration of dimethylamine is needed for the solution to have the same \(\mathrm{pH}\) as the solution in part (b)?

Step-by-Step Solution

Answer

Based on the calculations: a) The pH of a \(3.00 \times 10^{-4} \text{M}\) solution of trimethylamine is 10.17. b) The pH of a \(2.88 \times 10^{-3} \text{M}\) solution of methylamine is 11.06. c) The required concentration of dimethylamine to have the same pH as the solution in part b is \(2.22 \times 10^{-2} \text{M}\).

Step 1: Write the Kb expression for trimethylamine

The \(K_b\) expression for trimethylamine can be written as: \begin{equation} K_b = \frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]} \end{equation}

Step 2: Substitute the given values and solve for [OH-]

Given \(K_b\) for trimethylamine is \(6.5 \times 10^{-5}\), and the concentration of trimethylamine is \(3.00 \times 10^{-4} M\). Let \(x\) be the concentration of hydroxide ions (\([OH^-]\)). Since trimethylamine is a weak base, we can assume \(x << 3.00 \times 10^{-4} M\), and the concentration of trimethylamine will remain almost the same. The equation becomes: \begin{equation} 6.5 \times 10^{-5} = \frac{x^2}{3.00 \times 10^{-4}} \end{equation} Solving for \(x\), we get: \begin{equation} [OH^-] = x = 1.49 \times 10^{-4}\text{ M} \end{equation}

Step 3: Calculate pOH and pH

Now that we have the hydroxide ion concentration, we can calculate the pOH and pH using the following formulas: \begin{equation} pOH = -\log_{10} [OH^-] \end{equation} \begin{equation} pH = 14 - pOH \end{equation} Computing the values, we get pOH = 3.83 and pH = 10.17. The pH of the \(3.00 \times 10^{-4} \text{M}\) solution of trimethylamine is 10.17. ##Part b##

Step 1: Write the Kb expression for methylamine

The \(K_b\) expression for methylamine can be written as: \begin{equation} K_b = \frac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]} \end{equation}

Step 2: Substitute the given values and solve for [OH-]

Given \(K_b\) for methylamine is \(4.4 \times 10^{-4}\) and the concentration of methylamine is \(2.88 \times 10^{-3} \text{M}\). Let \(x\) be the concentration of hydroxide ions ([OH-]). Since methylamine is a weak base, we can assume \(x << 2.88 \times 10^{-3} \text{M}\) and the concentration of methylamine will remain almost the same. The equation becomes: \begin{equation} 4.4 \times 10^{-4} = \frac{x^2}{2.88 \times 10^{-3}} \end{equation} Solving for \(x\), we get: \begin{equation} [OH^-] = x = 1.14 \times 10^{-3}\text{ M} \end{equation}

Step 3: Calculate pOH and pH

Now that we have the hydroxide ion concentration, we can calculate the pOH and pH using the following formulas: \begin{equation} pOH = -\log_{10} [OH^-] \end{equation} \begin{equation} pH = 14 - pOH \end{equation} Computing the values, we get pOH = 2.94 and pH = 11.06. The pH of the \(2.88 \times 10^{-3} \text{M}\) solution of methylamine is 11.06. ##Part c##

Step 1: Write the Kb expression for dimethylamine

The \(K_b\) expression for dimethylamine can be written as: \begin{equation} K_b = \frac{[(CH_3)_2NH_2^+][OH^-]}{[(CH_3)_2NH]} \end{equation} We want dimethylamine concentration that results in the same pH as in part b. So, we need the solution of dimethylamine to have [OH-] = \(1.14 \times 10^{-3} \text{M}\).

Step 2: Substitute the given values and solve for required concentration of dimethylamine

Given \(K_b\) for dimethylamine is \(5.9 \times 10^{-4}\). We already know the concentration of hydroxide ions needs to be \(1.14 \times 10^{-3} \text{M}\). Let \(y\) be the concentration of dimethylamine. The equation becomes: \begin{equation} 5.9 \times 10^{-4} = \frac{(1.14 \times 10^{-3})^2}{y} \end{equation} Solving for \(y\), we get: \begin{equation} [(CH_3)_2NH] = y = 2.22 \times 10^{-2}\text{ M} \end{equation} The required concentration of dimethylamine to have the same pH as the solution in part b is \(2.22 \times 10^{-2} \text{M}\).

Key Concepts

Understanding Trimethylamine Kb and pH Calculation

When studying the base dissociation constant, or Kb, of trimethylamine (TMA), a compound associated with the smell of rotting fish, one must recognize that it's a measure of a base's strength. For TMA, this value is given as 6.5 x 10^-5. To calculate pH, you first need the hydroxide ion concentration [OH^-], which you can find using the formula \[\begin{equation}K_b = \frac{[OH^-]^2}{[TMA]_0 - [OH^-]}\right\right\right\right\right\right\right\right\right\end{equation}\]ignginolving will approximate the concentration of TMA as unchanged due to the small x. After finding [OH^-], you utilize pOH = -log10[OH^-] and then relate pH and pOH by pH = 14 - pOH. This is critical for students to grasp, as it forms the foundation for understanding how weak bases interact with water and affect the pH of a solution.

Calculating pH Using Methylamine Kb

The calculation of pH for a methylamine solution – a weaker base than trimethylamine – requires understanding its Kb value, which is 4.4 x 10^-4. Utilizing the same steps as for TMA, set up the equilibrium concentration equation and find the hydroxide ion concentration. However, methylamine has a different initial concentration and hence, a unique set of calculations to reach [OH^-]. Knowing the relationship between the base's strength, reflected by its Kb, and the resulting pH is pivotal. Students should take note of the magnitude differences in the Kb values and how they influence the hydroxide ion concentration and ultimately the pH of the solution.

Dimethylamine Kb in pH Matching

The concept of matching the pH of two different solutions, as shown with dimethylamine, revolves around manipulating the concentrations to achieve the same hydroxide ion levels. With a given Kb for dimethylamine (5.9 x 10^-4), we seek a concentration that yields the same pH as the methylamine solution from our previous example. It is an excellent exercise in understanding the direct relationship between base concentration, Kb, and pH – key aspects of acid-base chemistry. This reinforces the application of equilibrium concepts to achieve desired chemical conditions.

Hydroxide Ion Concentration and its Significance

In the realm of acid-base chemistry, the hydroxide ion concentration is a cornerstone value that dictates the alkalinity of a solution. Through equilibrium expressions, [OH^-] is calculated, setting the stage for further pH and pOH computations. Students should focus on how changes in [OH^-] directly impact these values. For weak bases like the amines discussed, understanding that the hydroxide ion concentration derived from their reaction with water is what defines their basicity and pH. Appreciating this relationship helps to cement the core principles of acid-base equilibrium.

The pOH and pH Relationship in Base Solutions

Connecting pOH and pH is vital for comprehending the acid-base properties of a solution. Given that pOH is defined as the negative logarithm of the hydroxide ion concentration, it has an inverse relationship with the concentration. Since pH is calculated by subtracting pOH from 14, any change in hydroxide ion concentration will reflect inversely on pH. It's essential for students to internalize the concept that when pOH decreases due to increased [OH^-], the pH increases, indicating a more basic solution. This mirrors the antithetical nature of the pOH and pH relationship and is paramount when doing pH calculations for weak base solutions.