Problem 28 - General Chemistry: Principles and Modern Applications

Question

What must be the molarity of an aqueous solution of trimethylamine, \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N},\) if it has a \(\mathrm{pH}=11.12 ?\) $$\begin{aligned} \left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+} &+\mathrm{OH}^{-} \\ K_{\mathrm{b}}=6.3 \times 10^{-5} \end{aligned}$$

Step-by-Step Solution

Answer

The initial molarity of the aqueous solution of trimethylamine is 0.00603 M.

Step 1: Calculate \( OH^- \) Concentration

First you need to calculate the concentration of \( OH^- \) ions using the given pH value. The formula connecting pH, pOH and \( OH^- \) is \( pOH = 14 - pH \), and concentration of \( OH^- \) ions is given by \( [OH^-] = 10^{-pOH} \). From the given pH=11.12, we have pOH = 14 - 11.12 = 2.88, so \( [OH^-] = 10^{-2.88} = 1.32 \times 10^{-3} \) M.

Step 2: Calculate \( \left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+} \) Concentration

Then, you need to calculate the concentration of \( \left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+} \) using the base dissociation constant \( K_{\mathrm{b}} \) and the calculated \( OH^- \) concentration. From \( K_{\mathrm{b}} = \frac{[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}][OH^-]}{[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}]}\) and assuming at equilibrium, \( [\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}] = [OH^-] \) and \( [\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}] = [\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}] - [OH^-] \), we substitute and obtain \( [\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}] = \sqrt{K_{\mathrm{b}} / [OH^-]} = \sqrt{6.3 \times 10^{-5} / 1.32 \times 10^{-3}} = 0.00735 \) M.

Step 3: Calculate Initial Molarity of Trimethylamine

Finally, you calculate the initial molarity of the trimethylamine solution before any dissociation took place. As \( [\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}] = [\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}] - [OH^-] \), we have initial concentration of trimethylamine \( [\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}]_0 = 0.00735 - 1.32 \times 10^{-3} = 0.00603 \) M.

Key Concepts

pH and pOH relationship

The pH and pOH of a solution are closely related as they represent the acidity and basicity of a solution respectively. Essentially, pH is a measure of the concentration of hydrogen ions \(H^+\), while pOH measures hydroxide ions \(OH^-\). Together, they provide a complete picture of a solution's chemical nature.
In pure water at 25°C, the sum of pH and pOH is always 14. This is because water undergoes autoionization, where \(H_2O \rightleftharpoons \ H^+ + OH^-\). In the case of the exercise, we know the pH, which allows us to use the equation:

• \( pOH = 14 - pH \)
• \( pOH = 14 - 11.12 = 2.88 \)

Knowing pOH helps us find the concentration of hydroxide ions using the formula: \( [OH^-] = 10^{-pOH} \). Calculating further gives us \( [OH^-] = 1.32 \times 10^{-3} \) M. This value is critical for determining the basic aspects of the solution.

Base dissociation constant

The base dissociation constant, denoted as \(K_b\), is a measure of the strength of a base in solution. It quantifies the extent to which a base can release hydroxide ions \(OH^-\) when dissolved in water, providing a way to estimate the degree of dissociation.
For the base trimethylamine \( (CH_3)_3N \), the dissociation in water is expressed as:

• \( (CH_3)_3N + H_2O \rightleftharpoons (CH_3)_3NH^+ + OH^- \)

The equilibrium expression for this reaction is derived as:
\[ K_b = \frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]} \]
Given \(K_b = 6.3 \times 10^{-5} \), it illustrates how relatively weak the base is, as low \(K_b\) values signify lower dissociation. When solving for equilibrium concentrations using \(K_b\), assumptions are often made that simplify calculations. Here, the assumption that at equilibrium, \( [OH^-] = [(CH_3)_3NH^+] \) effectively aids in simplifying the equation, making it feasible to solve for unknowns.

Equilibrium concentrations

Understanding equilibrium concentrations is crucial when analyzing how reactants and products interact dynamically in a solution. Equilibrium in terms of moles happens when reactants and products are formed at the same rate, leading to constant concentrations over time.
In our reaction involving trimethylamine, equilibrium is represented by balancing the concentrations of the base \( (CH_3)_3N \) and its ionized forms \( (CH_3)_3NH^+ \) and \( OH^- \). Beginning with initial concentrations, such as for \( (CH_3)_3N \), knowing \( [(CH_3)_3NH^+] \) helps calculate this using:

• Initial \( [CH_3)_3N]_0 = [(CH_3)_3NH^+] - [OH^-]\)

In this particular problem, substituting known values gives us:

Initial concentration \( [CH_3)_3N]_0 = 0.00735 \ \text{M} - 1.32 \times 10^{-3} \ \text{M} = 0.00603 \ \text{M}\)

This calculation is pivotal as it allows determination of how much base was initially present before any dissociation occurred, providing essential insight into concentration dynamics within the solution.