Problem 61 - Chemistry: An Atoms-Focused Approach

Question

The \(K_{\mathrm{b}}\) of aminoethanol, \(\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2},\) is \(3.1 \times 10^{-5}\) a. Is aminoethanol a stronger or weaker base than ethylamine, \(\mathrm{p} K_{\mathrm{b}}=3.36 ?\) b. Calculate the \(\mathrm{pH}\) of \(1.67 \times 10^{-2} \mathrm{M}\) aminoethanol. c. Calculate the [OH \(\left.^{-}\right]\) concentration of \(4.25 \times 10^{-4} M\) aminoethanol.

Step-by-Step Solution

Answer

Question: Compare the basic strength of aminoethanol and ethylamine using their Kb values and calculate the pH of a 1.67 x 10^-2 M aminoethanol solution. Also, find the OH- concentration for a 4.25 x 10^-4 M aminoethanol solution. Answer: Ethylamine is a stronger base than aminoethanol. The pH of a 1.67 x 10^-2 M aminoethanol solution is approximately 10.78. The OH- concentration for a 4.25 x 10^-4 M aminoethanol solution is approximately 1.84 x 10^-4 M.

Step 1: Part a: Comparing basic strength

We are given the \(K_{\mathrm{b}}\) value for aminoethanol and the $\mathrm{p} K_{\mathrm{b}}\( value for ethylamine. First, let's convert the \)\mathrm{p} K_{\mathrm{b}}\( value for ethylamine to \)K_{\mathrm{b}}$ value: \(\mathrm{p}K_{\mathrm{b}} = -\log K_{\mathrm{b}}\) \( K_{\mathrm{b}} = 10^{-\mathrm{p}K_{\mathrm{b}}}\) \( K_{\mathrm{b}} = 10^{-3.36} \approx 4.4 \times 10^{-4}\) Now we can compare the basic strength of aminoethanol and ethylamine: Aminoethanol: \(K_{\mathrm{b}} = 3.1 \times 10^{-5}\) Ethylamine: \(K_{\mathrm{b}} \approx 4.4 \times 10^{-4}\) Since the \(K_{\mathrm{b}}\) of ethylamine is greater than that of aminoethanol, ethylamine is a stronger base.

Step 2: Part b: Calculating the pH of aminoethanol

We will use the given concentration of aminoethanol, which is \(1.67 \times 10^{-2} \mathrm{M}\), along with the \(K_{\mathrm{b}}\) value to calculate the pH. The dissociation reaction of aminoethanol in water is: \(HOCH_{2}CH_{2}NH_{2} + H_{2}O \rightleftharpoons HOCH_{2}CH_{2}NH_{3}^{+} + OH^-\) Let x be the concentration of dissociated aminoethanol, hence: \(K_{\mathrm{b}} = \frac{x^2}{0.0167-x}\) Solve for x by assuming that x is very small compared to 0.0167 (in other words, \(0.0167-x \approx 0.0167\)). Now we can solve for x, which represents the \(OH^-\) concentration: \(x^2 \approx K_{\mathrm{b}} \times 0.0167\) \(x \approx \sqrt{3.1 \times 10^{-5} \times 0.0167} \approx 6.07 \times 10^{-4}\) Now, calculate the pOH: \(pOH = -\log[OH^-] = -\log(6.07 \times 10^{-4}) \approx 3.22\) Now, calculate the pH: \(pH = 14 - pOH = 14 - 3.22 \approx 10.78\)

Step 3: Part c: Calculating the [OH-] concentration of aminoethanol

We will use the given concentration of aminoethanol: \(4.25 \times 10^{-4} \mathrm{M}\). We have the \(K_{\mathrm{b}}\) value, so the equation remains the same as in part b: \(K_{\mathrm{b}} = \frac{x^2}{4.25 \times 10^{-4} - x}\) Solve for x by assuming that x is very small compared to \(4.25 \times 10^{-4}\): \(x^2 \approx K_{\mathrm{b}} \times 4.25 \times 10^{-4}\) \(x \approx \sqrt{3.1 \times 10^{-5} \times 4.25 \times 10^{-4}} \approx 1.84 \times 10^{-4}\) The concentration of \(OH^-\) ions is approximately \(1.84 \times 10^{-4} \mathrm{M}\).

Key Concepts

Basic Strength Comparison

When comparing the basic strength of two substances, such as aminoethanol and ethylamine, we look at their base dissociation constants, expressed as \(K_{\mathrm{b}}\). The \(K_{\mathrm{b}}\) value indicates how well a base can donate an electron pair to form a hydroxide ion, \(OH^-\), in solution. In general:

• A higher \(K_{\mathrm{b}}\) value means a stronger base.
• A lower \(K_{\mathrm{b}}\) value means a weaker base.

In our case, aminoethanol has a \(K_{\mathrm{b}}\) of \(3.1 \times 10^{-5}\), while ethylamine has a \(K_{\mathrm{b}}\) of approximately \(4.4 \times 10^{-4}\). Since ethylamine's \(K_{\mathrm{b}}\) is higher, it is the stronger base compared to aminoethanol. Understanding this concept can help predict how bases will react in different chemical environments.

pH Calculation

Calculating the pH of a solution involves determining the concentration of hydrogen ions \([H^+]\) or hydroxide ions \([OH^-]\). In aminoethanol's basic solution, the focus is on hydroxide ions. The steps for finding the pH are as follows:

• First, calculate the concentration of \([OH^-]\) using the \(K_{\mathrm{b}}\) value and the initial concentration of aminoethanol.
• Use this concentration to find the \(pOH\), where \(pOH = -\log([OH^-])\).
• Finally, convert \(pOH\) to pH using the relation: \(pH = 14 - pOH\).

For example, for a \(1.67 \times 10^{-2} \mathrm{M}\) aminoethanol solution, we calculated a \(pH\) of 10.78, indicating a basic solution. This step-by-step approach ensures accurate results and helps in understanding the behavior of bases in solutions.

OH- Concentration Calculation

To determine the hydroxide ion concentration \([OH^-]\) in a solution of aminoethanol, we use the equilibrium expression for \(K_{\mathrm{b}}\). The process involves the following:

• Using the equilibrium expression \(K_{\mathrm{b}} = \frac{x^2}{[initial\ concentration] - x}\), where \(x\) represents \([OH^-]\).
• Making an approximation that \(x\) is small compared to the initial concentration, simplifying the expression to \(x^2 \approx K_{\mathrm{b}} \times [initial\ concentration]\).
• Solving for \(x\), which gives us the \([OH^-]\) concentration.

For a \(4.25 \times 10^{-4} \mathrm{M}\) solution of aminoethanol, the calculation yields \([OH^-]\) approximately \(1.84 \times 10^{-4} \mathrm{M}\). This method is crucial for determining how basic a solution is, thereby influencing the pH and the related chemical properties.

Aminoethanol

Aminoethanol, also known as ethanolamine, is a compound with both alcohol and amine functional groups, represented as \(HOCH_2CH_2NH_2\). As an alkanolamine, it can participate in both acidic and basic reactions. The amine group can accept protons, while the hydroxyl group can slightly influence its basicity by forming hydrogen bonds.

• The \(K_{\mathrm{b}}\) value of aminoethanol indicates its effectiveness as a base.
• It can form complexes and act as a ligand, enhancing its applicability in various chemical processes.
• Despite being less basic than ethylamine, it finds use in cleaning products, pharmaceuticals, and gas treatment.

This multifaceted nature makes aminoethanol a versatile compound in different chemical contexts.

Acid-Base Equilibria

Understanding acid-base equilibria is crucial to predict solution behavior when mixing acids and bases. These equilibria involve the transfer of protons \(H^+\) between reactants.

• For bases like aminoethanol, equilibria focus on \([OH^-]\) production in water.
• The equilibrium constant \(K_{\mathrm{b}}\) helps to determine the extent of this production.
• Le Châtelier’s principle can be applied to predict shifts in equilibria based on concentration changes.

In the example of aminoethanol, this principle helps us understand its behavior in solution, as well as the resulting acidity or basicity, which are key for calculations like pH and \([OH^-]\) concentration. Recognizing these concepts aids in a deeper grasp of chemical reactions and solution dynamics.