Problem 106 - Chemistry: An Atoms-Focused Approach

Question

Calculate the indicated value based on the information given: a. What is the \(K_{\mathrm{b}}\) of the lactate ion? The \(K_{\mathrm{a}}\) of lactic acid is \(1.4 \times 10^{-4}\) b. What is the \(K_{\mathrm{b}}\) of the conjugate base of pyruvic acid? The \(K_{a}\) of pyruvic acid is \(2.8 \times 10^{-3} .\) c. What is the \(K_{2}\) of the conjugate acid of aniline? Aniline has a \(K_{\mathrm{b}}\) of \(5.9 \times 10^{-4}\) d. Quinine has two basic nitrogen atoms in its structure. The \(K_{\mathrm{b}}\) of the most basic nitrogen atom is \(3.3 \times 10^{-6} .\) What is the \(K_{a}\) of the HCl salt of that nitrogen atom?

Step-by-Step Solution

Answer

Based on the step-by-step solution above, we calculated the following ionization constants: a. The \(K_{\mathrm{b}}\) of the lactate ion is \(7.1 \times 10^{-11}\). b. The \(K_{\mathrm{b}}\) of the conjugate base of pyruvic acid is \(3.6 \times 10^{-12}\). c. The \(K_{2}\) of the conjugate acid of aniline is \(1.7 \times 10^{-11}\). d. The \(K_{\mathrm{a}}\) of the HCl salt of the most basic nitrogen atom in quinine is \(3.0 \times 10^{-9}\).

Step 1: Write down the given values

The \(K_{\mathrm{a}}\) of lactic acid is given as \(1.4 \times 10^{-4}\). Write down the ion product of water, \(K_\mathrm{w} = 1.0 \times 10^{-14}\).

Step 2: Calculate the \(K_\mathrm{b}\) for the lactate ion

Using the relationship \(K_\mathrm{w} = K_\mathrm{a} \times K_\mathrm{b}\), we can find the \(K_\mathrm{b}\) value of the lactate ion. Rearrange the equation to solve for \(K_\mathrm{b}\): \(K_\mathrm{b} = \frac{K_\mathrm{w}}{K_\mathrm{a}}\). Plug in the given values and calculate the result: \(K_\mathrm{b} = \frac{1.0 \times 10^{-14}}{1.4 \times 10^{-4}} = 7.1 \times 10^{-11}\) b. Calculate the \(K_{\mathrm{b}}\) of the conjugate base of pyruvic acid

Step 1: Write down the given values

The \(K_{\mathrm{a}}\) of pyruvic acid is given as \(2.8 \times 10^{-3}\). Write down the ion product of water, \(K_\mathrm{w} = 1.0 \times 10^{-14}\).

Step 2: Calculate the \(K_\mathrm{b}\) for the conjugate base

Using the relationship \(K_\mathrm{w} = K_\mathrm{a} \times K_\mathrm{b}\), we can find the \(K_\mathrm{b}\) value of the conjugate base of pyruvic acid. Rearrange the equation to solve for \(K_\mathrm{b}\): \(K_\mathrm{b} = \frac{K_\mathrm{w}}{K_\mathrm{a}}\). Plug in the given values and calculate the result: \(K_\mathrm{b} = \frac{1.0 \times 10^{-14}}{2.8 \times 10^{-3}} = 3.6 \times 10^{-12}\) c. Calculate the \(K_{2}\) of the conjugate acid of aniline

Step 1: Write down the given values

The \(K_{\mathrm{b}}\) of aniline is given as \(5.9 \times 10^{-4}\). Write down the ion product of water, \(K_\mathrm{w} = 1.0 \times 10^{-14}\).

Step 2: Calculate the \(K_{\mathrm{a}}\) for the conjugate acid

Using the relationship \(K_\mathrm{w} = K_\mathrm{a} \times K_\mathrm{b}\), we can find the \(K_\mathrm{a}\) value of the conjugate acid of aniline. Rearrange the equation to solve for \(K_\mathrm{a}\): \(K_\mathrm{a} = \frac{K_\mathrm{w}}{K_\mathrm{b}}\). Plug in the given values and calculate the result: \(K_\mathrm{a} = \frac{1.0 \times 10^{-14}}{5.9 \times 10^{-4}} = 1.7 \times 10^{-11}\). The \(K_{2}\) of the conjugate acid of aniline is the same as its \(K_\mathrm{a}\) value, so \(K_{2} = 1.7 \times 10^{-11}\). d. Calculate the \(K_{\mathrm{a}}\) of the HCl salt of the most basic nitrogen atom in quinine

Step 1: Write down the given values

The \(K_{\mathrm{b}}\) of the most basic nitrogen atom in quinine is given as \(3.3 \times 10^{-6}\). Write down the ion product of water, \(K_\mathrm{w} = 1.0 \times 10^{-14}\).

Step 2: Calculate the \(K_{\mathrm{a}}\) for the HCl salt

Using the relationship \(K_\mathrm{w} = K_\mathrm{a} \times K_\mathrm{b}\), we can find the \(K_\mathrm{a}\) value of the HCl salt of the most basic nitrogen atom in quinine. Rearrange the equation to solve for \(K_\mathrm{a}\): \(K_\mathrm{a} = \frac{K_\mathrm{w}}{K_\mathrm{b}}\). Plug in the given values and calculate the result: \(K_\mathrm{a} = \frac{1.0 \times 10^{-14}}{3.3 \times 10^{-6}} = 3.0 \times 10^{-9}\).

Key Concepts

Acid-Base Equilibrium

Acid-base equilibrium refers to the balance between the concentrations of acids and bases in a chemical system. It is crucial for understanding the behavior of chemicals like lactic acid and its corresponding base, lactate ion. Acid and base equilibrium is guided by the dissociation of acids into their ions in solution.

An acid, like lactic acid, donates protons (H+) to the solution, leaving behind its conjugate base. The equilibrium constant for acids, known as the acid dissociation constant \(K_a\), helps predict how completely an acid will dissociate in water.

The strength of an acid is inversely related to the strength of its conjugate base. Therefore, to find out the strength of the conjugate base, we often use equilibrium constants in our calculations.

Ion Product of Water

Water has an intrinsic slight ability to dissociate into ions: hydrogen \(H^+\) and hydroxide \(OH^-\). This gives rise to the ion product of water (\(K_w\)), a key constant in chemistry. It is represented by the equation \(K_w = [H^+][OH^-] = 1.0 \times 10^{-14}\).

This value is a cornerstone in acid-base calculations, especially in determining the equilibrium constants \(K_a\) and \(K_b\) for acids and bases, respectively. Knowing one can easily calculate the other using the relationship \(K_w = K_a \times K_b\).

Since the \(K_w\) is a constant at a given temperature (typically 25°C), it simplifies calculations involving acids and bases by allowing us to find either \(K_a\) or \(K_b\) when paired with its counterpart.

Conjugate Acids and Bases

Conjugate acids and bases are pairs that differ by only one proton (H+). In an acid-base reaction, the acid donates a proton and becomes a conjugate base, while the base accepts a proton becoming a conjugate acid. In these equilibria, understanding conjugate pairs is vital.

For example, lactic acid becomes lactate, a conjugate base, upon losing a proton. On the flip side, a base like ammonia (NH₃) forms its conjugate acid, ammonium (NH₄⁺), upon gaining a proton.

The strengths of acids and their corresponding bases are related but inverse. A strong acid will have a weak conjugate base, and vice versa. This relationship plays a pivotal role, particularly in buffer solutions which use conjugate pairs to maintain a stable pH level.

Chemical Equilibrium Calculations

Chemical equilibrium calculations are integral for determining the behavior and concentrations of chemical species in a reaction at equilibrium. These calculations utilize equilibrium constants, such as \(K_a\) for acids and \(K_b\) for bases, to help quantify the extent of a reaction.

The fundamental process involves setting up the equilibrium expression based on a balanced chemical reaction, inserting initial concentrations, and solving for the equilibrium concentrations.

• Determine initial concentrations.
• Write the equilibrium expression using \(K\) constants.
• Solve for unknowns using algebraic techniques or approximations.

Understanding these calculations allows chemists to predict reaction behaviors, making it easier to adjust conditions to favor production or suppression of particular products.