Problem 59 - Chemistry: Principles and Reactions

Question

A solution of an unknown weak acid at \(25^{\circ} \mathrm{C}\) has an osmotic pressure of \(0.878\) atm and a \(\mathrm{pH}\) of \(6.76 .\) What is \(K_{\mathrm{b}}\) for its conjugate base? (Assume that, in the equation for \(\pi\) [Chapter 10\(], i=1\).)

Step-by-Step Solution

Answer

Answer: The value of \(K_b\) for the conjugate base is approximately \(5.75 \times 10^{-8}\).

Step 1: Find the molarity of the weak acid solution using the given osmotic pressure

The formula relating osmotic pressure, molarity, and temperature is: \(\pi = i M R T\) Where \(\pi\) is the osmotic pressure in atm, \(i\) is the van't Hoff factor (given as 1), \(M\) is the molarity in mol/L, \(R\) is the gas constant (0.0821 Latm/(mol.K)), and \(T\) is the temperature in Kelvin. We are given, \(\pi = 0.878\) atm and \(T = 25^{\circ} \mathrm{C} = 298 \mathrm{K}\). First, convert the temperature to Kelvin: \(T_K = T_{\mathrm{C}} + 273 = 25 + 273 = 298 \mathrm{K}\). Now, solve for the molarity, \(M\): \(M = \frac{\pi}{iRT} = \frac{0.878}{1 \times 0.0821 \times 298} = 0.0361 \mathrm{M}\)

Step 2: Convert the given pH to the concentration of hydronium ions (\([\mathrm{H^+}]\))

Using the definition of pH, \([\mathrm{H^+}] = 10^{-\mathrm{pH}}\). We are given the pH as \(6.76\). Thus, \([\mathrm{H^+}] = 10^{-6.76} = 1.74 \times 10^{-7} \mathrm{M}\)

Step 3: Determine the concentration of hydroxide ions (\([\mathrm{OH^-}]\))

From the \(K_w\) expression, \( K_w = [\mathrm{H^+}] [\mathrm{OH^-}] = 1.0 \times 10^{-14}\). We have the value of \([\mathrm{H^+}]\). We can now find \([\mathrm{OH^-}]\) as follows: \([\mathrm{OH^-}] = \frac{K_w}{[\mathrm{H^+}]} = \frac{1.0 \times 10^{-14}}{1.74 \times 10^{-7}} = 5.75 \times 10^{-8} \mathrm{M}\)

Step 4: Calculate the value of \(K_b\) for the conjugate base

We have the molarity of the weak acid, \(M\), and the concentration of \(\mathrm{OH^-}\) ions. We can assume that all \(\mathrm{OH^-}\) ions have come from the dissociation of the conjugate base. Therefore, the initial concentration of the conjugate base is \(0.0361 \mathrm{M}\). Now, we can use the expression for \(K_b\) in terms of conjugate base (\([\mathrm{B^-}]\)) and hydroxide ions (\([\mathrm{OH^-}]\)) concentrations: \(K_b = \frac{[\mathrm{B^-}] [\mathrm{OH^-}]}{[\mathrm{HB}]}\) In this case, we assume that all the weak acid has dissociated into its conjugate base. Therefore, \([\mathrm{B^-}] = 0.0361 \mathrm{M}\) and \([\mathrm{OH^-}] = 5.75 \times 10^{-8} \mathrm{M}\). Now, we can calculate \(K_b\) as follows: \(K_b = \frac{(0.0361)(5.75 \times 10^{-8})}{0.0361 - 5.75 \times 10^{-8}} \approx 5.75 \times 10^{-8}\) So, the value of \(K_b\) for the conjugate base is approximately \(5.75 \times 10^{-8}\).

Key Concepts

Weak Acid

A weak acid is a type of acid that partially dissociates in solution. This means that when placed in water, only a fraction of the acid molecules will break down into ions. This is in contrast to a strong acid, which fully dissociates in solution.

• Weak acids have a higher pH compared to strong acids. This is because fewer hydronium ions \( \mathrm{H^+} \) are produced in the solution.
• The strength of a weak acid is quantified by its acid dissociation constant \( K_a \). A smaller \( K_a \) value indicates a weaker acid.
• Examples of weak acids include acetic acid and citric acid.

Understanding weak acids is important because they behave differently in chemical reactions and solutions. In practice, the equilibrium between the undissociated acid and its ions plays a crucial role in reaction mechanics.

Conjugate Base

The conjugate base is what remains after an acid has donated a proton ( \( \mathrm{H^+} \)). This concept is essential in understanding acid-base equilibria.

• A conjugate base can accept a proton to revert to its original acid form.
• The strength of a conjugate base is inversely related to the strength of the acid from which it is derived.
• In our exercise, when a weak acid partially dissociates, it forms its conjugate base.

For a weak acid like \( \mathrm{HA} \), in water, it partially converts to its conjugate base \( \mathrm{A^-} \), establishing an equilibrium. The position of this equilibrium helps in determining the \( K_b \), which is used to determine the strength of the conjugate base.

Molarity

Molarity \( (M) \) is a measure of the concentration of a solution, expressed as the number of moles of solute per liter of solution. It's a crucial concept in chemistry, especially for reactions in solutions like titrations and calculation of osmotic pressure.

• The formula for molarity is given by \( M = \frac{moles\ of\ solute}{liters\ of\ solution} \).
• It is important to use the correct units when calculating molarity to ensure accurate results in experiments and problem-solving.
• In the given exercise, the osmotic pressure formula \( \pi = i M R T \) helps us determine the molarity of the solution, linking it directly to the solute concentration.

Understanding molarity allows us to predict how substances will interact in a solution based on their concentrations, making it a foundational concept in chemistry.

Hydronium Ions

Hydronium ions \( \mathrm{H_3O^+} \) are formed when a hydrogen ion \( \mathrm{H^+} \) associates with a water molecule. These ions play a central role in the chemistry of acids and bases.

• The concentration of hydronium ions in a solution determines its acidity or pH. The lower the pH, the higher the \( \mathrm{H_3O^+} \) concentration.
• pH is calculated using the formula \( \mathrm{pH} = -\log{[H^+]} \), thereby giving insights into the acidity of the solution.
• In the exercise, converting the pH into the concentration of hydronium ions illustrates how acidic the unknown solution is, a necessary step for further calculations.

Hydronium ions not only help define the state of a solution but are also pivotal in understanding reaction mechanisms and equilibrium in acid-base chemistry.