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Chapter 14: Problem 38

A buffered solution is made by adding \(50.0 \mathrm{g} \mathrm{NH}_{4} \mathrm{Cl}\) to 1.00 \(\mathrm{L}\) of a 0.75-M solution of \(\mathrm{NH}_{3}\). Calculate the pH of the final solution. (Assume no volume change.)

Short Answer

Expert verified
The pH of the final buffered solution is approximately 9.155, calculated using the Henderson-Hasselbalch equation with concentrations of NH₄⁺ and NH₃ in the final solution.

Step by step solution

01

Calculate the concentration of NH₄⁺ and NH₃

First, we'll find the concentration of NH₄⁺ from the ammonium chloride added. We are given 50.0 g of NH₄Cl, and we need to convert it to moles and then find the concentration in the 1.00 L solution. Molar mass of NH₄Cl = 14.0067 (N) + 1.00784 (x4 Hydrogen) + 35.453 (Cl) = 53.491 g/mol Moles of NH₄Cl = mass / molar mass = 50.0 g / 53.491 g/mol = 0.9346 mol Concentration of NH₄⁺ = moles / volume = 0.9346 mol / 1.00 L = 0.9346 M Next, we're given the initial concentration of NH₃, which is 0.75 M. As we're assuming there is no volume change when NH₄Cl is added, this concentration remains the same in the final solution.
02

Use the Henderson-Hasselbalch equation

Now that we have the concentrations of NH₄⁺ and NH₃, we can use the Henderson-Hasselbalch equation to calculate the pH of the buffered solution. The equation is: pH = pKa + log ([A⁻]/[HA]) For NH₃/NH₄⁺, the pKa value is from the dissociation constant of NH₄⁺, which is: Ka = [H₃O⁺][NH₃] / [NH₄⁺] pKa = -log(Ka) For NH₄⁺, Ka = 5.6 x 10⁻¹⁰, so pKa = -log(5.6 x 10⁻¹⁰) ≈ 9.25. Plugging all values into the Henderson-Hasselbalch equation, we find: pH = 9.25 + log ([NH₃]/[NH₄⁺]) pH = 9.25 + log (0.75 M / 0.9346 M)
03

Calculate the pH

Now we can calculate the pH: pH = 9.25 + log (0.802) pH ≈ 9.25 + (-0.095) pH ≈ 9.155 The pH of the final buffered solution is approximately 9.155.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Henderson-Hasselbalch Equation
The Henderson-Hasselbalch equation is a formula that relates the pH of a solution to the pKa of the acid and the ratio of the concentrations of the deprotonated form (base) to the protonated form (acid). Specifically, it is expressed as:\[\begin{equation}\text{pH} = \text{pKa} + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)\end{equation}\]

This equation is particularly useful for buffered solutions, which are designed to resist changes in pH when small amounts of acid or base are added. A buffer typically consists of a weak acid and its conjugate base or a weak base and its conjugate acid. In the context of the exercise, ammonia (NH₃) represents the weak base, and ammonium (NH₄⁺) represents its conjugate acid.

To apply this equation, you need the pKa value, which comes from the acid’s dissociation constant (Ka). You also need the concentrations of the base \(A^-\) and the acid \(HA\). By plugging these values into the equation, you can solve for the pH of the solution. Understanding and applying the Henderson-Hasselbalch equation are essential for predicting the behavior of buffered solutions in various chemical contexts.

Acid-Base Equilibrium
The concept of acid-base equilibrium plays a central role in understanding how the Henderson-Hasselbalch equation works. Acid-base equilibrium refers to the state where the rates of the forward and reverse reactions of acid and base dissociation are equal, resulting in stable concentrations of the species involved.

For a simple acid dissociation, the equilibrium is given by:

\[\begin{equation}\text{HA} \leftrightarrow \text{H}^+ + \text{A}^-,\end{equation}\]

where HA is the weak acid, H⁺ is the hydrogen ion, and A⁻ is the conjugate base. The equilibrium constant for this reaction is the acid dissociation constant, Ka, and reflects the strength of the acid. The pKa is the negative logarithm of Ka and is used in the Henderson-Hasselbalch equation.

In the case of the buffered solution from the exercise, NH₄⁺ (the acid) is in equilibrium with NH₃ (the base) and H₃O⁺ (the hydronium ion). By carefully balancing the concentrations of NH₄⁺ and NH₃, the solution can minimize pH changes, making it an effective buffer.

Molar Mass Calculation
Molar mass calculation is a fundamental skill in chemistry which allows scientists and students to convert between the mass of a substance and the amount of substance in moles. The molar mass of a compound is determined by summing the atomic masses of each element present in the molecule, multiplied by the number of atoms of that element in the molecule.

Example Calculation from Exercise

In our given exercise with NH₄Cl, we calculate the molar mass as follows:

\[\begin{equation}\text{Molar mass of NH}_4\text{Cl} = \text{Molar mass of N} + 4 \times \text{Molar mass of H} + \text{Molar mass of Cl},\end{equation}\]

This calculation lets us convert the given mass of NH₄Cl into moles, which is crucial for subsequent calculations involving concentration. In real-world applications, knowing how to calculate molar mass is essential for preparing solutions and for stoichiometric calculations in chemical reactions.

Concentration of Solutions
The concentration of a solution is a measure of the amount of solute that is dissolved in a given quantity of solvent. The most common units of concentration are molarity (M), which is expressed as moles of solute per liter of solution. Understanding how to calculate and apply concentrations is critical for tasks such as mixing solutions and calculating pH values.

How to Calculate Concentration

The concentration is calculated using the formula:

\[\begin{equation}\text{Concentration (M)} = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}}\end{equation}\]

From our exercise, once the molar mass of NH₄Cl is known, we can find the moles of NH₄Cl, and then divide by the volume of the solution to get the concentration.

Applying these calculations to the buffer solution, you can determine the concentrations of both the weak base NH₃ and the conjugate acid NH₄⁺, which are necessary inputs for the Henderson-Hasselbalch equation to calculate the pH. This process highlights the interconnectivity between the concepts of molar mass, concentration, and acid-base equilibrium in the study of chemistry.

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Most popular questions from this chapter

When a diprotic acid, \(\mathrm{H}_{2} \mathrm{A}\), is titrated with \(\mathrm{NaOH}\), the protons on the diprotic acid are generally removed one at a time, resulting in a pH curve that has the following generic shape: a. Notice that the plot has essentially two titration curves. If the first equivalence point occurs at \(100.0 \mathrm{mL}\) NaOH added, what volume of NaOH added corresponds to the second equivalence point? b. For the following volumes of NaOH added, list the major species present after the OH \(^{-}\) reacts completely. i. \(0 \mathrm{mL} \mathrm{NaOH}\) added ii. between 0 and \(100.0 \mathrm{mL}\) NaOH added iii. \(100.0 \mathrm{mL}\) NaOH added iv. between 100.0 and \(200.0 \mathrm{mL}\) NaOH added v. \(200.0 \mathrm{mL} \mathrm{NaOH}\) added vi. after \(200.0 \mathrm{mL}\) NaOH added c. If the \(\mathrm{pH}\) at \(50.0 \mathrm{mL}\) NaOH added is \(4.0,\) and the \(\mathrm{pH}\) at \(150.0 \mathrm{mL} \mathrm{NaOH}\) added is \(8.0,\) determine the values \(K_{\mathrm{a}_{1}}\) and \(K_{\mathrm{a}_{2}}\) for the diprotic acid.

You have a solution of the weak acid HA and add some HCl to it. What are the major species in the solution? What do you need to know to calculate the \(\mathrm{pH}\) of the solution, and how would you use this information? How does the \(\mathrm{pH}\) of the solution of just the HA compare with that of the final mixture? Explain.

Consider a buffer solution where [weak acid] \(>\) [conjugate base]. How is the pH of the solution related to the \(\mathrm{p} K_{\mathrm{a}}\) value of the weak acid? If [conjugate base] \(>\) [weak acid], how is pH related to \(\mathrm{p} K_{\mathrm{a}} ?\)

Mixing together solutions of acetic acid and sodium hydroxide can make a buffered solution. Explain. How does the amount of each solution added change the effectiveness of the buffer?

Consider the following four titrations (i-iv): i. \(150 \mathrm{mL}\) of \(0.2 \mathrm{M} \mathrm{NH}_{3}\left(K_{\mathrm{b}}=1.8 \times 10^{-5}\right)\) by \(0.2 \mathrm{M} \mathrm{HCl}\) ii. \(150 \mathrm{mL}\) of \(0.2 \mathrm{M}\) HCl by \(0.2 \mathrm{M} \mathrm{NaOH}\) iii. \(150 \mathrm{mL}\) of \(0.2 \mathrm{M} \mathrm{HOCl}\left(K_{\mathrm{a}}=3.5 \times 10^{-8}\right)\) by \(0.2 \mathrm{M} \mathrm{NaOH}\) iv. \(150 \mathrm{mL}\) of \(0.2 \mathrm{M} \mathrm{HF}\left(K_{\mathrm{a}}=7.2 \times 10^{-4}\right)\) by \(0.2 \mathrm{M} \mathrm{NaOH}\) a. Rank the four titrations in order of increasing \(\mathrm{pH}\) at the halfway point to equivalence (lowest to highest \(\mathrm{pH}\) ). b. Rank the four titrations in order of increasing \(\mathrm{pH}\) at the equivalence point. c. Which titration requires the largest volume of titrant (HCl or \(\mathrm{NaOH}\) ) to reach the equivalence point?
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