/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 39 Calculate the \(\mathrm{pH}\) af... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 39

Calculate the \(\mathrm{pH}\) after 0.010 mole of gaseous \(\mathrm{HCl}\) is added to \(250.0 \mathrm{mL}\) of each of the following buffered solutions. a. \(0.050 M \mathrm{NH}_{3} / 0.15 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\) b. \(0.50 M \mathrm{NH}_{3} / 1.50 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\) Do the two original buffered solutions differ in their pH or their capacity? What advantage is there in having a buffer with a greater capacity?

Short Answer

Expert verified
The pH of both buffered solutions after adding 0.010 mol of HCl is approximately 8.98. The two original buffered solutions have the same pH but differ in capacity. The 0.50 M NH鈧� / 1.50 M NH鈧凜l buffered solution has a greater capacity compared to the 0.050 M NH鈧� / 0.15 M NH鈧凜l buffered solution, which makes it more efficient and resistant to pH changes when more acid or base is added to the system.

Step by step solution

01

Calculate the initial moles of NH鈧� and NH鈧凜l

In a 250 mL solution: Moles of NH鈧� = Molarity 脳 Volume = 0.050 M 脳 0.250 L = 0.0125 mol Moles of NH鈧凜l = Molarity 脳 Volume = 0.15 M 脳 0.250 L = 0.0375 mol
02

Reaction between HCl and NH鈧�

When 0.010 mol of gaseous HCl is added, it will react with NH鈧� to form NH鈧勨伜 ions. As the amount of HCl is less than NH鈧�, 0.010 mol of NH鈧� will react with 0.010 mol of HCl to form 0.010 mol of NH鈧勨伜. Moles of NH鈧� after reaction = 0.0125 - 0.010 = 0.0025 mol Moles of NH鈧勨伜 after reaction = 0.0375 + 0.010 = 0.0475 mol
03

Use the Henderson-Hasselbalch equation to find pH

The Henderson-Hasselbalch equation is: pH = pK_a + log ([NH鈧僝/[NH鈧勨伜]) The pK_a of NH鈧勨伜 is 9.25. Hence, the pH = 9.25 + log (0.0025/0.0475) = 9.25 + log(0.05263) 鈮� 8.98 #b. 0.50 M NH鈧� / 1.50 M NH鈧凜l buffered solution#
04

Calculate the initial moles of NH鈧� and NH鈧凜l

In a 250 mL solution: Moles of NH鈧� = Molarity 脳 Volume = 0.50 M 脳 0.250 L = 0.125 mol Moles of NH鈧凜l = Molarity 脳 Volume = 1.50 M 脳 0.250 L = 0.375 mol
05

Reaction between HCl and NH鈧�

When 0.010 mol of gaseous HCl is added, it will react with NH鈧� to form NH鈧勨伜 ions. As the amount of HCl is less than NH鈧�, 0.010 mol of NH鈧� will react with 0.010 mol of HCl to form 0.010 mol of NH鈧勨伜. Moles of NH鈧� after reaction = 0.125 - 0.010 = 0.115 mol Moles of NH鈧勨伜 after reaction = 0.375 + 0.010 = 0.385 mol
06

Use the Henderson-Hasselbalch equation to find pH

The Henderson-Hasselbalch equation is: pH = pK_a + log ([NH鈧僝/[NH鈧勨伜]) Hence, the pH = 9.25 + log (0.115/0.385) = 9.25 + log(0.2987) 鈮� 8.98
07

Comparison of pH and capacity between the two buffered solutions

The two original buffered solutions have the same pH values (8.98), but their capacities are different. A buffer with greater capacity contains more moles of NH鈧� and NH鈧勨伜 ions, allowing it to maintain the pH even when more acid or base is added. The advantage of having a buffer with greater capacity is to maintain the pH under a broader range of conditions, making it more efficient and resistant to pH changes when more acid or base is added to the system. In this case, the 0.50 M NH鈧� / 1.50 M NH鈧凜l buffered solution has a greater capacity compared to the 0.050 M NH鈧� / 0.15 M NH鈧凜l buffered solution.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

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 fundamental tool in understanding buffer solutions. It's used to calculate the pH of a buffer solution. The equation is: \[\text{pH} = \text{p}K_a + \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right)\] In our problem, ammonia \((\text{NH}_3)\) acts as the base, while ammonium chloride \((\text{NH}_4^+\)) represents the acid. Here,
  • \(\text{p}K_a\) is a constant that depends on the weak acid involved. For ammonium, \(\text{p}K_a = 9.25\).
  • The equation helps you find the pH by using the concentrations (or moles) of the base and acid in the buffer.
The Henderson-Hasselbalch equation simplifies the calculations, making it easier to predict how the buffer will respond to the addition of acids or bases.
pH Calculation
Calculating pH is an essential part of understanding how buffers work. In the given example, we used the initial moles of ammonia \((\text{NH}_3)\) and ammonium \((\text{NH}_4^+)\) after adding \(\text{HCl}\).
  • The purpose of calculating pH is to see how close to neutral (pH = 7) a solution is.
  • By using the Henderson-Hasselbalch equation, we determined that the final pH in both buffer solutions is 8.98.
The pH calculation helps in assessing how well a buffer can maintain a stable pH even when external acids or bases are added.
Buffer Capacity
Buffer capacity tells us how well a buffer can resist changes in pH. A buffer with a high capacity contains more moles of the buffering components (like \(\text{NH}_3\) and \(\text{NH}_4^+\)).
  • A higher buffer capacity implies that the solution can handle larger amounts of added acid or base before the pH changes significantly.
  • In the example, the 0.50 M \(\text{NH}_3\) / 1.50 M \(\text{NH}_4\text{Cl}\) solution has a greater capacity than the 0.050 M \(\text{NH}_3\) / 0.15 M \(\text{NH}_4\text{Cl}\) solution.
Having a buffer with a greater capacity is advantageous when you need to maintain stable pH conditions in a solution, especially during chemical reactions or biological processes.
Acid-Base Reactions
Acid-base reactions play a crucial role in buffer solutions. When an acid like \(\text{HCl}\) is added to the buffer containing ammonia \((\text{NH}_3)\), it reacts to form ammonium ions \((\text{NH}_4^+)\). This reaction showcases:
  • The buffer鈥檚 ability to neutralize acids, thanks to the presence of \(\text{NH}_3\).
  • How the balance between \(\text{NH}_3\) and \(\text{NH}_4^+\) is restored, maintaining an almost constant pH.
Understanding these reactions helps in grasping how buffers protect against drastic pH changes, ensuring the stability of the environment in which they are used.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider the titration of \(100.0 \mathrm{mL}\) of \(0.100 \mathrm{M} \mathrm{H}_{2} \mathrm{NNH}_{2}\) \(\left(K_{\mathrm{b}}=3.0 \times 10^{-6}\right)\) by \(0.200 M\) HNO \(_{3}\). Calculate the pH of the resulting solution after the following volumes of \(\mathrm{HNO}_{3}\) have been added. a. \(0.0 \mathrm{mL}\) b. \(20.0 \mathrm{mL}\) c. \(25.0 \mathrm{mL}\) d. \(40.0 \mathrm{mL}\) e. \(50.0 \mathrm{mL}\) f. \(100.0 \mathrm{mL}\)

Malonic acid \(\left(\mathrm{HO}_{2} \mathrm{CCH}_{2} \mathrm{CO}_{2} \mathrm{H}\right)\) is a diprotic acid. In the titration of malonic acid with NaOH, stoichiometric points occur at \(\mathrm{pH}=3.9\) and \(8.8 .\) A 25.00 -mL sample of malonic acid of unknown concentration is titrated with 0.0984 \(M \mathrm{NaOH},\) requiring \(31.50 \mathrm{mL}\) of the NaOH solution to reach the phenolphthalein end point. Calculate the concentration of the initial malonic acid solution. (See Exercise \(113 .\) )

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.

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.)

Which of the following can be classified as buffer solutions? a. \(0.25 M\) HBr \(+0.25 M\) HOBr b. \(0.15 M \mathrm{HClO}_{4}+0.20 \mathrm{M} \mathrm{RbOH}\) c. \(0.50 M\) HOCl \(+0.35 M\) KOCl d. \(0.70 M \mathrm{KOH}+0.70 \mathrm{M} \mathrm{HONH}_{2}\) e. \(0.85 M \mathrm{H}_{2} \mathrm{NNH}_{2}+0.60 \mathrm{M} \mathrm{H}_{2} \mathrm{NNH}_{3} \mathrm{NO}_{3}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Organic Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Kinetics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.