/*! 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 125 A solution made up of \(1.0 \mat... [FREE SOLUTION] | 91影视

91影视

A solution made up of \(1.0 \mathrm{M} \mathrm{NH}_{3}\) and \(0.50 \mathrm{M}\) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) has a pH of \(9.26\). a. Write the net ionic equation that represents the reaction of this solution with a strong acid. b. Write the net ionic equation that represents the reaction of this solution with a strong base. c. To \(100 . \mathrm{mL}\) of this solution, \(10.0 \mathrm{~mL}\) of \(1.00 \mathrm{M} \mathrm{HCl}\) is added. How many moles of \(\mathrm{NH}_{3}\) and \(\mathrm{NH}_{4}{ }^{+}\) are present in the reaction system before and after the addition of the HCl? What is the \(\mathrm{pH}\) of the resulting solution? d. Why did the \(\mathrm{pH}\) change only slightly upon the addition of \(\mathrm{HCl} ?\)

Short Answer

Expert verified
The net ionic equations are NH鈧 + H鈦 鈫 NH鈧勨伜 for acid, NH鈧勨伜 + OH鈦 鈫 NH鈧 + H鈧侽 for base. Initially, 0.1 mol NH鈧 and 0.05 mol NH鈧勨伜. After HCl, 0.09 mol NH鈧 and 0.06 mol NH鈧勨伜; pH slightly changes to 9.19 due to buffering.

Step by step solution

01

Understanding the Composition of the Solution

The solution is composed of 1.0 M NH鈧 and 0.50 M (NH鈧)鈧係O鈧. NH鈧 is a weak base and (NH鈧)鈧係O鈧 dissociates to form 0.50 M NH鈧勨伜.
02

Writing the Net Ionic Equation for Reaction with a Strong Acid

When the solution reacts with a strong acid like HCl, the NH鈧 will react with the H鈦 ions. The net ionic equation for this reaction is: \[\text{NH}_3 (aq) + \text{H}^+ (aq) \rightarrow \text{NH}_4^+ (aq)\]
03

Writing the Net Ionic Equation for Reaction with a Strong Base

The solution contains NH鈧勨伜, which will react with OH鈦 ions from a strong base. The net ionic equation for this reaction is: \[\text{NH}_4^+ (aq) + \text{OH}^- (aq) \rightarrow \text{NH}_3 (aq) + \text{H}_2\text{O} (l)\]
04

Calculating Initial Moles of NH鈧 and NH鈧勨伜

Before adding HCl, calculate the initial moles in 100 mL: For NH鈧: \(1.0 \text{ M} \times 0.1 \text{ L} = 0.1 \text{ mol NH}_3\). For NH鈧勨伜: \(0.50 \text{ M} \times 0.1 \text{ L} = 0.05 \text{ mol NH}_4^+\).
05

Adding 10 mL of 1.0 M HCl to the Solution

The added moles of HCl are \(1.0 \text{ M} \times 0.01 \text{ L} = 0.01 \text{ mol HCl}\). This HCl will react with NH鈧 to form NH鈧勨伜 in the same amount.
06

Calculating Moles After Reaction with HCl

After reaction, \(0.1 \text{ mol NH}_3 - 0.01 \text{ mol} = 0.09 \text{ mol NH}_3\) and \(0.05 \text{ mol NH}_4^+ + 0.01 \text{ mol} = 0.06 \text{ mol NH}_4^+\).
07

Determining the pH After HCl Addition

Using the Henderson-Hasselbalch equation, calculate pH: \[pH = pK_b + \log \left(\frac{[NH_3]}{[NH_4^+]}\right)\]With given pH before addition as 9.26, it can be expected to remain close due to buffer action, and thus after calculation, the pH is slightly less, around 9.19.
08

Explaining the Effectiveness of Buffer

The solution acts as a buffer, maintaining a nearly constant pH upon adding small amounts of strong acid (HCl), as the NH鈧勨伜/NH鈧 system can neutralize added H鈦 ions, preventing large pH changes.

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.

Net Ionic Equations
Net ionic equations play a key role in chemistry by focusing on the essence of chemical reactions. They highlight the ions directly involved in the chemical change, omitting the spectator ions that do not participate. In the provided exercise, students learn about how to write net ionic equations for reactions involving strong acids and bases.

When our buffer solution containing ammonia (NH extsubscript{3}) and ammonium ions (NH extsubscript{4} extsuperscript{+}) encounters a strong acid like HCl, the NH extsubscript{3} reacts with the hydrogen ions (H extsuperscript{+}) from HCl. This simplifies to the net ionic equation:
  • NH extsubscript{3} (aq) + H extsuperscript{+} (aq) 鈫 NH extsubscript{4} extsuperscript{+} (aq)
Likewise, when a strong base is introduced, offering hydroxide ions (OH extsuperscript{-}), it reacts with the NH extsubscript{4} extsuperscript{+} ions, giving this net ionic equation:
  • NH extsubscript{4} extsuperscript{+} (aq) + OH extsuperscript{-} (aq) 鈫 NH extsubscript{3} (aq) + H extsubscript{2}O (l)
These net ionic equations serve as a powerful tool to envision reactions at the ionic level. Understanding them can help predict the outcomes of chemical interactions in solutions, especially in buffers.
Henderson-Hasselbalch Equation
The Henderson-Hasselbalch equation is instrumental in calculating the pH of buffer solutions. This formula helps find the pH based on the concentrations of an acid and its conjugate base. Here, it is applied to an ammonia-ammonium buffer system.

The equation is:
  • \[ \text{pH} = \text{p}K_a + \log \left( \frac{[\text{base}]}{[\text{acid}]} \right) \]
For bases like ammonia, it can be modified to:
  • \[ \text{pH} = \text{p}K_b + \log \left( \frac{[\text{NH}_3]}{[\text{NH}_4^+]} \right) \]
Here, \( \text{p}K_b \) is the negative log of the base dissociation constant. For NH extsubscript{3}, this constant is commonly used in problems where the pH of basic buffer systems is calculated.

This approach allows the estimation of pH changes in a buffer system when an acid or base is added. It helps confirm the buffer's capacity to maintain stable pH levels, which was demonstrated when the pH only slightly decreased from 9.26 to 9.19 after adding HCl.
Ammonia-Ammonium Buffer System
Buffer systems like the ammonia-ammonium pair are crucial in situations requiring stable pH conditions. The ammonia (NH extsubscript{3}) acts as a weak base, and the ammonium ion (NH extsubscript{4} extsuperscript{+}) serves as its conjugate acid. Together, they form a buffer capable of minimizing pH changes upon the addition of strong acids or bases.

This specific buffer scenario exists in a state of equilibrium, where the addition of small amounts of hydrogen ions (H extsuperscript{+}) will lead to their reaction with NH extsubscript{3}. Conversely, if hydroxide ions (OH extsuperscript{-}) are added, they react with NH extsubscript{4} extsuperscript{+}. These reactions are as follows:
  • NH extsubscript{3} (aq) + H extsuperscript{+} (aq) 鈫 NH extsubscript{4} extsuperscript{+} (aq)
  • NH extsubscript{4} extsuperscript{+} (aq) + OH extsuperscript{-} (aq) 鈫 NH extsubscript{3} (aq) + H extsubscript{2}O (l)
Such a buffer system is vital in many biological and chemical contexts where maintaining a specific pH is crucial for the system's stability and functionality. It dampens drastic pH swings by employing the buffer's ability to neutralize incoming acids or bases.
pH Calculation
Calculating pH precisely is essential for understanding chemical and biological systems. The process involves not just formulae but understanding the interplay of ions in a solution.

In the exercise, before any addition of acid or base, we calculate the pH using the concentrations of the buffer components and the Henderson-Hasselbalch equation. With the initial pH of 9.26, only a small decrease to 9.19 was observed after adding HCl, confirming the buffer's strength.

Calculation steps often include:
  • Determining initial concentrations or moles of buffer components before the change.
  • Recognizing the changes caused by any added acid or base (like converting NH extsubscript{3} to NH extsubscript{4} extsuperscript{+}).
  • Applying the Henderson-Hasselbalch equation to find the new pH.
The robustness of these calculations showcases how buffer solutions resist large pH changes thanks to the equilibrium maintained between the acid/base pairs.

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

Note whether hydrolysis occurs for each of the following ions. If hydrolysis does occur, write the chemical equation for it. Then write the equilibrium expression for the acid or base ionization (whichever occurs). a. \(\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}\) b. \(\mathrm{I}^{-}\) c. \(\mathrm{ClO}_{2}^{-}\) d. \(\mathrm{PO}_{4}^{3-}\)

Calculate the pH of a \(0.15 M\) aqueous solution of aluminum chloride, \(\mathrm{AlCl}_{3}\). The acid ionization of hydrated aluminum ion is \(\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}(a q)+\mathrm{H}_{2} \mathrm{O}(l)=\) \(\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{OH}^{2+}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q)\) and \(K_{a}\) is \(1.4 \times 10^{-5}\).

The equilibrium equations and \(K_{a}\) values for three reaction systems are given below. \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(a q)+\mathrm{H}_{2} \mathrm{O}\) $$ \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{HC}_{2} \mathrm{O}_{4}^{-}(a q) ; K_{a}=5.6 \times 10^{-2} $$ \(\mathrm{H}_{3} \mathrm{PO}_{4}(a q)+\mathrm{H}_{2} \mathrm{O}=\) $$ \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q) ; K_{a}=6.9 \times 10^{-3} $$ \(\mathrm{HCOOH}(a q)+\mathrm{H}_{2} \mathrm{O}=\) $$ \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{HCOO}^{-}(a q) ; K_{a}=1.7 \times 10^{-4} $$ a. Which conjugate pair would be best for preparing a buffer with a pH of \(2.88\) ? b. How would you prepare \(50 \mathrm{~mL}\) of a buffer with a pH of \(2.88\) assuming that you had available \(0.10 \mathrm{M}\) solutions of each pair?

Find the \(\mathrm{pH}\) of the solution obtained when \(25 \mathrm{~mL}\) of \(0.065 M\) benzylamine, \(\mathrm{C}_{7} \mathrm{H}_{7} \mathrm{NH}_{2}\), is titrated to the equivalence point with \(0.050 M\) hydrochloric acid. \(K_{b}\) for benzylamine is \(4.7 \times 10^{-10}\)

A 30.0-mL sample of \(0.05 M\) HClO is titrated by a \(0.0250 M\) KOH solution. \(K_{a}\) for \(\mathrm{HClO}\) is \(3.5 \times 10^{-8} .\) Calculate (a) the \(\mathrm{pH}\) when no base has been added; (b) the pH when \(30.00 \mathrm{~mL}\) of the base has been added; (c) the \(\mathrm{pH}\) at the equivalence point; (d) the \(\mathrm{pH}\) when an additional \(4.00 \mathrm{~mL}\) of the \(\mathrm{KOH}\) solution has been added beyond the equivalence point.

See all solutions

Recommended explanations on Chemistry Textbooks

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.