/*! 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 25 The measured \(\mathrm{pH}\) of ... [FREE SOLUTION] | 91影视

91影视

The measured \(\mathrm{pH}\) of a \(0.100-\mathrm{M}\) solution of \(\mathrm{NH}_{3}(a q)\) at \(25^{\circ} \mathrm{C}\) is \(11.12 .\) Calculate \(K_{\mathrm{b}}\) for \(\mathrm{NH}_{3}(a q)\) at \(25^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The value of \(K_b\) for \(\mathrm{NH}_3(aq)\) at \(25^{\circ} \mathrm{C}\) is \(1.74 \times 10^{-5}\).

Step by step solution

01

Determine the hydroxide ion concentration

Since the pH of the solution is given as 11.12, we can determine the pOH using the relation: \[\text{pOH} = 14 - \text{pH} = 14 - 11.12 = 2.88\]Using the pOH, calculate the hydroxide ion concentration \[\text{[OH}^-\text{]} = 10^{-\text{pOH}} = 10^{-2.88}\]
02

Calculate [OH鈦籡

Utilizing the expression from Step 1, \[\text{[OH}^-\text{]} = 10^{-2.88} \approx 1.32 \times 10^{-3} \, \text{M}\]Now we have the concentration of hydroxide ions.
03

Use the equilibrium expression for Kb

For the equilibrium of ammonia in water:\[\text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^-\]The equilibrium expression is:\[K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]}\]Assuming the change in concentration of ammonia, \([\text{NH}_3]\), is small, \[[\text{NH}_3] \approx 0.100 - x \approx 0.100\, \text{M}\]where \(x\) is the concentration of \([\text{OH}^-]\) at equilibrium.
04

Solve for Kb

Now plug the values into the equilibrium expression, assuming \([\text{NH}_4^+] = [\text{OH}^-] = 1.32 \times 10^{-3} \, \text{M}\):\[K_b = \frac{(1.32 \times 10^{-3})(1.32 \times 10^{-3})}{0.100} \approx 1.74 \times 10^{-5}\]
05

Conclusion: Determine Kb

The base dissociation constant, \(K_b\), for \(\mathrm{NH}_3(aq)\) at \(25^{\circ} \mathrm{C}\) is \(1.74 \times 10^{-5}\).

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.

pH and pOH relationship
Understanding the relationship between pH and pOH is vital for aqueous solutions, especially when discussing bases and acids. The two values are interconnected and can be expressed by the equation \( \text{pH} + \text{pOH} = 14 \). This relationship underlines that as pH increases, indicating a more basic solution, the pOH value must decrease, and vice versa.
For instance, in our exercise with ammonia, the pH is given as 11.12, signaling a basic solution. To find the pOH, we could subtract the pH from 14, leading us to a pOH of 2.88. Thus, knowing either pH or pOH allows us to easily determine the other by using the formula above.
Recognizing this relationship is fundamental when calculating the equilibrium concentrations or constants, such as the base dissociation constant, \(K_b\), in chemical reactions involving bases like ammonia.
Ammonia equilibrium in water
When ammonia dissolves in water, it establishes an equilibrium reaction. This reaction can be represented by: \( \text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^- \). In this reaction, ammonia (\(\text{NH}_3\)) acts as a base and accepts a proton from water.
The equilibrium is characterized by the formation of ammonium ions (\(\text{NH}_4^+\)) and hydroxide ions (\(\text{OH}^-\)). The equilibrium constant for this reaction is known as the base dissociation constant, or \(K_b\).
The value of \(K_b\) offers insight into the strength of the base in water. In simpler terms, it tells us how easily the ammonia will dissociate into its ions in an aqueous solution. Calculating \(K_b\) for ammonia involves using the concentrations of the products and reactants at equilibrium, specifically by plugging these values into the expression: \( K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]} \). This calculation helps us understand both the behavior and characteristics of ammonia as it reacts in water.
Hydroxide ion concentration
The concentration of hydroxide ions in a solution is key to determining the solution's basicity, which is the opposite of acidity. For bases, like ammonia in water, the hydroxide ion concentration directly results from the dissociation process.
Once you know the pOH, with a value such as 2.88, you can find the hydroxide ion concentration using scientific notation: \( \text{[OH}^-\text{]} = 10^{-\text{pOH}} \). For the given case of ammonia, this calculation gives us approximately \( 1.32 \times 10^{-3} \text{M} \).
Having this concentration is crucial when you proceed to calculate other equilibrium properties, such as \(K_b\), making the hydroxide ion concentration a pivotal piece of the puzzle. Remember that concentration levels are always temperature-dependent and contribute to understanding the nature of the chemical reaction in an aqueous environment.

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

Given that \(K_{w}\) for water is \(2.40 \times 10^{-14} \mathrm{M}^{2}\) at \(37^{\circ} \mathrm{C}\), compute the \(\mathrm{pH}\) of a neutral aqueous solution at \(37^{\circ} \mathrm{C}\), which is the normal human body temperature. Is a \(\mathrm{pH}=7.00\) solution acidic or basic at \(37^{\circ} \mathrm{C} ?\)

Give the conjugate base for each of the following acids: (a) \(\mathrm{HClO}(a q)\) (b) \(\mathrm{NH}_{4}^{*}(a q)\) (c) \(\mathrm{HN}_{3}(a q)\) (d) \(\mathrm{HS}^{-}(a q)\)

Identify which of the following species are Br酶nsted-Lowry acids and which are Br酶nsted-Lowry bases in water. In each case give the chemical formula for the conjugate member of the conjugate acid-base pair: (a) \(\mathrm{HCNO}(a q)\) (b) \(\mathrm{OBr}^{-}(a q)\) (c) \(\mathrm{HClO}_{3}(a q)\) (d) \(\mathrm{CH}_{3} \mathrm{NH}_{3}^{*}(a q)\) (e) \(\mathrm{ClNH}_{2}(a q)\) (f) \(\mathrm{HONH}_{2}(a q)\)

The neutralization of an acid with a base yields a salt and water in aqueous solution. (a) Calculate the resulting \(\mathrm{pH}\) when \(50.0\) milliliters of \(0.250\) -M hydrochloric acid solution is mixed with \(50.0\) milliliters of \(0.250\) -M sodium hydroxide solution at \(25^{\circ} \mathrm{C}\). (b) Calculate the resulting \(\mathrm{pH}\) when \(50.0\) milliliters of \(0.250-\mathrm{M}\) acetic acid solution is mixed with \(50.0\) milliliters of \(0.250-\mathrm{M}\) sodium hydroxide solution at \(25^{\circ} \mathrm{C}\). (c) Explain why the \(\mathrm{pH}\) of the resulting solution in part (b) is not \(7.00\).

Determine whether each of the following substances is an Arrhenius acid, a Br贸nsted-Lowry acid, or a Lewis acid (it is possible for each to be of more than one type): (a) \(\mathrm{HCl}(a q)\) (b) \(\mathrm{AlCl}_{3}(a q)\) (c) \(\mathrm{BCl}_{3}(a q)\)

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.