Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 13: Problem 175

Calculate the \(\mathrm{pH}\) of a \(0.200-M\) solution of \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHF}\). Hint: \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHF}\) is a salt composed of \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}\) and \(\mathrm{F}^{-}\) ions. The principal equilibrium in this solution is the best acid reacting with the best base; the reaction for the principal equilibrium is $$\begin{aligned} \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}(a q)+\mathrm{F}^{-}(a q) & \rightleftharpoons \\ \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}(a q) &+\mathrm{HF}(a q) \quad K=8.2 \times 10^{-3}\end{aligned}$$

Short Answer

Expert verified
The pH of the 0.200-M solution of $\mathrm{C}_{5}\mathrm{H}_{5}\mathrm{NHF}$ is approximately 1.74.

Step by step solution

01

Write the Principal Equilibrium Reaction

Write the principal equilibrium reaction between C5H5NH+ and F-. \( C5H5NH^+(aq) + F^-(aq) \rightleftharpoons C5H5N(aq) + HF(aq) \)
02

Write Down the Equilibrium Constant

Write down the equilibrium constant (K) given for the reaction. We have, \( K = 8.2 \times 10^{-3} \)
03

Set Up an ICE Table

Set up an Initial, Change, and Equilibrium (ICE) table for the reaction. \( \begin{array}{c|ccc} & C5H5NH^+ & +F^- & \rightleftharpoons & C5H5N & +HF \\ \hline I & 0.200 & 0.200 & & 0 & 0 \\ C & -x & -x & & +x & +x \\ E & 0.200-x & 0.200-x & & x & x \\ \end{array} \)
04

Write the Equilibrium Expression

Write the equilibrium expression (using K) and substitute the values from the ICE table. \( K = \frac{[C5H5N][HF]}{[C5H5NH^+][F^-]} \) Plug in the values from the ICE table: \( 8.2 \times 10^{-3} = \frac{x^2}{(0.200-x)^2} \)
05

Solve the Equation for x

Assuming that the reaction proceeds to an equilibrium where x is much less than 0.200, we can solve the equation for x. \( x^2 = (8.2 \times 10^{-3}) (0.200)^2 \) We can now find the x value: \( x = \sqrt{(8.2 \times 10^{-3})(0.200)^2} \) \( x = 0.01807\)
06

Calculate the pH

Now that we have the concentration of HF (x), we can calculate pH using the following equation: \(\phantom{xx} pH = -\log_{10}(\mathrm{H}^+ )\) We know that the concentration of H+ is equal to the concentration of HF because we have a 1:1 relationship between them: \(\phantom{xx} [\mathrm{H}^+] = [HF] = 0.01807 \) Therefore, the pH of the solution will be: \(\phantom{xx} pH = -\log_{10}(0.01807) \) Calculating the pH, \(\phantom{xx} pH \approx 1.74 \) The pH of the 0.200-M solution of C5H5NHF is approximately 1.74.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Quinine \(\left(\mathrm{C}_{20} \mathrm{H}_{24} \mathrm{N}_{2} \mathrm{O}_{2}\right)\) is the most important alkaloid derived from cinchona bark. It is used as an antimalarial drug. For quinine, \(\mathrm{p} K_{\mathrm{b}_{1}}=5.1\) and \(\mathrm{p} K_{\mathrm{b}_{2}}=9.7\left(\mathrm{p} K_{\mathrm{b}}=-\log K_{\mathrm{b}}\right) .\) Only 1 g quinine will dissolve in \(1900.0 \mathrm{mL}\) of solution. Calculate the \(\mathrm{pH}\) of a saturated aqueous solution of quinine. Consider only the reaction \(\mathrm{Q}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{QH}^{+}+\mathrm{OH}^{-}\) described by \(\mathrm{p} K_{\mathrm{b}_{1}},\) where \(\mathrm{Q}=\) quinine.

A \(2.14-\mathrm{g}\) sample of sodium hypoiodite is dissolved in water to make 1.25 L of solution. The solution \(\mathrm{pH}\) is \(11.32 .\) What is \(K_{\mathrm{b}}\) for the hypoiodite ion?

Write the reaction and the corresponding \(K_{\mathrm{b}}\) equilibrium expression for each of the following substances acting as bases in water. a. \(\mathrm{NH}_{3}\) b. \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}\)

A \(0.100-\mathrm{g}\) sample of the weak acid HA (molar mass \(=\) \(100.0 \mathrm{g} / \mathrm{mol}\) ) is dissolved in \(500.0 \mathrm{g}\) water. The freezing point of the resulting solution is \(-0.0056^{\circ} \mathrm{C}\). Calculate the value of \(K_{\mathrm{a}}\) for this acid. Assume molality equals molarity in this solution.

An aqueous solution contains a mixture of 0.0500 \(M\) HCOOH \(\left(K_{\mathrm{a}}=1.77 \times 10^{-4}\right)\) and \(0.150 M \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\left(K_{\mathrm{a}}=1.34 \times\right.\) \(10^{-5}\) ). Calculate the \(p\) H of this solution. Because both acids are of comparable strength, the \(\mathrm{H}^{+}\) contribution from both acids must be considered.
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Chemical Reactions

Read Explanation

Nuclear Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemistry Branches

Read Explanation

Making Measurements

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.

Sign-up for free