Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 14: Problem 67

Monochloroacetic acid, \(\mathrm{HC}_{2} \mathrm{H}_{2} \mathrm{ClO}_{2}\), is a skin irritant that is used in "chemical peels" intended to remove the top layer of dead skin from the face and ultimately improve the complexion. The value of \(K_{\mathrm{a}}\) for monochloroacetic acid is \(1.35 \times 10^{-3}\). Calculate the \(\mathrm{pH}\) of a \(0.10 \mathrm{M}\) solution of monochloroacetic acid.

Short Answer

Expert verified
The pH of a 0.10 M solution of monochloroacetic acid is approximately 1.94.

Step by step solution

01

Write the chemical equilibrium equation for monochloroacetic acid

Monochloroacetic acid will dissociate into hydrogen ions and its conjugate base in water: \[HC_2 H_2 ClO_2(aq) \rightleftharpoons H^+(aq) + C_2 H_2 ClO_2^-(aq)\]
02

Write the Ka expression for the equilibrium

For the above equilibrium, the Ka expression is given by: \[K_a = \frac{[H^+][C_2 H_2 ClO_2^-]}{[HC_2 H_2 ClO_2]}\]
03

Establish the initial and equilibrium concentrations

Let \(x\) be the change in concentration of \(H^+\). Therefore, the concentration of \(C_2 H_2 ClO_2^-\) will also change by \(x\). Initial Concentrations: \([HC_2 H_2 ClO_2] = 0.10\:M,\: [H^+] = 0,\: [C_2 H_2 ClO_2^-] = 0\) Equilibrium Concentrations: \([HC_2 H_2 ClO_2] = 0.10 - x\:M,\: [H^+] = x\:M,\: [C_2 H_2 ClO_2^-] = x\:M\)
04

Substitute concentrations in the Ka expression and solve for x

Substituting the equilibrium concentrations into the Ka expression, we get: \[1.35 \times 10^{-3} = \frac{(x)(x)}{(0.10 - x)}\] For weak acids, the value of \(x\) is usually much smaller than the initial concentration so we can simplify the equation to: \[1.35 \times 10^{-3} = \frac{x^2}{0.10}\] Now, solving for \(x\), which represents the hydrogen ion concentration ([H+]): \[x = \sqrt{1.35 \times 10^{-3} \times 0.10} = \sqrt{1.35 \times 10^{-4}} \]
05

Calculate the pH

Using the hydrogen ion concentration found in step 4, we can calculate the pH using the formula: \[pH = -\log [H^+]\] \[pH = -\log(\sqrt{1.35 \times 10^{-4}})\] Now, using a logarithmic calculator, the pH of the 0.10 M solution of monochloroacetic acid can be calculated: \(\)pH \(\approx 1.94\) Therefore, the pH of a 0.10 M solution of monochloroacetic acid is approximately 1.94.

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

Calculate the percentage of pyridine \(\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\right)\) that forms pyridinium ion, \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}\), in a \(0.10 \mathrm{M}\) aqueous solution of pyridine \(\left(K_{\mathrm{b}}=1.7 \times 10^{-9}\right)\)

Rank the following \(0.10 \mathrm{M}\) solutions in order of increasing \(\mathrm{pH}\). a. HI, HF, NaF, NaI b. \(\mathrm{NH}_{4} \mathrm{Br}, \mathrm{HBr}, \mathrm{KBr}, \mathrm{NH}_{3}\) c. \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{NO}_{3}, \mathrm{NaNO}_{3}, \mathrm{NaOH}, \mathrm{HOC}_{6} \mathrm{H}_{5}, \mathrm{KOC}_{6} \mathrm{H}_{5}, \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\), \(\mathrm{HNO}_{3}\)

What are the major species present in \(0.015 M\) solutions of each of the following bases? a. \(\mathrm{KOH}\) b. \(\mathrm{Ba}(\mathrm{OH})_{2}\) What is \(\left[\mathrm{OH}^{-}\right]\) and the \(\mathrm{pH}\) of each of these solutions?

Papaverine hydrochloride (abbreviated papH \(^{+} \mathrm{Cl}^{-} ;\) molar mass \(=\) \(378.85 \mathrm{~g} / \mathrm{mol}\) ) is a drug that belongs to a group of medicines called vasodilators, which cause blood vessels to expand, thereby increasing blood flow. This drug is the conjugate acid of the weak base papaverine (abbreviated pap; \(K_{\mathrm{b}}=8.33 \times 10^{-9}\) at \(35.0^{\circ} \mathrm{C}\) ). Calculate the \(\mathrm{pH}\) of a \(30.0 \mathrm{mg} / \mathrm{mL}\) aqueous dose of papH \(^{+} \mathrm{Cl}^{-}\) prepared at \(35.0^{\circ} \mathrm{C} . K_{\mathrm{w}}\) at \(35.0^{\circ} \mathrm{C}\) is \(2.1 \times 10^{-14}\).

The equilibrium constant \(K_{\mathrm{a}}\) for the reaction \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}(a q)+\mathrm{H}_{2} \mathrm{O}(l)=\) \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}(\mathrm{OH})^{2+}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q)\) is \(6.0 \times 10^{-3}\). a. Calculate the \(\mathrm{pH}\) of a \(0.10 \mathrm{M}\) solution of \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\). b. Will a \(1.0 M\) solution of iron(II) nitrate have a higher or lower \(\mathrm{pH}\) than a \(1.0 \mathrm{M}\) solution of iron(III) nitrate? Explain.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Making Measurements

Read Explanation

Kinetics

Read Explanation

Chemical Reactions

Read Explanation

Inorganic Chemistry

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