Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 15: Problem 105

A solution contains a weak monoprotic acid HA and its sodium salt NaA both at \(0.1 M\) concentration. Show that \(\left[\mathrm{OH}^{-}\right]=K_{\mathrm{w}} / K_{\mathrm{a}}\).

Short Answer

Expert verified
\[\left[\mathrm{OH}^{-}\right]= \frac{K_{\mathrm{w}}}{ K_{\mathrm{a}}}\]. Therefore, in a buffer solution of a weak acid and its salt, the hydroxide ion concentration equals the ion product of water divided by the acid dissociation constant.

Step by step solution

01

Write the equilibrium expression of the acid

Start by writing the chemical reactions of the weak acid and its salt in water. \[ HA <=> H^{+} + A^{-} \] The equilibrium constant expression is \( K_a = \frac{{[H^{+}][A^{-}]}}{{[HA]}} \)
02

Consider the common ion effect

Given that the sodium salt NaA also dissociates to produce the common ion A-, we have 0.1 M concentration of A- since the salt is a strong electrolyte. The weak acid and base exist in the buffer solution, thus the concentration of A- will be the same in the equilibrium situation as initially 0.1 M. As a result, A- is excluded from the expression, resulting in \( K_a = [H^{+}] \)
03

Relate \(Ka\) to \(Kw\)

Note that \(K_w = [H^{+}][OH^{-}]\) (the ion product of water), so the \( [H^{+}] = \frac{K_w}{[OH^{-}]}\). Substitute this into the equation from step 2, leading to \( K_a = \frac{K_w}{[OH^{-}]}\)
04

Solve for \( [OH^{-}] \)

Rearranging the latter equation gives our desired solution. So, \([OH^{-}] = \frac{K_w}{K_a}\)

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

Malonic acid \(\left[\mathrm{CH}_{2}(\mathrm{COOH})_{2}\right]\) is a diprotic acid. Compare its two \(K_{\mathrm{a}}\) values with that of acetic acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\left(K_{\mathrm{a}}\right),\) and account for the differences in the three \(K_{\mathrm{a}}\) values.

The pH of a \(0.30 M\) solution of a weak base is 10.66 . What is the \(K_{\mathrm{b}}\) of the base?

Calculate the number of moles of \(\mathrm{KOH}\) in \(5.50 \mathrm{~mL}\) of a \(0.360 \mathrm{M}\) KOH solution. What is the \(\mathrm{pOH}\) of the solution?

The ion product of \(\mathrm{D}_{2} \mathrm{O}\) is \(1.35 \times 10^{-15}\) at \(25^{\circ} \mathrm{C}\). (a) Calculate \(\mathrm{pD}\) where \(\mathrm{pD}=-\log \left[\mathrm{D}^{+}\right]\). (b) For what values of \(\mathrm{pD}\) will a solution be acidic in \(\mathrm{D}_{2} \mathrm{O} ?\) (c) Derive a relation between \(\mathrm{pD}\) and \(\mathrm{pOD}\).

Which of the following solutions has the highest pH: (a) 0.40 M HCOOH, (b) \(0.40 M \mathrm{HClO}_{4}\), (c) \(0.40 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH} ?\)
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Chemistry Branches

Read Explanation

Organic Chemistry

Read Explanation

Making Measurements

Read Explanation

Nuclear Chemistry

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