   Chapter 16, Problem 16.155QP General Chemistry - Standalone boo...

11th Edition
Steven D. Gammon + 7 others
ISBN: 9781305580343

Solutions

Chapter
Section General Chemistry - Standalone boo...

11th Edition
Steven D. Gammon + 7 others
ISBN: 9781305580343
Textbook Problem

A chemist needs a buffer with pH 4.35. How many milliliters of pure acetic acid (density = 1.049 g/mL) must be added to 465 mL of 0.0941 M NaOH solution to obtain such a buffer?

Interpretation Introduction

Interpretation:

The amount of pure acetic acid (in milliliters) needed to be added to 465 mL of 0.0941 M NaOH to prepare a buffer with pH 4.35 has to be explained.

Concept Introduction:

Acid ionization constant Ka :

The ionization of a weak acid HA can be given as follows,

HA(aq)H+(aq)+A-(aq)

The equilibrium expression for the above reaction is given below.

Ka=[H+][A-][HA]

Where,

Ka is acid ionization constant,

[H+]   is concentration of hydrogen ion

[A-]   is concentration of acid anion

[HA] is concentration of the acid

To Explain: The amount of pure acetic acid (in milliliters) needed to be added to 465 mL of 0.0941 M NaOH to prepare a buffer with pH 4.35

Explanation

Given data:

Density of pure  acetic acid = 1.049 g/mL

The volume of NaOH is 465 mL

The concentration of NaOH is 0.0941 M

Amount of acetic acid needed:

The concentration of hydronium ion is:

[H3O+] =10-pH =10-4.35 =4.46×105 M

0.465 L of 0.0941 M NaOH will produce 0.043756 mol of acetate ion.

Use equilibrium expression for acetic acid.

[HAc][Ac-] =[H+]Ka=4.46×1051.7×105=2

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