   Chapter 16, Problem 16.45QP 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

What is the hydronium-ion concentration of a 2.00 M solution of 2,6-dinitrobenzoic acid, (NO2)2C6H3COOH, for which Ka = 7.94 × 10−2?

Interpretation Introduction

Interpretation:

The hydronium ion concentration of a 2.00 M solution of 2,6-dinitrobenzoic acid has to be calculated

Concept Information:

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

pH definition:

The pH of a solution is defined as the negative base-10 logarithm of the hydrogen or hydronium ion concentration.

pH=-log[H+]

On rearranging, we get

[H+]=10pH

Explanation

To Calculate: The hydronium ion concentration of a 2.00 M solution of 2,6-dinitrobenzoic acid

Given data:

The concentration of the 2,6-dinitrobenzoic acid solution = 2.00 M

The Ka for the given acid is 7.94×102

Ionization of 2,6-dinitrobenzoic acid:

Let us denote 2,6-dinitrobenzoic acid ((NO2)2C6H5COOH) as HDin and its conjugate anion ((NO2)2C6H5COO-) as Din-

Set up the equilibrium table for the given acid solution.

Let x be the unknown concentration.

 HDin(aq)+H2O(l)     ⇄        H3O+(aq)  +   Din-(aq) Initial (M) 2.00 -x 2.00-x 0.00 0.00 Change (M) +x +x Equilibrium (M) x x

The Ka value for 2,6-dinitrobenzoic acid = 7

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