   Chapter 16, Problem 16.7QP 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

H2A is a weak diprotic acid having Ka2 = 4.5 × 10−4 and Ka2 = 6.7 × 10−8. If the concentration of H2A is 0.45 M, what is the concentration of A2−?

Interpretation Introduction

Interpretation:

The concentration of A2 of a weak diprotic acid H2A has to be calculated

Concept Information:

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 Calculate: The concentration of A2 of a weak acid HA

Explanation

Given data:

The concentration of a weak diprotic acid H2A = 0.45 M

The value of Ka1 = 4.5×104

The value of Ka2 = 6.7×108

Ionizations of weak diprotic acid H2A :

First ionization:

H2AHA- + H+

Set up an equilibrium table for the first ionization.

Let x be the unknown

 H2A    ⇌    HA-   +   H+ Initial (M) 0.45 -x 0.45-x 0.00 0.00 Change (M) +x +x Equilibrium (M) x x

Ka1 =[H+][HA-][H2A]   4.5×104 =x20.45xAssuming x is smaller than 0.45 and neglecting it, we get x20.45 x =(4

Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started 