   Chapter 16, Problem 16.47QP 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

Phthalic acid, H2C8H4O4, is a diprotic acid used in the synthesis of phenolphthalein indicator. Ka1 = 1.2 × 10−3, and Ka2 = 3.9 × 10−6. a Calculate the hydronium-ion concentration of a 0.015 M solution. b What is the concentration of the C8H4O42− ion in the solution?

Interpretation Introduction

Interpretation:

The concentration of hydronium ion and concentration of C8H4O42- ion of a 0.015 M phthalic acid solution 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

Diprotic and polyprotic acids:

Acids having two or more hydrogen atoms are termed as diprotic or polyprotic acids.  These acids lose one proton at a time by undergoing successive ionizations.

For diprotic acids, the successive ionization constants are designated as Ka1andKa2

For triprotic acids, the successive ionization constants are designated as Ka1,Ka2andKa3

To Calculate: The concentration of hydronium ion and concentration of C8H4O42- ion of a 0.015 M phthalic acid solution

Explanation

Given data:

Phthalic acid (H2C8H4O4)   is a diprotic acid

The concentration of the phthalic acid solution = 0.015 M

The value of Ka1 = 1.2×103

The value of Ka2 = 3.9×106

Ionizations of Phthalic acid:

Phthalic acid is a diprotic acid

The conjugate base resulting from the first ionization is the acid for the second ionization, and its starting concentration is the equilibrium concentration from the first ionization.

Let us represent phthalic acid (H2C8H4O4) as H2Ph and its first ionization conjugate anion (HC8H4O4) as HPh- and second ionization conjugate anion (C8H4O4) as Ph2-

The ionizations of phthalic acid is as follows,

H2Ph(aq)H+(aq)+ HPh-(aq)HPh-(aq)H+(aq)+ Ph2-(aq)

The ionization constant for first ionization is: Ka1 = 1.2×103

The ionization constant for second ionization is: Ka2 = 3.9×106

First ionization:

Construct an equilibrium table for first ionization.

Let x be the unknown in the first ionization.

 H2Ph(aq)⇄     H+(aq)+   HPh-(aq) Initial (M) 0.015 −x 0.015-x 0.00 0.00 Change (M) +x +x Equilibrium (M) x x

Write the equilibrium expression and substitute the equilibrium concentrations into it.

Ka1 =[H+][HPh-][H2Ph]   1.2×103 =x20.015xx cannot be ignored. So, solve x using quadratic equationx2+(0.0012x)1.80×105=0             x =0.0012±(0.0012)24(1.80×105)2 =0

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