   Chapter 16, Problem 108GQ

Chapter
Section
Textbook Problem

The hydrogen phthalate ion, C8HsO4−, is a weak acid with Ka = 3.91 × 10−6. C 8 H 5 O 4 − (aq) + H 2 O( l )  ⇄  C 8 H 4 O 4 2 − (aq) + H 3 O + (aq) What is the pH of a 0.050 M solution of potassium hydrogen phthalate. KC8H5O4? Note: To find the pH for a solution of the anion, we must take into account that the ion is amphiprotic. It can be shown that, for most cases of amphiprotic ions, the H3O+ concentration is [H 3 O + ] =  K a1   ×   K a2 For phthalic acid, C8H6O4 is Ka1 is 1.12 × 10−3, and Ka2 is 3.91 × 10−6.

Interpretation Introduction

Interpretation:

pH has to be determined for the 0.050 M potassium hydrogen phthalate solution.

Concept introduction:

Amphoteric solutions:

The solutions behaves like acids and bases.That means it donate protons as

Well as accepts protons from the water molecule.

The equilibrium expression for the amphoteric solution:

[H3O+] = K1×K2

Explanation

KC8H5O4 dissociates into K+ and C8H5O4- ions into the solution. K+ ions does not affect pH of the solution. C8H5O4- ions abstract protons from water and producing [OH-] ions into the solution and can also donate proton to water for producing H3O+ ions into the solution.

Therefore, C8H5O4- is an amphoteric nature.

Let’s calculate the [H3O+]

For amphoteric ions.

[H3O+] = K1×K2

The values of K1 and K2 of C8H6O4 are 1

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