   # Ascorbic acid (vitamin C, C 6 H 8 O 6 ) is a diprotic acid ( K a 1 = 6.8 × 10 −5 and K a 2 = 2.7 × 10 −12 ). What is the pH of a solution that contains 5.0 mg of acid per millilitre of solution? Ascorbic acid ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 16, Problem 72PS
Textbook Problem
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## Ascorbic acid (vitamin C, C6H8O6) is a diprotic acid (Ka1 = 6.8 × 10−5 and Ka2 = 2.7 × 10−12). What is the pH of a solution that contains 5.0 mg of acid per millilitre of solution? Ascorbic acid

Interpretation Introduction

Interpretation:

The pH of a solution has to be calculated when 5.0 mg of acid per millilitre of solution of ascorbic acid.

Concept introduction:

If a molecule donate only one hydrogen atom then the acid is called a monoprotic acid.

Example: HCl, HBr and HClO3

If each molecule can donate two or more hydrogen ions, the acid is a polyprotic acid.

A diprotic acid, such as sulfuric acid, H2SO4 has two acidic hydrogen atoms.

Some acids, such as phosphoric acid, H3PO4 are triprotic acids.

Polyprotic bases:

Can accept more than on proton (H+). So, it can react with water many times to produce many hydroxide ions (OH).

Diprotic bases:

Can accept two protons.

Example: Sulphate ion SO4(aq)2-

### Explanation of Solution

Let’s calculate the concentration of ascorbic acid

Given amount = 5.0mg

Molecular mass of ascorbic acid = 176g/mol

Molarity of ascorbic acid = Mass of ascorbic acid(g)Molar mass of ascorbic acid(g/mol)×1000Volume of solution(mL)=   [5.0mgC6H8O6×1g1000mg]176.g/molC6H8O6×(10001mL)=   2.84×10-2M

Ascorbic acid is a diprotic acid. The ionization steps of ascorbic are as follows.

First ionization:C6H8O6(aq) + H2O(aq)C6H7O6-(aq) + H3O+(aq)Equilibrium expression:Ka1 = [C6H7O6-][H3O+][C6H8O6]Ka1= 6.8×10-5Second ionization:C6H7O6-(aq) + H2O(aq)C6H6O62-(aq) + H3O+(aq)Equilibrium expression:Ka2[C6H6O62-][H3O+][C7H7O6-]Ka2= 2.7×10-12

From the Ka1 and Ka2 values,  Ka2 is smaller than the  Ka1.

Therefore, H3O+ is almost produced entirely from  Ka1.

Let’s calculate the H3O+ from  Ka1.

Enter the ICE table the concentrations before equilibrium is established, the change that occurs as the reaction proceeds to equilibrium and the concentrations when equilibrium has been achieved

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