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Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

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BuyFindarrow_forward

Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

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?

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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

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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Chapter 16 Solutions

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Sect-16.10 P-1.2ACPSect-16.10 P-1.3ACPSect-16.10 P-2.1ACPSect-16.10 P-2.2ACPSect-16.10 P-2.3ACPSect-16.10 P-2.4ACPSect-16.10 P-2.5ACPCh-16 P-1PSCh-16 P-2PSCh-16 P-3PSCh-16 P-4PSCh-16 P-5PSCh-16 P-6PSCh-16 P-7PSCh-16 P-8PSCh-16 P-9PSCh-16 P-10PSCh-16 P-11PSCh-16 P-12PSCh-16 P-13PSCh-16 P-14PSCh-16 P-15PSCh-16 P-16PSCh-16 P-17PSCh-16 P-18PSCh-16 P-19PSCh-16 P-20PSCh-16 P-21PSCh-16 P-22PSCh-16 P-23PSCh-16 P-24PSCh-16 P-25PSCh-16 P-26PSCh-16 P-27PSCh-16 P-28PSCh-16 P-29PSCh-16 P-30PSCh-16 P-31PSCh-16 P-32PSCh-16 P-33PSCh-16 P-34PSCh-16 P-35PSCh-16 P-36PSCh-16 P-37PSCh-16 P-38PSCh-16 P-39PSCh-16 P-40PSCh-16 P-41PSCh-16 P-42PSCh-16 P-43PSCh-16 P-44PSCh-16 P-45PSCh-16 P-46PSCh-16 P-47PSCh-16 P-48PSCh-16 P-49PSCh-16 P-50PSCh-16 P-51PSCh-16 P-52PSCh-16 P-53PSCh-16 P-54PSCh-16 P-55PSCh-16 P-56PSCh-16 P-57PSCh-16 P-58PSCh-16 P-59PSCh-16 P-60PSCh-16 P-61PSCh-16 P-62PSCh-16 P-63PSCh-16 P-64PSCh-16 P-65PSCh-16 P-66PSCh-16 P-67PSCh-16 P-68PSCh-16 P-69PSCh-16 P-70PSCh-16 P-71PSCh-16 P-72PSCh-16 P-73PSCh-16 P-74PSCh-16 P-75PSCh-16 P-76PSCh-16 P-77PSCh-16 P-78PSCh-16 P-79PSCh-16 P-80PSCh-16 P-81PSCh-16 P-82PSCh-16 P-83PSCh-16 P-84PSCh-16 P-85GQCh-16 P-86GQCh-16 P-87GQCh-16 P-88GQCh-16 P-89GQCh-16 P-90GQCh-16 P-91GQCh-16 P-92GQCh-16 P-93GQCh-16 P-94GQCh-16 P-95GQCh-16 P-96GQCh-16 P-97GQCh-16 P-98GQCh-16 P-99GQCh-16 P-100GQCh-16 P-101GQCh-16 P-102GQCh-16 P-103GQCh-16 P-104GQCh-16 P-105GQCh-16 P-106GQCh-16 P-107GQCh-16 P-108GQCh-16 P-109GQCh-16 P-110GQCh-16 P-111ILCh-16 P-112ILCh-16 P-113ILCh-16 P-114ILCh-16 P-115ILCh-16 P-116ILCh-16 P-117ILCh-16 P-118ILCh-16 P-119SCQCh-16 P-120SCQCh-16 P-121SCQCh-16 P-122SCQCh-16 P-123SCQCh-16 P-124SCQCh-16 P-125SCQCh-16 P-126SCQCh-16 P-127SCQCh-16 P-128SCQCh-16 P-129SCQ

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