   # A typical vitamin C tablet (containing pure ascorbic acid, H 2 C 6 H 6 O 6 ) weighs 500. mg. One vitamin C tablet is dissolved in enough water to make 200.0 mL of solution. Calculate the pH of this solution. Ascorbic acid is a diprotic acid. ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 13, Problem 107E
Textbook Problem
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## A typical vitamin C tablet (containing pure ascorbic acid, H2C6H6O6) weighs 500. mg. One vitamin C tablet is dissolved in enough water to make 200.0 mL of solution. Calculate the pH of this solution. Ascorbic acid is a diprotic acid.

Interpretation Introduction

Interpretation: The pH of the given solution, prepared by dissolving one vitamin C tablet in enough water to make 200.0mL of solution, is to be calculated.

Concept introduction: The pH of a solution is defined as a figure that expresses the acidity of the alkalinity of a given solution.

The pH of a solution is calculated by the formula, pH=log[H+]

At equilibrium, the equilibrium constant expression is expressed by the formula,

Ka=ConcentrationofproductsConcentrationofreactants

### Explanation of Solution

Explanation

To determine: The pH of the given solution, prepared by dissolving one vitamin C tablet in enough water to make 200.0mL of solution.

The equilibrium constant expression for the initial dissociation reaction is,

Ka=[H+][HC6H6O6][H2C6H6O6]

The initial dissociation reaction is,

H2C6H6O6(aq)H+(aq)+HC6H6O6(aq)

At equilibrium, the equilibrium constant expression is expressed by the formula,

Ka=ConcentrationofproductsConcentrationofreactants

Where,

• Ka is the acid dissociation constant.

The equilibrium constant expression for the given reaction is,

Ka=[H+][HC6H6O6][H2C6H6O6] (1)

The number of moles of H2C6H6O6 is 2.84×10-3mol_ .

The mass of one vitamin C tablet is 500mg(0.500g) .

The molar mass of H2C6H6O6 =6C+8H+6O=((6×12)+(8×1)+(6×16))g/mol=176g/mol

The number of moles of a substance is calculated by the formula,

Numberofmoles=GivenmassMolarmass

Substitute the value of the given mass and the molar mass of H2C6H6O6 in the above expression.

Numberofmoles=0.500g176g/mol=2.84×10-3mol_

The initial [H2C6H6O6] is 0.014M_ .

The calculated number of moles of H2C6H6O6 is 2.84×103mol .

The total volume is 200mL(0.200L) .

The concentration is calculated by the formula,

Concentration=NumberofmolesVolume(L)

Substitute the value of the number of moles of H2C6H6O6 and the volume in the above expression.

Concentration=2.84×103mol0.200L=0.014M_

The [H+] from first reaction is 1.051×10-3M_ .

The change in concentration of H2C6H6O6 is assumed to be x .

The ICE table for the stated reaction is,

H2C6H6O6(aq)H+(aq)+HC6H6O6(aq)Inititialconcentration0.01400Changex+x+xEquilibriumconcentration0.014xxx

The equilibrium concentration of [H2C6H6O6] is (0.014x)M .

The equilibrium concentration of [H+] is xM .

The equilibrium concentration of [HC6H6O6] is xM .

The Ka for H2C6H6O6 is 7.9×105 .

Substitute the value of Ka , [H2C6H6O6] , [H+] and [HC6H6O6] in equation (1).

7.9×105=[x][x][0.014x]7.9×105=[x]2[0.014x]

The value of x will be very small as compared to 0.014 . Hence, it is ignored from the term [0.014x] .

Simplify the above expression.

7.9×105=[x]2[0.014][x]2=(1.106×106)[x]=1.051×10-3M_

Therefore, the [H+] from the first reaction is 1

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