   Chapter 16, Problem 99GQ

Chapter
Section
Textbook Problem

Saccharin (HC7H4NO3S) is a weak acid with pKa = 2.32 at 25 °C. It is used in the form of sodium saccharide, NaC7H4NO3S. What is the pH of a 0.01M solution of sodium saccharide at 25 °C? Interpretation Introduction

Interpretation:

The pH of a 0.10M solution of sodium saccharide at 25oC has to be determined.

Concept introduction:

Equilibrium constants:

The equilibrium constant is used to quantitative measurement of the strength of the acid and bases in the water.

Ka is an acid constant for equilibrium reactions.

HA + H2OH3O++ A-Ka[H3O+][A-][HA]

Kb is base constant for equilibrium reaction.

BOH + H2OB++ OH-Ka[B+][OH-][BOH]

Ion Product constant for water

Kw= [H3O+][OH-]       =1.00×10-14 pH = -log[H3O+]pOH= -log[OH-]

Relation between pH and pOH

pH + pOH =14

Explanation

The given pKa of saccharine HC7H4NO3S is 2.32.

Pka = -logKa2.32 = -logKaa= 10-2.32Ka = 4.8×10-3

Let’s calculate the Kb of saccharine.

Kw= Ka×Kb

Kb of C7H4NO3S- = KwC7H4NO3S-                              = 1.0×10-144.8×10-3                              = 2.1×10-12Kb of C7H4NO3S- is 2.1×10-12

The equilibrium chemical reaction is as follows.

C7H4NO3S-(aq) + H2O(l)HC7H4NO3S(aq) + OH-(aq)

The equilibrium expression:

Kb[HC7H4NO3S][OH-][C7H4NO3S-]

The initial concentration =0.10

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

C7H4NO3S-(aq) + H2O(l)HC7H4NO3S(aq) + OH-(aq)I         0.10M                  --                    --                      --C           -x                      --                  + x                   + xE        (0

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