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Chemistry & Chemical Reactivity

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

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Section
BuyFindarrow_forward

Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 17.1, Problem 1CYU
Textbook Problem
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You have a 0.30 M solution of formic acid (HCO2H) and have added enough sodium formate (NaHCO2) to make the solution 0.10 M in the salt. Calculate the pH of the formic acid solution before and after adding solid sodium formate.

Interpretation Introduction

Interpretation:

For the given solution 0.30molL1 formic acid, the value of pHhas to be calculated before and after the addition of 0.10M sodium formate solution.

Concept introduction:

In aqueous solution an acid undergoes ionization. The ionization of an acid is expressed in terms of the equilibrium constant. The quantitative measurement tells about the strength of the acid. Higher the value of Kastronger will be the acid. The acid dissociation can be represented as following equilibrium

  HAaq+ H2OlH3O+aq+ A1aq

A weak acid undergoes partial dissociation, there is a relation between dissociation constantKa (or equilibrium constant) and the concentration of reactants and products.

  Ka=H3O+eqAeqHAeq

Here,

  • H3O+eqis the equilibrium concentration of hydronium ion.
  • Aeqis the equilibrium concentration of conjugate base of the acid.
  • HAeqis the equilibrium concentration of acid.

The ICE table gives the relationship between the concentrations of species at equilibrium.

  EquationHAaq+H2OaqH3O+aq+AInitialMc00ChangeMx+x+xEquilibriumMcxxx

The acid-dissociation constant will be

  Ka=H3O+eqAeqHAeq

From the ICE table (1),

  H3O+eq=xAeq=xHAeq=cx

Substitute x for H3O+eq, x for Aeqand cx for HAeq.

The acid-dissociation constant will be

  Ka=xxcx=x2cx

  Ka=x2cx                                                                   (1)

This relation can be modified if one of the ion is already present before the ionization of acid. Then there will be some extent of suppression for the acid. This can be explained on the basis of Le-Chatelier’s principle. According to which reaction will be more on the left side rather than right if one the ion from product side is already present before equilibrium. This suppression of ionization of acid is called as “Common Ion effect”.

Therefore an ICE table is used to give the concentration relationships between ions. For example, if anion Ay M is already present in the solution before equilibrium,

EquationHAaq+H2OaqH3O+aq+AaqInitialMc0yChangeMx+x+xEquilibriumMcxxy+x

The acid-dissociation, Ka, constant for such cases will be written as

  Ka=H3O+eqAeqHAeq

From ICE table

  H3O+eq=xAeq=x+yHAeq=cx

Substitute forKa, x forH3O+eq,x+y forAeq and cx forHAeqin equation (3).

The acid-dissociation constant will be

  Ka=xx+ycx

There is an assumption for common ion effect, according to which the value of “x” is very small on comparing to the initial concentration of acidHAc”and initial concentration of anionA,“y”. Thus it can be neglected and the Ka can be written as

  Ka=xycx

The final relation for Ka in presence of a common ion and neglectingx with respect to cis written as follows,

  Ka=xyc                                                                                                 (2

The pHof the solution is calculated by using the relation,

  pH=logH3O+                                                                                         (3)

Explanation of Solution

Given data is as follows:

The value of dissociation constantKa, for formic acid is 1.8×104.

The initial concentration of formic acid is0.30molL1.

The initial concentration of sodium formate is0.10molL1

The ICE table (1), for the solution when only formic acid is present and it undergoes dissociation.

EquationHCOOHaq+H2OaqH3O+aq+HCOOaqInitialM0.3000ChangeMx+x+xEquilibriumM0.30xxx

From the ICE table (1),

  cx=0.30x

The acid dissociation constant for the formic acid will be written as,

  Ka=x2cx

Substitute, 0.30x for cx and 1.8×104 for Ka.

  1.8×104=x20.30x

Apply the approximation, that the value of x is very small in comparison to 0.30 so it can be neglected with respect to 0.30.

  1.8×104=x20.30

Rearrange for x

  x=1.8×1040.30=0.54102=0.73×102

Therefore the hydronium ion concentrationH3O+ is 0.73×102molL1.

Now using the equation (3) the value of pH is calculated as follows,

  pH=logH3O+

Substitute,0.73×102for H3O+.

  pH=log0.73×102=2.13=2

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Chapter 17 Solutions

Chemistry & Chemical Reactivity
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