What mass of sodium acetate, NaCH₃CO₂, must he added to 1.00 L of 0.10 M acetic acid to give a solution with a pH of 4.50?

Interpretation:

Mass of sodium acetate that must be added to acetic acid solution such that the resulting solution has pH 4.50 has to be determined.

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 Ka stronger will be the acid. The acid dissociation can be represented as following equilibrium

  HA(aq.)+ H2O(l)⇌H3O+(aq.) + A−1(aq.)

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

Ka=[H3O+](eq)[A−](eq)[HA](eq) (1)

Here,

• [H3O+](eq) is the equilibrium concentration of hydronium ion.
• [A−](eq) is the equilibrium concentration of conjugate base of the acid.
• [HA](eq) is the equilibrium concentration of acid.

The ICE table (1) gives the relationship between the concentrations of species at equilibrium.

EquationHA(aq)+H2O(aq)⇌H3O+(aq)+A−Initial(M)c00Change(M)−x+x+xEquilibrium(M)c−xxx

From the ICE table (1),

[H3O+](eq)=x[A−](eq)=x[HA](eq)=c−x

Substitute x for [H3O+](eq), x for [A−](eq) and c−x for [HA](eq) in equation (1).

The acid-dissociation constant will be

Ka=(x)(x)(c−x)=x2(c−x)

Ka=x2(c−x) (2)

This table can be modified if one of the ion is already present before the ionization of acid. Then there will be some extent of suppression of dissociation of the weak 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 weak acid in presence of strong electrolyte having common ion is called as “Common Ion effect”.

Therefore a modified ICE table (2) is used to give the concentration relationships between ions. For example, if anion A− is already present in the solution before equilibrium,

EquationHA(aq)+H2O(aq)⇌H3O+(aq)+A−(aq)Initial(M)c0yChange(M)−x+x+xEquilibrium(M)c−xxy+x

Here,

• y is the initial concentration of the anion A− (common ion coming from strong electrolyte) present in the solution before the dissociation of a weak acid HA.

From the ICE table (2),

[H3O+](eq)=x[A−](eq)=x+y[HA](eq)=c−x

Substitute x for [H3O+](eq), x+y for [A−](eq) and c−x for [HA](eq) in equation (1).

The expression for  acid dissociation constant, Ka, will be given as,

Ka=(x)(x+y)(c−x)

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 acid (HA)“c”and initial concentration of anion (A−), “y”. Thus x can be neglected with respect to y and c.

Then Ka can be written as,

Ka=(x)(y)c (3)

The pH of the solution is calculated by using the relation,

pH=−log[H3O+]

Above equation can be rearranged for [H3O+]. Therefore expression for [H3O+] is given as,

[H3O+]=10−pH (4)

The mass of sodium acetate that must be added to acetic acid so that resulting solution has pH 4.50 is calculated as below.

Given:

Refer to the table 16.2 in the textbook for the value of Ka.

The value of dissociation constant, Ka, for acetic acid is 1.8×10−5.

The initial concentration of acetic acid is 0.10 M.

The volume of solution is 1 L.

pH of the resulting solution is 4.50.

Acetic acid undergoes dissociation in aqueous solution and the reaction is given as,

    CH3COOH(aq)+H2O(aq)⇌H3O+(aq)+CH3COO−(aq)

Sodium acetate is a strong electrolyte and dissociates to give acetate ion and sodium ion in aqueous solution.

    NaCH3CO2(aq)+H2O(l)⇌CH3COO−(aq)+Na+(aq)

In presence of sodium acetate having common acetate ion, due to the common ion effect suppression of dissociation of acetic acid will occur.

The expression for the dissociation constant Ka for acetic acid given as,

Ka=[H3O+](eq)[CH3COO−](eq)[CH3COOH](eq) (5)

Calculate the concentration of [H3O+] present in the solution using equation (4).

Substitute 4.50 for pH in equation (4).

[H3O+]=10−4.50=3.16×10−5 (6)

The ICE table (3) is given as follows,

EquationCH3COOH(aq)+H2O(aq)⇌H3O+(aq)+CH3COO−(aq)Initial(M)0.100yChange(M)−x+x+xEquilibrium(M)0.10−xxy+x

Here,

• x is the number of moles of acetic acid dissociated per liter and is equal to the concentration of hydronium ion, [H3O+] and acetate ion, [CH3COO−] coming from acetic acid in presence of sodium acetate.
• y is the initial concentration of the acetate ions present in the solution coming from strong electrolyte sodium acetate.
• y+x is the total concentration of acetate ions coming from acetic acid as well as sodium acetate.

From equation (6) and ICE table (3),

 x=3.16×10−5

From the ICE table (3),

[CH3COOH]=0