Question

Consider the titration of 233 mL of 0.855 M propanoic acid, CH3CH2COOH, Ka = 1.3 x 10^-5 with 1.125 M KOH for the following question.

What is the pH of the solution at the equivalence point? (answer is somewhere around 9)

Answer 1

Consider the titration of 233 mL of 0.855 M propanoic acid, CH3CH2COOH, Ka = 1.3 x 10^-5 with 1.125 M KOH for the following question.

What is the pH of the solution at the equivalence point? (answer is somewhere around 9)

This is a weak acid + storng base titration...

In Equivalence Point of an Acid; there will be Hydrolysis

hydrolysis is the process in which the conjguate base is hydrolysed with water to form once again the weak acid .

the salt fromed

CH3CH2COOH + NaOH = CH3CH2COONa + H2O

now,

CH3CH2COONa ionizes = CH3CH2COO- + Na+

this will form hydrolysis

CH3CH2COO- + H2O = CH3CH2COOH + OH-

OH- is formed, therefore this will be basic

assume

HA = propanoic acid, A- propanoate

then:

A- is formed

the next equilibrium is formed, the conjugate acid and water

A- + H2O <-> HA + OH-

Kb by definition since it is an base:

Kb = [HA ][OH-]/[A-]

Ka can be calcualted as follows:

Kb = Kw/Ka = (10^-14)/(1.3*10^-5) = 7.69*10^-10

and;

assume x = [OH-]

[OH-] = x= [HA] due to equilibrium

Find HA in equilbrium

mmol of A- = Macid*Vacid = 0.855*233 = 199.215 mmol of A-

mmol of base = 199.215 neutralization

V base = mmol of base / M base = 199.215/1.125 = 177.08 mL of base used

V total = 233 + 177.08 = 410.08 mL

[A-] = mmol of A- / Total V = 199.215/ 410.08 = 0.485795 M

[A-] = M-x = 0.485795-x

Kb = [HA ][OH-]/[A-]

7.69*10^-10 = x*x/(0.485795-x)

solve for x with quadratic equation

x = OH- = 1.933*10^-5

[OH-]  =  1.933*10^-5 M

pOH = -log(OH-) = -log( 1.933*10^-5) = 4.713768

pH = 14-4.713768 = 9.2862

pH = 9.2862