Question

Question 1.

* What is the pH of 30 ml of a 0.35 M acetic acid solution?(pKa = 4.74)

*What is the pH if you titrate the above 0.35 M acetic acid solution with 15 mL of a 0.30 M NaOH solution? (in what region would this be in)

*How many mL of 0.30 M NaOH do you have to add to reach the equivalence point? What is the pH of this equivalence point?

*What is the pH of this solution after you add 60 mL of 0.30 M NaOH Solution?

Answer 1

1) pH of weak acid = 1/2[ pKa -log C] = 1/2[ 4.74 -log 0.35] =2.14

2)

CH3COOH + NaOH -----------------> CH3COONa + H2O

30x0.35=10.5 15x0.30= 4.5 0 ------ initial mmol

6.0 0 4.5 ----- equilibrium

The pH of formed buffer is given by Hendersen equation

pH = pKa + log [conjugate base]/[acid]

= 4.74 + log 4.5/6.0 = 4.61

3) At equivalence

mmoles of acid = mmoles of base

30x0.35 = V x0.30

Thus Volume of base = 35 mL

CH3COOH + NaOH -----------------> CH3COONa + H2O

30x0.35=10.5 35x0.30= 10.5 0 ------ initial mmol

0 0 10.5 ----- equilibrium

This is a salt of weak acid and strong base , basic in solution due to anionic(basic) hydrolysis.

[salt] = mmole/ volume = 10.5(30+35) =0.1615 M

It pH is given by

pH = 1/2[pKw +pKa+logC]

= 12[ 14 + 4.74 + log 0.1615

=8.974

4)

CH3COOH + NaOH -----------------> CH3COONa + H2O

30x0.35=10.5 60x0.30= 18 0 ------ initial mmol

0 7.5 10.5 ----- equilibrium

Thus the solution has excess of strong base , so basic and the

[OH-] = [NaOH] = mmol/volume = 7.5/90 =0.0833 M

and pOH = -log [OH-] = -log 0.0833= 1.079

pH = 14 -pOH = 14-1.079 =12.92