Question

Titration of 50.0mL of 0.100M HX (Ka=1.5x10^-5) with 0.100M NaOH. Calculate the pH of:

- buffer formed at the addition of 12.5mL NaOH

- buffer formed at the addition of 25.0mL NaOH

- buffer formed at the addition of 37.5mL NaOH

- solution obtained at the endpoint

Please do the endpoint especially!

Answer 1

pKa of acid = -log (1.5x10-5) = 4.823

HX + NaOH ----------------------------> NaX + H2O

50x0.1 0 0 ----- intial mmoles

Q1)

HX + NaOH ----------------------------> NaX + H2O

50x0.1 =5 0 0 ----- intial mmoles

----------- 12.5x0.1= 1.25 --------- change

3.75 0 1.25 --------- equilibrium

The pH of this buffer = pKa + log [conjugate base]/[acid]

= 4.823 + log 1.25/3.75 =

=4.346

Q2)

HX + NaOH ----------------------------> NaX + H2O

50x0.1 =5 0 0 ----- intial mmoles

----------- 25x0.1= 2.5 --------- change

2.5 0 2.5 --------- equilibrium

The pH of this buffer = pKa + log [conjugate base]/[acid]

= 4.823 + log 2.5/2.5 = 4.823

Q3)

HX + NaOH ----------------------------> NaX + H2O

50x0.1 =5 0 0 ----- intial mmoles

----------- 37.5x0.1= 3.75 --------- change

1.25 0 3.75 --------- equilibrium

The pH of this buffer = pKa + log [conjugate base]/[acid]

= 4.823 + log 3.75/1.25

= 5.300

Q4)

At equivalence

HX + NaOH ----------------------------> NaX + H2O

50x0.1 =5 0 0 ----- intial mmoles

----------- 50x0.1= 5 --------- change

0 0 5 --------- equilibrium

The solution has salt of weak acid and strong base andis basic .

[salt] = mmols/volume = 5/(50+50) = 0.05

It pH given by

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

= 1/2[ 14 + 4.823 + log 0.05]

= 8.76