Question

A 25.0 mL sample of a 0.115 M solution of acetic acid is titrated with a 0.144 M solution of NaOH. Calculate the pH of the titration mixture after 10.0, 20.0, and 30.0 mL of base have been added. (The Ka for acetic acid is 1.76 x 10^-5).

10.0 mL of base =

20.0 mL of base =

30.0 mL of base =

Answer 1

let the volume of NaOH required for neutralisation be v.so,

25*0.115=0.144*v

or v=19.96 mL

a)when 10mL of the base is added, there will be a buffer.so,

pH=pKa+log(salt/acid)

=4.76 + log(10*0.144/(25*0.115-10*0.144))

=4.7615

b) When 20 mL of the base is added, there is excess Oh=H0

[OH-]=(20-19.96)*0.144/(25+20)

=1.38*10^-4

so pOH=3.86

so pH=14-3.86

=10.14

when 30 mL of the base is added, there is excess of OH-. so,

[OH-]=(30-19.96)*0.144/(30+25)

=0.0262

so pOH=1.58

so, pH=14-1.58

=12.42