Question

a 25.0mL solution containing 0.60 M HC2H3O2(acetic acid) was titrated with 0.50 M NaOH. Ka of acetic acid is 1.8*10^-5

Calculate pH at initial volume(0mL added)

Calculate pH at 1/2 equivalence point.

Calculate pH at equivalence point.

Calculate pH at 20.00 mL

Answer 1

Ka = 1.8 x 10^-5

pKa = 4.74

millimoles of acid = 25 x 0.60 = 15

1) before the titration

pH = 1/2 (pKa -logC)

pH = 1/2 (4.74-log(0.6))

pH = 2.48

2) half -equivalence point :

at half equivalence point moles of acid = moles of salt

here pH = pKa

[salt/acid] = 1

pH = pKa + log[salt/acid]

pH = 4.74 + log(1)

pH = 4.74

3) at Equivalence point

at equivalence point M1V1= M2 V2

                                 25 x 0.60 = 0.5 x V2

                               V2 = 30ml

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

mixture contain slat CH3COONa and now we have to use salt hydrolysis

[salt] =C= 12 / (25+30) = 0.218 M

pH = 7 + 1/2 [pKa +logC]

    = 7 + 1/2 [4.74 +log(0.218)]

pH = 9.04

4) 20 ml 0.05M NaOH

    NaOH millimoles = 20 x0.5 = 10

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

15                10                                     0                       0 ---------------initial

5                   0                                 10                    10----------------equilibrium   

mixture contain weak acid and salt so it forms buffer

For acidic buffer

Henderson-Hasselbalch equation

pH = pKa + log[salt/acid]

pH = 4.74 + log[10/5]

pH = 5.04