Question

25 mL of Water were added to 15 mL of the solution that was formed from 35 mL of 0.40 M HOAc combined with 10 mL of 1.0 M NaOAc. Calculate the following when the pH was found to be 4.22

a) H+

b) OAc-

c) HOAc

d) Ka

e) pKa

Answer 1

a) [H+] = 10^ -pH = 10^ -4.22

       = 6 x 10^ -5 M

b) we use dilution law M1V1 = M2V2 to find final conc of each

M1 = 1 M for NaOAC , V1 = 10 ml , V2 = 10+35+15 = 60 ml

we find M2

1 x 10 = M2 x 60

M2 = [NaOAc] = 0.167 = [OAc-]                     ( since 1 NaOAc gives 1 Na+ and 1 AcO-)

c)   for HOAc ,     M1 = 0.4 , V1 =35 ml , V2 = 60 ml , we find M2

0.4 x 35 = M2 x 60

M2 = 0.233 = [HOAc]

e) we find pka from Henderson eq of buffer

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

where acid is HOAc and its conjugate base is OAc-

4.22 = pka + log ( 0.167 /0.233)

pka = 4.37

d) Ka = 10^ -pka = 10^ -4.37

= 4.27 x 10^ -5

e)