Question

a) In 2-3 senteces describe how propanic acid and propanoate anion can be used to make a buffer solution if the ka of propanic acid is 1.34 x 10^-5

b)determine the ph if the concetration of propanic acid was 1.3 x10^-3 M and the concentration of propanoate was 1.8x10^-2 M. Is this pH in the rage of the buffer?

c) explain if the pH of the solution in part b is in the range of thie buffer?

Answer 1

a) A buffer can be made from a weak acid and its conjugate base

Propanoic aicid is ak acid ( Ka = 1.34 x 10^-5 is low indicates weak acid) . and conjugate base of propanoic acid is propanoate anion . Hence these combination makes a buffer

b)   pH = pka + log [ conjugate base] /[acid]    where pka = -log Ka = -log ( 1.34 x 10^-5) = 4.87

pH = 4.87 + log ( 1.8 x 10^-2) / ( 1.3 x 10^-3)

= 6

c) pH of solution is not in range of buffer

since pH range of buffer is +1 or -1 of pH i.e   3.87 to 5.87 is pH range . But we had 6 which is above the pH range