Question

If a buffer solution is 0.130 M in a weak acid (K) = 7.2 x 10-5) and 0.580 M in its conjugate base, what is the pH? pH =

Answer 1

1)

Ka = 7.2*10^-5

pKa = - log (Ka)
= - log(7.2*10^-5)
= 4.143

use:
pH = pKa + log {[conjugate base]/[acid]}
= 4.143+ log {0.58/0.13}
= 4.792

Answer: 4.79

2)

mol of HCl added = 0.07 mol

C2H5COO- will react with H+ to form C2H5COOH

Before Reaction:
mol of C2H5COO- = 0.165 M *1.5 L
mol of C2H5COO- = 0.2475 mol

mol of C2H5COOH = 0.328 M *1.5 L
mol of C2H5COOH = 0.492 mol

after reaction,
mol of C2H5COO- = mol present initially - mol added
mol of C2H5COO- = (0.2475 - 0.07) mol
mol of C2H5COO- = 0.1775 mol

mol of C2H5COOH = mol present initially + mol added
mol of C2H5COOH = (0.492 + 0.07) mol
mol of C2H5COOH = 0.562 mol


Ka = 1.34*10^-5

pKa = - log (Ka)
= - log(1.34*10^-5)
= 4.873

since volume is both in numerator and denominator, we can use mol instead of concentration
use:
pH = pKa + log {[conjugate base]/[acid]}
= 4.873+ log {0.1775/0.562}
= 4.372

Answer: 4.37