Question

What is the pH of a buffer that is prepared by mixing 30.0 mL of 1.0 M HF and 20.0 mL of 2.0 M KF? (Ka for HF 7.2 x 10−4)

What is the pH of a solution that contains 0.40 M CH3COOH and 0.30 M CH3COONa at 25°C (Ka=1.8 x 10−5)?

Answer 1

1)

Concentration after mixing = mol of component / (total volume)

M(F-) after mixing = M(F-)*V(F-)/(total volume)

M(F-) after mixing = 2.0 M*20.0 mL/(20.0+30.0)mL

M(F-) after mixing = 0.8 M

Concentration after mixing = mol of component / (total volume)

M(HF) after mixing = M(HF)*V(HF)/(total volume)

M(HF) after mixing = 1.0 M*30.0 mL/(30.0+20.0)mL

M(HF) after mixing = 0.6 M

Ka = 7.2*10^-4

pKa = - log (Ka)

= - log(7.2*10^-4)

= 3.143

use:

pH = pKa + log {[conjugate base]/[acid]}

= 3.143+ log {0.8/0.6}

= 3.268

Answer: 3.27

2)

Ka = 1.8*10^-5

pKa = - log (Ka)

= - log(1.8*10^-5)

= 4.745

use:

pH = pKa + log {[conjugate base]/[acid]}

= 4.745+ log {0.3/0.4}

= 4.62

Answer: 4.62