Question

0.30 moles of NaOH is added to 0.80 mole of weak acid HX. The resultant solution has pH 5.25. Calculate Ka for HX.

A. 3.4 x 10^(-6)

B. 9.4 x 10^(-6)

C. 6.5 x 10^(-5)

D. 2.4 x 10^(-5)

E. 1.6 x 10^(-6)

Answer 1

0.30 mol of NaOH and HX reacts to form 0.30 mol of NaX
moles of HX remaining = 0.80 - 0.30 = 0.50 mol

use:
pH = pKa + log {[conjugate base]/[acid]}
pH = pKa + log {[X-]/[HX]}
5.25 = pKa + log {0.30/0.50}
pKa = 5.472

use:
pKa = -log Ka
5.472 = -log Ka
Ka = 3.4*10^-6
Answer: A