If 50.0 ml of .25M HX (Ka = 7.5 x 10^-5) is neutraliuzed with .75M NaOH calculate the pH at the endpoint.

Answer 1

The titration of a weak acid (HX) with a strong base (NaOH) creates a buffer during the titration, and the pH will NOT be 7 at the endpoint. The pH will be determined by the hydrolysis of the resulting salt (NaX) and the Kb for the conjugate base (X^-).

moles HX = 0.050 L x 0.25 mol/L = 0.0125 moles

mls 0.75 M NaOH needed to neutralize HX: (x L)(0.75 mol/L) = 0.0125 moles and x = 0.0167 L = 16.7 mls

Final volume after neutralization = 50.0 mls + 16.7 mls = 66.7 mls = 0.0667 L

Final concentration of NaX = 0.0125 moles NaX/0.0667 L = 0.1874 M

Hydrolysis of NaX:

X^- + H₂O = HX + OH^-

Kb = Kw/Ka = 1x10^-14/7.5x10^-5 = 1.33x10^-10

Kb = [HX][OH^-]/[X^-] = 1.33x10^-10

1.33x10-10 = (x)(x)/0.25

x² = 3.3x10^-11

x = 5.7x10^-6 M = [OH^-]

pOH = 5.2

pH = 8.8