Question

Topic: "Ka Weak Acid Calculations"

1)The pH of an aqueous solution of 0.420 M hydrofluoric acid is_______ .

2)The pOH of an aqueous solution of 0.581 M hydrocyanic acid is_______ .

Answer 1

1)

HF ------------------------> H+ + F-

0.420                              0      0 -------->I

-x                                  +x       +x -------- >C

0.420-x                         x         x --------------> E

Ka = [H+][F-]/[HF]

6.3 x 10^-4 = x^2 / 0.420 -x

x^2 + 6.3 x 10^-4 x - 2.65 x 10^-4 = 0

x = 0.0160 M

[H+] = 0.0160 M

pH = -log [H+]

pH = -log (0.0160)

pH = 1.80

2)

pH = 1/2 [pKa - log C]

pH = 1/ 2 [9.21 -log 0.581]

pH = 4.72

pH + pOH = 14

pOH = 9.28