Question

The pH of a 0.650 M HQ solution is 1.809. Calculate the ionization constant for HQ, the pH of the solution which results when 15.0 mL of 0.420 M NaOH are added to 25.0 mL of 0.650 M HQ, and the pH of the solution which results when 25.0 mL of 0.650 M HQ is reacted with an equivalent number of moles of 0.150 M NaOH.

Answer 1

pH = 1.809

[H+] = 0.0155 M

[H+] = sqrt (Ka x C)

0.0155 = sqrt (Ka x 0.650)

Ka = 3.71 x 10^-4

ionization constant for HQ = 3.71 x 10^-4

millimoles of NaOH = 15 x 0.420 = 6.3

millimoles of HQ = 25 x 0.650 = 16.25

HQ + NaOH ----------------> NaQ + H2O

16.25 6.3 0 0

9.95 0 6.3 6.3

here weak acid and conjugate base remained.

pH = pKa + log [salt / acid]

= 3.43 + log [6.3 / 9.95]

= 3.23

pH = 3.23

25.0 mL of 0.650 M HQ is reacted with an equivalent number of moles of 0.150 M NaOH.

millimoles of HQ = millimoles of NaOH

25 x 0.650 = 0.150 x V

volume of NaOH = 108.3 mL

it is equivalence point only salt formed.

salt concentration = 16.25 / (25 + 108.3)

= 0.122 M

pH = 7 + 1/2 (pKa + log C)

= 7 + 1/2 (3.43 + log 0.122)

= 8.26

pH = 8.26