Question

1) What is the pH of a 0.187 M aqueous solution of potassium fluoride, KF?

2) What is the pH of a 7.42×10-2 M aqueous solution of ammonium nitrate, NH4NO3 ?

Please show work, I'm confused on how to solve when NO Ka or Kb value is given in the problem!

Answer 1

1)

use:

Kb = Kw/Ka

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

Kb = (1.0*10^-14)/Ka

Kb = (1.0*10^-14)/6.6*10^-4

Kb = 1.515*10^-11

F- dissociates as

F- + H2O -----> HF + OH-

0.187 0 0

0.187-x x x

Kb = [HF][OH-]/[F-]

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((1.515*10^-11)*0.187) = 1.683*10^-6

since c is much greater than x, our assumption is correct

so, x = 1.683*10^-6 M

use:

pOH = -log [OH-]

= -log (1.683*10^-6)

= 5.7739

use:

PH = 14 - pOH

= 14 - 5.7739

= 8.2261

Answer: 8.23

Only 1 question at a time please