Question

a) A solution consists of 0.25 M hydrofluoric acid (HF) and 0.28 M sodium fluoride (NaF). The K of hydrofluoric acid acid is 6.8 x 10"", Calculate the pH of the solution. b) To one liter of the solution from part (a) is added 0.0200 moles of perchloric acid (HCIO4). There is NO change in the total volume. Calculate the pH after the addition of the perchloric acid.

Answer 1

a) pH=3.2 b) pH=1.7

Step-by-step explanation:

HF is a weak acid and its Ka=
`6.8x10^(-4)` so you can write the equation:

HF⇄
`H^(+) + F^(-)`

NaF is a ionic compound and completely dissociates, also HCl
`O_(4)` is a strong acid and completely dissociates.

in the case of HF, when you have a 1L of 0.25M solution you really have 0.25moles, but this acid dissociates and at equilibium you have 0.25-xmolesHF and produces xmoles of
`H^(+)` and x moles of
`F^(-)` and you can write K like this:
`K=([H^(+)][F^(-)]  )/([HF])`. note that
`F^(-)` meanly comes from dissociation of NaF and
`F^(-)`=0.28M

`K=(x.0.28)/(0.25-x)` and solving for x, x=6.09x
`10^(-4)`M

this x=
`[H^(+) ]` and you are abble to find pH with
`pH=-log[H^(+)]=-log(6.09x10^(-4) )=3.2`

Now you entered 0.0200 moles of HCl
`O_(4)` whose dissociation produces 0.0200 moles of
`H^(+)`. Due to the volume is the same, total concentration of
`H^(+)` is the sum of both, initial and added
`pH=-log(0.0200+6.09x10^(-4) )= 1.7`