Answer:
a) pH=3.2 b) pH=1.7
Explanation:
HF is a weak acid and its Ka=[tex]6.8x10^{-4}[/tex] so you can write the equation:
HF⇄[tex]H^{+} + F^{-}[/tex]
NaF is a ionic compound and completely dissociates, also HCl[tex]O_{4}[/tex] 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 [tex]H^{+}[/tex] and x moles of [tex]F^{-}[/tex] and you can write K like this: [tex]K=\frac{[H^{+}][F^{-}] }{[HF]}[/tex]. note that [tex]F^{-}[/tex] meanly comes from dissociation of NaF and [tex]F^{-}[/tex]=0.28M
[tex]K=\frac{x.0.28}{0.25-x}[/tex] and solving for x, x=6.09x [tex]10^{-4}[/tex]M
this x=[tex][H^{+} ][/tex] and you are abble to find pH with [tex]pH=-log[H^{+}]=-log(6.09x10^{-4} )=3.2[/tex]
Now you entered 0.0200 moles of HCl[tex]O_{4}[/tex] whose dissociation produces 0.0200 moles of [tex]H^{+}[/tex]. Due to the volume is the same, total concentration of [tex]H^{+}[/tex] is the sum of both, initial and added [tex]pH=-log(0.0200+6.09x10^{-4} )= 1.7[/tex]
