Question #64787

1 Answer
Apr 12, 2017

Answer:

Alrighty

Explanation:

I will walk you through two methods for HF (because typing is a pain :)). The first method will be your ICE table, which is the tougher approach, because, if you follow all the conventions, you have to do plenty of extra steps. The second one is the conventional way of doing this...

Enjoy

ICE Table
enter image source here

Finding X:
#Ka=([Sal t^(-)]*[H^+])/([Acid])#

Because #[Sal t^(-)]=[H^+]# I can rewrite this as #[H^+]^2#

We also know that #[H^+]# will later be x, hence:

#6.8*10^-4=([H^+]^2)/(0.15-x)#
#6.8*10^-4=(x^2)/(0.15-x)#

#1.02*10^-4-6.8*10^-4x=x^2#

#x^2-6.8*10^-4x+1.02*10^-4=0#

x=0.0101 (and a negative value, which we do not care about)

Therefore #[H^+]=0.0101mol#

Because #pH=-log([H^+])#

#pH=-log(0.0101)#
#pH=2.00#

Now the conventional way of doing this

You know that what the Ka value is, is essentially:

#Ka=([Sal t^(-)]*[H^+])/([Acid])#

Because #[Sal t^(-)]=[H^+]# I can rewrite this as #[H^+]^2#

Hence, we end up with:
#Ka=([H^+]^2)/([Acid])#

Now we have to make the assumption, that the concentration of the acid will initially bearly change, so we will keep it at 0.15M.
I.e.
#6.8*10^-4=([H^+]^2)/(0.15)#
#1.02*10^-4=[H^+]^2#
#[H^+]=0.0101#

Because #pH=-log([H^+])#

#pH=-log(0.0101)#
#pH=2.00#

Now we can do the same for NaF

First we have to determine the Kb for NaF:

#Kb=(Kw)/(Ka)=(10^-14)/(6.8×10^(-4))=1.47*10^-11#

From here we can go through the same process as above:

#Kb=([HF]⋅[OH^-])/([F^-])#

#1.47*10^-11=[OH^-]^2/0.15#

#[OH^-]=1.48*10^-6#

Hence, the pOH is:

#pOH=-log(1.48*10^-6)#
#pOH=5.83#

Hence #pH=14-pOH#

#pH=8.17#

Lastly, we can find the pH of the mixture

I made an ICE table for you again:

enter image source here

Note how the #F^-# already has a 0.15 mol concentration due to the dissociated NaF.

Now solve in the same way that we did for the first part:
#6.8*10^-4=([H^+]*0.15+x)/(0.15-x)# <-- you can also replace #[H^+]# by x

#6.8*10^-4=(x*(0.15+x))/(0.15-x)# <-- the denominator will bearly change, so we will leave it at 0.15

#6.8*10^-4=(x*(0.15+x))/(0.15)#

#1.02*10^-4=x^2+0.15x#

#x^2+0.15x-1.02*10^-4=0#

x=0.149 (and a negative value we do not care about)

Therefore #[H^+]=0.149mol#

Hence, the pH is:

#pH=-log(0.149)#
#pH=0.827#