# chemistry - exam practice help!!

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i'm not exactly sure how to do this question :/
i don't even know where to start.

"Determine the equilibrium [F-] of the following solution with initial concentrations od [HF]= 1.296 M and [NaF] = 1,045 M (ka for HF is 6.6 * 10 ^ -4)

options are

a) 1.046
b) 0.251
c) 2.344
d) 8.2 * 10^-4
e) 5.3*10^-4

all i know is that HF is a weak acid
and
that HF dissociates into H+ and F-

:S

any help at all is appreciated!!

• chemistry - exam practice help!! - ,

I assume the (NaF) = 1.045M and not 1,045M.
..........HF ==> H^+ + F^-
I........1.296...0......0
C.........-x.....x......x
E.......1.296-x...x.....x

.........NaF ==> Na^+ + F^-
I.......1.045.....0......0
C.....-1.045.....1.045..1.045
E........0......1.045..1.045

Ka for HF = (H^+)(F^-)/(HF)
(H^+) = x
(F^-) = x + 1.045
(HF) = 1.296-x

The easy of solving this is to substitute concns I've listed above, assume x + 1.045 = 1.045 and 1.296-x = 1.296; then solve for x in the Ka expression. When you obtain that, then total (F^-) = 1.045+x. If you want to do it up brown, set up the Ka expression and solve the quadratic equation.

• chemistry - exam practice help!! - ,

thank you! and yes the 1,045 was a typo ,

but how did you know that the change row in the ice table for NaF under Na+ and F- were gonna be 1.045 ? wouldn't they be x as well ? that's just the only part that messes me up. :S

• chemistry - exam practice help!! - ,

NaF is a salt, it is completely soluble in water, it is an ionic compound, and it is 100% ionized in aqueous solution. There is no unionized NaF left over as there is in HF, a weak acid and weak electrolyte.

• chemistry - exam practice help!! - ,

wow, i don't know how i missed that connection before. of course, its simple. i must be getting too nervous for my test lol. Thank you so much for clearing that up for me!

• chemistry - exam practice help!! - ,

It's ok to get nervous; just don't thing crooked, think straight.

• chemistry - exam practice help!! - ,

swag

• chemistry - exam practice help!! - ,

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