Dissociation Constant Homework Check

[yellowduck]

Question:
A solution of hydrofluoric acid contains 2.0g of HF per litre and has a pH of 2.2. What is the dissociation constant for HF?

My Answer:
HF + H2O <-> H3O+ + F-

HF = 2.0g
H30+ = 10^-2.2 = .0063g
F- = H3O = .0063

Ka = [H3O+][F-] / HF
= (.0063)(.0063) / 2.0
=1.98 x 10^-5

Therefore the dissociation constant for HF is 2.0 x 10^-5

Can anyone tell me if I am on the right track?
Thanks

 

[PPonte]

Question:
HF = 2.0g

To find [tex]K_a[/tex] you need the concentration of [tex]HF[/tex] expressed in [tex]mol.dm^{-3}[/tex]. Convert [tex]2.0 g.dm^{-3}[/tex] to [tex]mol.dm^{-3}[/tex] using molar mass.

Question:
H30+ = 10^-2.2 = .0063g

The concentration of [tex]H_3O^+[/tex] that you calculated using that expression is molar concentration, so the units are [tex]mol.dm^{-3}[/tex].

 

[yellowduck]

To find [tex]K_a[/tex] you need the concentration of [tex]HF[/tex] expressed in [tex]mol.dm^{-3}[/tex]. Convert [tex]2.0 g.dm^{-3}[/tex] to [tex]mol.dm^{-3}[/tex] using molar mass.



The concentration of [tex]H_3O^+[/tex] that you calculated using that expression is molar concentration, so the units are [tex]mol.dm^{-3}[/tex].

My Text mentions nothing about [tex]mol.dm^{-3}[/tex]
it does say to calculate [tex][H_3O^{+}] = 10^{-x}[/tex]
So that's how I got .0063 mol/L

I also have, from the text the formula for calculating the acid dissociation constant [tex]K_{a}[/tex]

Ka = [H30+][X-] / [HX]

 

[PPonte]

So that's how I got .0063 mol/L

That's right you got [tex]0.0063 mol/L[/tex]! But you indicated on your previous post:

Question:
H30+ = 10^-2.2 = .0063g

I am just correcting the units. It is not grams but mol/dm³ or mol/L, since 1 L = 1 dm³.
You, maybe, ask why the units are those. Because you find the pH of a solution from the molar concentration of [tex]H_3O^+[/tex]. The expression is:
[tex]pH = - \log\left[ H_3O^{+}_{(aq)} ][/tex]

Then, using that expression, knowing the value of pH, you can just find the molar concentration of [tex]H_3O^+[/tex].
--------

I also have, from the text the formula for calculating the acid dissociation constant [tex]K_{a}[/tex]
Ka = [H30+][X-] / [HX]

The formula is correct.

 

[yellowduck]

Hmm... yes I seem to mess up on the units sometimes.

I would like to do this all in mol/L if possible. (we have been taught to this point only in mol/L)

So does this look better?
HF + H2O <-> H3O+ + F-

HF = 1+19=20 g/mol
2g HF x 1 mol HF / 20g HF = .1 mol HF
So HF = .1 mol/L

H30+ = 10^-2.2 = .0063 mol/L
F- = H3O = .0063 mol/L

Ka = [H3O+][F-] / HF
= (.0063)(.0063) / .1
=4.0 x 10^-4

Therefore the dissociation constant for HF is 4.0 x 10^-4

Can you please give me your feedback on this?

Thank you very much for all your help. This forum and people who help have been a valuable resource.

 

[Hootenanny]

Hmm... yes I seem to mess up on the units sometimes.

I would like to do this all in mol/L if possible. (we have been taught to this point only in mol/L)

So does this look better?
HF + H2O <-> H3O+ + F-

HF = 1+19=20 g/mol
2g HF x 1 mol HF / 20g HF = .1 mol HF
So HF = .1 mol/L

H30+ = 10^-2.2 = .0063 mol/L
F- = H3O = .0063 mol/L

Ka = [H3O+][F-] / HF
= (.0063)(.0063) / .1
=4.0 x 10^-4

Therefore the dissociation constant for HF is 4.0 x 10^-4

Can you please give me your feedback on this?

Thank you very much for all your help. This forum and people who help have been a valuable resource.

Looks good to me

 

[yellowduck]

Thank you PPonte and Hootenanny for taking the time to review my work and help me out.

Thanks Again.

 

[PPonte]

My pleasure!