# chemistry

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calculate the pH of the solution when 100mL of .05M KF is mixed with 200mL of .05M HF; Ka (HF) = 3.467x10^-4. I tried to use the Henderson Hasselbach equation to get the correct answer (pH=3.16) but can't seem to get it correct. Any help would be AMAZING! Thank you in advance!

• chemistry -

Why not show your work with the HH equation and let me find the error.

• chemistry -

I tried to find the numbers I need for the HH equation, and I can't seem to find the correct way to use my chemical equation to find them.
H+ + KF -> HF + K+
.01 .005 0 0
-.005 -.005 +.005 +.005
.005 0 .005 .005 <moles left at equilibrium
KF <-> F- + K+
0 0 .005
^I believe there is an error with the chemical equations I used because the equilibrium should shift to my products in the <-> equation.

• chemistry -

the formatting is a little off, I'm so sorry. Let me know if you need me to re-type it.

• chemistry -

Oh! I think I just caught my mistake...HF is a weak acid, so that reaction does not go to completion....

• chemistry -

And then the KF reaction will go to completion because it is completely soluable!

• chemistry -

I got it! Nevermind! I just was assuming the wrong equation went to completion! Thank you for asking me to type it out! It definitely helped me! :)

• chemistry -

You may be ok by now but all you need to do is to substitute HF and KF. There is no reaction.
(KF) = 0.05 x (100/300) = ?
(HF) = 0.05 x (200/300) = ?
Then HF goes in for the acid concn and KF goes in for the base concn.

Actually, there is an easier way to do it I think. Work with millimoles.
millimols HF = 0.05 x 200 mL = ?
millimoles KF = 0.05 x 100 = ?
Then mmols KF substitutes for the base
mmols HF substitutes for the acid.
Technically, substituting mmols is not substituting concn BUT mmols/mL = concn and since you divide both acid and base by the same number (300 in this case), the 300 cancels and you never have to bother with it.

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