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!
Why not show your work with the HH equation and let me find the error.
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.
the formatting is a little off, I'm so sorry. Let me know if you need me to re-type it.
Oh! I think I just caught my mistake...HF is a weak acid, so that reaction does not go to completion....
And then the KF reaction will go to completion because it is completely soluable!
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! :)
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.