Acid/Base Mixtures?

Question

I'm having a terrible time with this. Please help so that I will begin to understand! Please show your steps! I know it looks long so any help at all would be appreciated!

1. What is the value of the equilibrium constant (three sig figs) for the reaction of hydrofluoric acid and ammonia?
K =
Fill in the following reaction table with mmol (to the nearest 0.01 mmol) for the reaction occurring when 44.7 mL of 0.222-M ammonia are mixed with 27.6 mL of 0.292-M hydrofluoric acid.
HF + NH3 ↔ F1- + NH41+
initial
delta
final
What is the final concentration of ammonium ion to three significant figures?

How many mmols (three sig figs) of the limiting reactant are actually present at equilibrium? This value is not based on the final amount shown in the table because the table assumes complete reaction (not equilibrium) of the limiting reactant.

Link to acid/base table: http://www.webassign.net/wertz/caqs/html/resource/acidbase_pka.html

Answer 1

When I click on the link to acid/base table, it pops up this message: "We're sorry. You do not have access to this page."
That is alright. I still can help you. But the reaction constants I use might be slightly different from those in the acid/base table.
Now, we have to use the following constants:
HF <==> H+ + F-, pKa = 3.15 or Ka = 10^(-3.15)...(1)
NH4+ <==> H+ + NH3, pKa = 9.25 or Ka=10^(-9.25)..(2)
If you can solve the problem now, do not read what I write below-------:
For reaction (Use [X] to represent the equilibrium concentration of X):
HF + NH3 <==> F- + NH4+
K = [F-]*[NH4+] / [HF]*[NH3]
= {[H+]*[F-] / [HF]}*{[NH4+] / [H+]*[NH3]}
= 10^(-3.15)/10^(-9.25) = 10^(6.1)
= 1.26x10^6
Initial:
NH3: 44.7*0.222 = 9.92 (mmole)
HF: 27.6*0.292 = 8.06 (mmole)
F- and NH4+: 0.00 mmole
Since HF is the limiting reactant, and since the equilibrium constant K is really big, approximately all HF would be consumed.
Delta:
NH3: -8.06 mmole
HF: -8.06 mmole
F-: +8.06 mmole
NH4+: +8.06 mmole
Final:
NH3: 1.86 mmole
F-: 8.06 mmole
NH4+: 8.06 mmole
HF: Very small but not completely zero. it can be calculated as:
K = 1.26x10^6 = [F-]*[NH4+] / [HF]*[NH3]
= 8.06*8.06 / {(HF)*1.86}
Therefore: (HF) = 8.06*8.06 / {(1.26x10^6)*1.86} = 2.77x10^-5 (mmole)

Answer 2

Calculate the molarity of the factors of the answer: HCOOH : M1V1 = M2V2 M1 * one thousand = 5.0*2 hundred M1 = one thousand/one thousand = a million.0M HCOONa: Molar mass HCOONa = sixty 8.00g/mol 34.00g = 0.50mol answer is 0.50 M in HCOONa this could be a buffer answer and pH is set by way of use of the Henderson - Hasselbalch equation: pH = pKa + log ([A-]/[HA]) pKa = -log a million.8*10^-4 pKa = 3.seventy 4 pH = 3.seventy 4 + log (0.50/a million.00) pH = 3.seventy 4 + log 0.5 pH = 3.seventy 4 + (-0.30) pH = 3.seventy 4-0.30 pH = 3.40 4