Given the following information hydrofluoric acid HF trimeth

Given the following information: hydrofluoric acid HF trimethylamine Ka 7.2x10-4 6.3 10 CH3)3N (1) Write the net ionic equation for the reaction that occurs when equal volumes of 0.165 M aqueous hydrofluoric acid and trimethylamine are mixed. It is not necessary to include states such as (aq) or (s). (2) At equilibrium the will be favored. seven. (3) The pH of the resulting solution will be

Solution

part 1)

1)Net ionic equation:

HF+(CH₃)₃N ----->(CH₃)₃NH⁺ +F^-

2) Let the V volume of the reactants be mixed

So,mol of (CH₃)₃N=mol of HF=0.165 mol/L*V=equal (they react in 1:1 molar ratio),so the reaction goes to completion or is at its equivalence point.

Thus no HF or (CH₃)₃N will be left over in the solution but the salt (CH₃)₃NH⁺ .This weak acid-weak base neutralization involves negative enthalpy change,(delta H=-ve ) so forward rxn is favored.The salt formed hydrolyses and forms a newequilibrium

HF+(CH₃)₃N ----->(CH₃)₃NH⁺ +F^-

(CH₃)₃NH⁺ +H2O <----->(CH₃)₃N +H3O+

3) (CH₃)₃NH⁺ +H2O <----->(CH₃)₃N +H3O+

ka((CH₃)₃NH⁺ )=kw/kb=10^-14/(6.3*10^-5)=1.587*10^-10

ka=1.587*10^-10=[(CH₃)₃N][H3O+]/[(CH₃)₃NH+]

[(CH₃)₃NH+]=0.165V/2V=0.0825M

ICE table

ka=1.587*10^-10=x^2/(0.0825M-x)

0.0825>>x as very low dissociation of salt takes place

1.587*10^-10=x^2/(0.0825M)

x=[H3O+]=0.362*10^-5M

pH=-log[H3O+]=-log (0.362*10^-5)=5.4

pH=5.4

	[(CH3)3NH+]	[(CH3)3N ]	[H3O+]
initial	0.0825M	0	0
change	-x	+x	+x
equilibrium	0.0825M-x	x	x