Let's call fluoroacetic acid HFAc. Then
.........HFAc ==> H^+ + FAc^-
I.......1.11......0......0
C.....-0.055....0.055...0.055
E......you do it..0.055..0.055
Ka = (H^+)(FAc^-)/(HFAc)
Substitute the E line into Ka expression and solve for Ka, then convert to pKa.
Note: The acid is 5% dissociated, therefore, every 100 mols dissociates to give 5 of each ion and leaves 95 undissociated molecules. Therefore, M x 0.05% gives you the ion M and 1.11- that gives you the undissociated part.
A 1.11 M solution of fluoroacetic acid, FCH2CO2H, is 5% dissociated in water. Calculate the value of the pKa of FCH2CO2H.
pKa:
?? I dn't even know how to strt with this
2 answers
got it!
ka= 3.26*10^-3
so pKa=2.49
tnx
ka= 3.26*10^-3
so pKa=2.49
tnx