Question

when 0.20 mole HF is dissolved in water to a volume of 1.00 L, 5.8% of the HF dissociates to form F- (aq). What is the equilibrium constant for the reaction ?

HF(aq) + H2O (I) ----- F- (aq)+H3O+(aq)

Answer 1

0.2mol/L -= 0.2M
0.20 x 0.00.058 = 0.0116
..........HF ==> H^+ + F^-
initial...0.2.....0.....0
change.-0.0116...0.0116..0.0116
equil.....?........?......?

Ka = (H^+)(F^-)/(HF)
Substitute from the ICE chart above and solve for Ka.

Answer 2

Answer is 7.14x10^-4