In: Chemistry
Calculate the concentration of all species in a 0.17 M KF solution. Ka(HF)=6.3×10−4
Express your answer using two significant figures. Enter your answers numerically separated by commas. [K+], [F−], [HF], [OH−], [H3O+] =
i tried . --->
0.17,0.17,2.2⋅10−6,2.2⋅10−6,4.5⋅10−9 |
M and it was wrong. |
please help.
Ans. Being salt of weak acid (HF) and strong base (KOH), KF is a strong electrolyte. It completely dissociates into constituent ions K+ and F- when dissolved in water.
The fluoride ion (F-) is a conjugate base of weak acid. Such anions turn the solution basic by accepting a proton from water and thus increasing [OH-].
F-(aq) + H2O(l) ----------> HF(aq) + OH-(aq)
K+ simply acts as a spectator ion. [K+] = 0.17 M
# Following complete dissociation of KF in water, the initial [F-] = 0.17M
Since F- acts as conjugate base, we need to calculate base dissociation (Kb) constant from the given acid dissociation constant (Ka).
Ka x Kb = 10-14
Or, Kb = 10-14 / Ka = 10-14 / (6.3 x 10‑4) = 1.587 x 10-11
# Create an ICE table with initial [F-] = 0.17 M as shown in figure.
Now,
Base dissociation constant, Kb = [HF] [OH-] / [F-] - all concentrations at equilibrium
Or, 1.587 x 10-11 = (X) (X) / (0.17 - X)
Or, 1.587 x 10-11 = X2 / (0.17) ; assuming 0.17 >> X
Or, 1.587 x 10-11 x 0.17 = X2
Or, X = (2.6984 x 10-12)½
Hence, X = 1.643 x 10-6
# Now, The equilibrium concentrations-
[OH-] at equilibrium = X = 1.643 x 10-6 M
[HF] at equilibrium = X = 1.643 x 10-6 M
[F-] at equilibrium = 0.17 - X = 1.6999836 x 10-1 M = 0.17 M (approx..)
[H3O+] = 10-14 / [OH-] = 6.088 x 10-9 M
[K+] = 0.17 - remains unaffected