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Calculate the concentration of all species in a 0.17 M KF solution. Ka(HF)=6.3×10−4 Express your answer...

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.

Solutions

Expert Solution

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


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