Calculate the [NH₃] in a solution of 0.250 M ammonium fluoride (NH₄F).

Question

Calculate the [NH₃] in a solution of 0.250 M ammonium fluoride (NH₄F). Hint: Use the Ka of HF (6.8 × 10⁻⁴) and the Kb of NH₃ (1.8 × 10⁻⁵) to calculate Kc for the reaction NH₄⁺(aq) + F⁻(aq) ⇌ NH₃(aq) + HF(aq).

Answer 1

Hi Colby! Let's break this problem down:

Here is the information we have been given:

NH4+(aq)+F−(aq)⟶NH3(aq)+HF(aq)

Concentration of solution [NH4F]=0.250M

Acid dissociation constant for HF, Ka=6.8x 10−4

Base dissociation constant for ammonia (NH3), Kb=1.8×10−5

First, dissociation reactions are obtained using acid/base dissociation constants:

For, HF+H2O (l)⟶H3O+(aq)+F−(aq)

The dissociation constant is:

Ka=([H3O+][F−])/ [HF]

For, NH3+H2O (l)⟶NH4+(aq)+OH−(aq)

The dissociation constant is:

Kb=([NH4+][OH−])/[NH3]

The equilibrium expression for the given reaction is:

NH4+(aq)+F−(aq)⟶NH3(aq)+HF(aq)

We use Kc= ([A^a][B^b])/( [C^c][D^d])

(Where A, B,C, D are the concentrations from the balanced equation and a,b,c,d are the stoichiometric coefficients, which in this case a = 1, b = 1, c = 1, d = 1)

to get:

Kc=([NH3][HF]) / ([NH4+][F−])

We can arrange Kb and Ka like this

1/((Kb)(Ka))=[HF]/([H3O+][F−]) × [NH3]/([NH4+][OH−])

Which can also be represented as:

1/(Kb)(Ka)=KC × 1/([H3O+][OH−])

and, 1/([H3O+][OH−])=1/K_w

So the equation then becomes 1/((Kb)(Ka))= K_c × 1/Kw

which is the same as 1/((Kb)(Ka))=K_c/K_w

we will rewrite this as:

K_c=K_w/ (K_a×K_b)

K_w is a constant value: 1 x 10^-14

so now we can fill in our equation to obtain the equilibrium constant, K_c for our reaction:

K_c= (10^-14)/ (6.8x10^-4 x 1.8x10^-5)=8.17 x10^-7

We can then make our "ICE" chart for the reaction:

NH₄⁺ + F^- <--> NH₃ + HF

I 0.25 0.25 ___ ___

C -X -X X X

E 0.25-X 0.25-X X X

**this is because 0.25M of NH₄F will fully dissociate in solution to produce 0.25M of NH₄⁺ and 0.25M of F^-

^K_c^{= ([NH}₃^{][HF]) / ([NH}₄^+][F-])

8.17 x10^-7= X²/ (0.25-x)²

**because Ka is very small, we know X will also be very small so we can assume that 0.25-X is approximately equal to 0.25

This leaves us with: 8.17 x10^-7= X²/ (0.25)²

Solving for X: √(8.17 x10^-7x 0.0625)

X=2.24 x 10^-4

So the concentration of [NH3] in the solution is 2.24 x 10^-4M

Hope that helps, best of luck with your studies!