16. HX is a weak acid with pKa = 6.0. Calculate the equilibrium constant of the

ID: 1047909 • Letter: 1

Question

16. HX is a weak acid with pKa = 6.0. Calculate the equilibrium constant of the following reactions

So, what this means is that when HX + H2O => X + H30 , Ka = 10^(-6) and when HX + H2O => H2X + OH, Kb = 10^(-8) right?

a) HX + OH- H2O + X Answer: 10^8

I don't understand this because it is with OH, not H2O

b) HX + H2O H3O+ + X Answer: 10^-6

I understand this!

c) X- + H2O HX + OH Answer: 10^-8

I don't understand this!

d) X- + H3O+ HX + H2O Answer: 10^6

I understand this, because you switched the sides you take the inverse of Ka.

Cansomeone please explain A and C though because I don't understand how you can do that without having HX + H2O for the reaction... Thankyou!

(b): Given pKa = 6.0

=> - logKa = 6.0

=> Ka = 10-6

Ka is the equilibrium constant for the dissociation of HX to H3O+(aq) and X-(aq).

Hence for the dissociation reaction:

HX + H2O ----- > H3O+(aq) + X-(aq) : Ka = Ka = 10-6

Hence Ka = 10-6 = [H3O+(aq)]x[X-(aq)] / [HX] -------- (1)

(a): For the dissociation of water:

H2O + H2O -------- > H3O+(aq) + OH-(aq): Kw = 10-14

Hence Kw = 10-14 = [H3O+(aq)]x[OH-(aq)] ------ (2)

Now dividing equation-(2) by equation-(1):

Kw / Ka = 10-14 / 10-6 = [H3O+(aq)]x[OH-(aq)] / ([H3O+(aq)]x[X-(aq)] / [HX]

=> 10-8 = [OH-(aq)]x[HX] / [X-(aq)] ------- (3)

If we reverse the above equation-(3) we get

[X-(aq)] / [OH-(aq)]x[HX] = 1 / 10-8 = 108 ------(4)

The above calculated equation-(4) is the equilibrium constant for the reaction given in (a), which is:

HX + OH-(aq) H2O + X-(aq) : K = 108 (answer)

(c): The above calculated equation-(3) is the equilibrium constant for the reaction given in (c) which is

X-(aq) + H2O HX + OH-(aq): K = 10-8 = [OH-(aq)]x[HX] / [X-(aq)] (answer)

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