Question

The acid disassociation constant, Ka, for hydrofluoric acid is 6.8 x 10−4. Given a 0.020 M solution of hydrofluoric acid, answer the following questions. HF(aq)  H+(aq) + F−(aq)

A. Write the equilibrium expression.

B. What is the concentration of acid, H+, in the equilibrium solution?

Answer 1

A)

HF -----> H+ + F-

K = [H+][F-]/[HF]

B)

HF dissociates as:

HF -----> H+ + F-

2*10^-2 0 0

2*10^-2-x x x

Ka = [H+][F-]/[HF]

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((6.8*10^-4)*2*10^-2) = 3.688*10^-3

since x is comparable c, our assumption is not correct

we need to solve this using Quadratic equation

Ka = x*x/(c-x)

6.8*10^-4 = x^2/(2*10^-2-x)

1.36*10^-5 - 6.8*10^-4 *x = x^2

x^2 + 6.8*10^-4 *x-1.36*10^-5 = 0

This is quadratic equation (ax^2+bx+c=0)

a = 1

b = 6.8*10^-4

c = -1.36*10^-5

Roots can be found by

x = {-b + sqrt(b^2-4*a*c)}/2a

x = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 5.486*10^-5

roots are :

x = 3.363*10^-3 and x = -4.043*10^-3

since x can't be negative, the possible value of x is

x = 3.363*10^-3

so,

[H+] = x = 3.363*10^-3 M

Answer: 3.36*10^-3 M