Question

What is the actual concentration of the molecular form of HF in a 1.0 M HF solution given that K_a of HF is 6.8 ´ 10^-4?

Answer 1

Lets write the dissociation equation of HF

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

1 0 0

1-x x x

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

Ka = x*x/(c-x)

Assuming small x approximation, that is lets assume that 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)*1) = 2.608*10^-2

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/(1-x)

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

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

Let's solve this quadratic equation

Comparing it with general form: (ax^2+bx+c=0)

a = 1

b = 6.8*10^-4

c = -6.8*10^-4

solution of quadratic equation is found by below formula

x = {-b + √(b^2-4*a*c)}/2a

x = {-b - √(b^2-4*a*c)}/2a

b^2-4*a*c = 2.72*10^-3

putting value of d, solution can be written as:

x = {-6.8*10^-4 + √(2.72*10^-3)}/2

x = {-6.8*10^-4 - √(2.72*10^-3)}/2

solutions are :

x = 2.574*10^-2 and x = -2.642*10^-2

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

x = 2.574*10^-2

[HF] = 1.0 - x

= 1.0 - 2.574*10^-2

= 0.974 M

Answer: 0.974 M