Question

What is the pH of a 1.0 M aqueous solution of HF?

Answer 1

Ka of HF = 6.6*10^-4

Lets write the dissociation equation of HF

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

1.0 0 0

1.0-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 ((6.6*10^-4)*1.0) = 2.569*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)

6.6*10^-4 = x^2/(1-x)

6.6*10^-4 - 6.6*10^-4 *x = x^2

x^2 + 6.6*10^-4 *x-6.6*10^-4 = 0

Let's solve this quadratic equation

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

a = 1

b = 6.6*10^-4

c = -6.6*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 = 2.64*10^-3

putting value of d, solution can be written as:

x = {-6.6*10^-4 + √(2.64*10^-3)}/2

x = {-6.6*10^-4 - √(2.64*10^-3)}/2

solutions are :

x = 2.536*10^-2 and x = -2.602*10^-2

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

x = 2.536*10^-2

So, [H+] = x = 2.536*10^-2 M

we have below equation to be used:

pH = -log [H+]

= -log (2.536*10^-2)

= 1.6

Answer: 1.6