Question

Calculate the [H ] and pH of a 0.000183 M acetic acid solution. Keep in mind that the Ka of acetic acid is 1.76 × 10^-5.

Answer 1

CH3COOH dissociates as:

CH3COOH -----> H+ + CH3COO-

1.83*10^-4 0 0

1.83*10^-4-x x x

Ka = [H+][CH3COO-]/[CH3COOH]

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 ((1.76*10^-5)*1.83*10^-4) = 5.675*10^-5

since x is comparable c, our assumption is not correct

we need to solve this using Quadratic equation

Ka = x*x/(c-x)

1.76*10^-5 = x^2/(1.83*10^-4-x)

3.221*10^-9 - 1.76*10^-5 *x = x^2

x^2 + 1.76*10^-5 *x-3.221*10^-9 = 0

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

a = 1

b = 1.76*10^-5

c = -3.221*10^-9

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 = 1.319*10^-8

roots are :

x = 4.863*10^-5 and x = -6.623*10^-5

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

x = 4.863*10^-5

so.[H+] = x = 4.863*10^-5 M

we have below equation to be used

pH = -log [H+]

= -log (4.863*10^-5)

= 4.31

Answer: 4.31