Question

Calculate the pH of a 0.040 M solution of CH3CO2H.

Answer 1

Lets write the dissociation equation of CH3COOH

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

4*10^-2 0 0

4*10^-2-x x x

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

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 ((1.8*10^-5)*4*10^-2) = 8.485*10^-4

since x is comparable c, our assumption is not correct

we need to solve this using Quadratic equation

Ka = x*x/(c-x)

1.8*10^-5 = x^2/(4*10^-2-x)

7.2*10^-7 - 1.8*10^-5 *x = x^2

x^2 + 1.8*10^-5 *x-7.2*10^-7 = 0

Let's solve this quadratic equation

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

a = 1

b = 1.8*10^-5

c = -7.2*10^-7

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 = 2.88*10^-6

putting value of d, solution can be written as:

x = {-1.8*10^-5 + √(2.88*10^-6)}/2

x = {-1.8*10^-5 - √(2.88*10^-6)}/2

solutions are :

x = 8.396*10^-4 and x = -8.576*10^-4

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

x = 8.396*10^-4

So, [H+] = x = 8.396*10^-4 M

we have below equation to be used:

pH = -log [H+]

= -log (8.396*10^-4)

= 3.08

Answer: 3.08