Question

What is the pH of a 0.135 M CH3CO2H solution? CH3CO2H Ka = 1.75 x 10-5

Answer 1

Lets write the dissociation equation of CH3COOH

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

0.135 0 0

0.135-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 ((1.75*10^-5)*0.135) = 1.537*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)

1.75*10^-5 = x^2/(0.135-x)

2.362*10^-6 - 1.75*10^-5 *x = x^2

x^2 + 1.75*10^-5 *x-2.362*10^-6 = 0

Let's solve this quadratic equation

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

a = 1

b = 1.75*10^-5

c = -2.362*10^-6

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 = 9.45*10^-6

putting value of d, solution can be written as:

x = {-1.75*10^-5 + √(9.45*10^-6)}/2

x = {-1.75*10^-5 - √(9.45*10^-6)}/2

solutions are :

x = 1.528*10^-3 and x = -1.546*10^-3

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

x = 1.528*10^-3

So, [H+] = x = 1.528*10^-3 M

we have below equation to be used:

pH = -log [H+]

= -log (1.528*10^-3)

= 2.82

Answer: 2.82