Question

Calculate the pH of a .30 M solution of acetic acid (CH3COOH, Ka= 1.8 x 10^-5) at 25 degrees Celsius.

Answer 1

Lets write the dissociation equation of CH3COOH

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

0.3 0 0

0.3-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)*0.3) = 2.324*10^-3

since c is much greater than x, our assumption is correct

so, x = 2.324*10^-3 M

so.[H+] = x = 2.324*10^-3 M

we have below equation to be used:

pH = -log [H+]

= -log (2.324*10^-3)

= 2.63

Answer: 2.63

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