Question

what is pH of 1M of acetic acid ka acetic acid 1.8x10^- 5 to what volume must be 1 liter solution of it diluted so that pH of resulting solution twice the original value?

Answer 1

pKa = -log(10)[Ka]

Ka for acetic acid is [H+][CH3COO-]/[CH3COOH] where [] indicates the concentration.

pKa = log(10)[CH3COOH] - log(10)[H+] - log(10)[CH3COO-]

as [H+] = [CH3COO-]

pKa = log(10)[CH3COOH] - 2*log(10)[H+]

For acetic acid pKa = 4.76 and -log(10)[H+] = pH

4.76 = log(10)(0.1) + 2*pH

=> 2*pH = 4.76 + 1

=> pH = 5.76/2

=> pH = 2.88

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Answer 2

hey here is your answer

What is the pH of a 0.1M CH3COOH solution where ka=1.8*10^-5?










I




CH3-COOH--------->CH3-COO- + H+

Acetic acid partially dissociates as above.

If 'x' is the degree of dissociation of Acetic acid, then the concetration of H+ ion is Cx we can find (H+) as described below.

Ka=Cx^2/1-x

Ka =1.8 x 10^-5

C = 0.1

If we consider x is negligible ie., x~0

Then x=(Ka/c)^0.5

={1.8 x 10^-5/0.1)^0.5 =0.01342

So concetration of H+ ion =Cx

=0.1 x 0.01342 =0.001342

pH=- log(0.001342) =2.87

However it is only an approximate method. Without using approximation, if we solve equation

Ka =Cx^2/1-x , for value of 'x', where C=0.1 then

We get x = 0.01273

So concetration of H+ ion = Cx

=0.1 x 0.01273 =0.001273

pH = - log (0.001273)

= 2.89.

hope you understand