Question

The ka value for acetic acid, ch3cooh(aq), is 1.8× 10–5 m. calculate the ph of a 2.00 m acetic acid solution.

Answer 1

Hello!

The dissociation reaction of Acetic Acid is the following:

CH₃COOH + H₂O ⇄ CH₃COO⁻ + H₃O⁺

The ka value is expressed as follows:


`Ka= ([CH_3COO^(-)]*[H_3O^(+)] )/([CH_3COOH])`

After the dissociation, we can clear for the concentration of H₃O⁺ (x) in the following way (Considering that x is small).


`Ka= (x*x)/([CH_3COOH]_i-x) \\ \\ x= √([CH_3COOH]_i*Ka)= \sqrt{2M*1,8*10^(-5) } \\ \\ x=0,006M=[H_3O^(+) ]`

So, to finish, we apply the definition of pH


`pH=-log[H_3O^(+)] =-log (0,006M)=2,22`

So the pH of this solution is 2,22

Have a nice day!