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:

$Ka= \frac{[CH_3COO^{-}]*[H_3O^{+}] }{[CH_3COOH]}$

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

$Ka= \frac{x*x}{[CH_3COOH]_i-x} \\ \\ x= \sqrt{[CH_3COOH]_i*Ka}= \sqrt{2M*1,8*10^{-5} } \\ \\ x=0,006M=[H_3O^{+} ]$

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!