Hello!
The chemical reaction for the dissociation of Acetic Acid is the following:
CH₃COOH + H₂O ⇄ CH₃COO⁻ + H₃O⁺
To solve the problem we are going to use the Henderson-Hasselbach Equation, as follows, but first we need to know the pKa:
[tex]pKa=-log(Ka)=-log( 1,8*10^{-5} )=4,74[/tex]
[tex]pH=pKa + log( \frac{[CH_3COO^{-} ]}{[CH_3COOH]} )[/tex]
[tex]pH=4,74+log( \frac{\frac{20 g CH_3COONa }{0,480 L}* \frac{1 mol CH_3COONa}{82,03 g CH_3COONa} }{0,160 M} )=5,24[/tex]
So, the pH of this solution is 5,24
Have a nice day!
