Question

A buffer is prepared by adding 20.0 g of sodium acetate to 480 ml of a 0.160 m acetic acid solution. determine the ph of the buffer. the ka of acetic acid is 1.8 x 10-5

Answer 1

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:

$pKa=-log(Ka)=-log( 1,8*10^{-5} )=4,74$

$pH=pKa + log( \frac{[CH_3COO^{-} ]}{[CH_3COOH]} )$

$pH=4,74+log( \frac{\frac{20 g CH_3COONa }{0,480 L}* \frac{1 mol CH_3COONa}{82,03 g CH_3COONa} }{0,160 M} )=5,24$

So, the pH of this solution is 5,24

Have a nice day!