Question

Vant hoff factor for 0.01 m aqueuos solution acetic acid is 1.04, the ph of the solution

Answer 1

Explanation:

Hope this helps you.if helped plz do follow and mark as brainliest ans.

Answer 2

The pH of the solution is 3.39

Explanation:

The relationship between Van't Hoff factor and degree of dissociation is given as:

$\alpha=\frac{i-1}{n-1}$

where,

n = number of ions = 2 (for acetic acid)

i = Van't Hoff factor = 1.04

Putting values in above equation, we get:

$\alpha=\frac{1.04-1}{2-1}\\\\\alpha=0.04$

The chemical equation for the dissociation of acetic acid follows:

                    $CH_3COOH\rightarrow CH_3COO^-+H^+$

Initial:              c              

At eqllm:      $c-c\alpha$                     $c\alpha$            

We are given:

Concentration of acetic acid (c) = 0.01 M

To calculate the pH of the solution, we use the equation:

$pH=-\log [H^+]$

where,

$[H^+]$ = concentration of hydrogen ions = $c\alpha=0.01\times 0.04=4\times 10^{-4}M$

Putting values in above equation, we get:

$pH=-\log(4\times 10^{-4)\\\\pH=3.39$

Learn more about pH and degree of dissociation:

https://brainly.com/question/13617035

https://brainly.in/question/2457575

#learnwithbrainly