Question

What is the final pH of a solution when 0.1 moles of acetic acid is added to water to a final volume of 1L?

Answer 1

Answer:

The pH of solution is 2.88 .

Explanation:

The reaction is :

$CH_3COOH-->CH_3COO^-+H^+$

We know, $K_a$ for this reaction is = $1.76\times 10^{-5}.$

Also, since volume of water is 1 L.

Therefore, molarity of solution is equal to number of moles.

Also, $K_a=\dfrac{[CH_3COO^-][H^+]}{[CH_3COOH]}$

Let, amount of $CH_3COO^- and\ H^+$ produce is x.

So,

$K_a=\dfrac{[x][x]}{[0.1]}\\1.76\times 10^{-5}=\dfrac{[x][x]}{[0.1]}$

$x=0.0013\ mol.$

We know, $pH=-log{x}=-log(0.0013)=2.88$

Therefore, pH of solution is 2.88 .

Hence, this is the required solution.