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

The pH of solution is 2.88 .

Step-by-step explanation:

The reaction is :

`CH_3COOH-->CH_3COO^-+H^+`

We know,
`K_a` for this reaction is =
`1.76* 10^(-5).`

Also, since volume of water is 1 L.

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

Also,
`K_a=([CH_3COO^-][H^+])/([CH_3COOH])`

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

So,

`K_a=([x][x])/([0.1])\\1.76* 10^(-5)=([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.