Question

A 0.01 M acetic acid solution is 1% ionised. An another acetic acid is 10% ionised. What will be the concentration of another acetic acid?

Answer 1

The concentration will be `9 × 10^"-5"color(white)(l) "mol/L"`.

Explanation:

Let `c` be the initial concentration of acetic acid.

The equation for the equilibrium is

`color(white)(m)"HA" + "H"_2"O" ⇌ "A"^"-" + "H"_3"O"^"+"`
`c-αc color(white)(mmmmm) αc color(white)(mml) αc`

The equilibrium concentration of `"HA"` will decrease by `αc`, and those of `"A"^"-"` and `"H"_3"O"^"+"` will be `αc`.

`K = (αc)^2/(c-αc) = (α^2c^2)/(c(1-α)) = (α^2c)/(1-α)`

For the first solution

`K = ((0.01)^2× 0.01)/(1-0.01) = (1 × 10^"-6")/0.99 = 1.0 × 10^"-6"`

For the second solution,

`α^2c = K(1-α)`

`c= (K(1-α))/α^2`

`c = (1.0 × 10^"-6"(1-0.10))/(0.10)^2 = 9 × 10^"-5"color(white)(l) "mol/L"`