# 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?

Jan 7, 2018

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

$\textcolor{w h i t e}{m} \text{HA" + "H"_2"O" ⇌ "A"^"-" + "H"_3"O"^"+}$
c-αc color(white)(mmmmm) αc color(white)(mml) αc

The equilibrium concentration of $\text{HA}$ will decrease by αc, and those of $\text{A"^"-}$ and $\text{H"_3"O"^"+}$ 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"