# A 0.100 M solution of chloroacetic acid (ClCH_2COOH) is 11.0% ionized. Using this information, calculate [ClCH_2COO^−] and [H^+]?

Dec 7, 2017

[ClCH_2COO^−]=[H^+]=0.011M=0.011molcolor(white)(l)dm^(-3)

#### Explanation:

$\left[C l C {H}_{2} C O O H\right] = 0.100 M$

As 11% is ionised, the concentration of $C l C {H}_{2} C O O H$ is now 89% of its original value. $0.1 \cdot 0.89 = 0.089 M$

Being a weak acid, we shall assume that [ClCH_2COO^−]=[H^+]

So, the concentrations for ClCH_2COO^− is the same as ${H}^{+}$.

The concentration which is ionised is $0.011 M$, which is the concentration of the two ions.