Question

A 0.100 M solution of bromoacetic acid (BrCH2COOH) is 13.2% ionized. Calculate [H+]? Calculate BrCH2COO^ - ? Calculate BrCH2COOH. (looking for Molarity)

Answer 1

Answer: The concentration of hydrogen ion and bromoacetate ion is 0.0132 M and 0.0132 M resepectively and that of bromoacetic acid is 0.0868 M

Explanation:

We are given:

Molarity of bromoacetic acid = 0.100 M

Percent of ionization = 13.2 %

The chemical equation for the ionization of bromoacetic acid follows:

$BrCH_2COOH\rightarrow BrCH_2COO^-+H^+$

1 mole of bromoacetic acid produces 1 mole of bromoacetate ion and 1 mole of hydrogen ion

Molarity of hydrogen ion = 13.2 % of 0.100 = $\frac{13.2}{100}\times 0.100=0.0132M$

Molarity of bromoacetate ion = molarity of hydrogen ion = 0.0132 M

Molarity of bromoacetic acid = Molarity of solution - Molarity of ionized substance

Molarity of bromoacetic acid = 0.100 - 0.0132 = 0.0868 M

Hence, the concentration of hydrogen ion and bromoacetate ion is 0.0132 M and 0.0132 M resepectively and that of bromoacetic acid is 0.0868 M

Answer 2

The concentration of hydrogen ion, bromoacetate ion and bromoacetic acid are 0.0132 M , 0.0132 M and 0.0868 M respectively.

Given:

Molarity of bromoacetic acid = 0.100 M

Percent of ionization = 13.2 %

The chemical equation for the ionization of bromoacetic acid follows:

$BrCH_2COOH----> BrCH_2COO^ -+H^+$

How to calculate Molarity?

1 mole of bromoacetic acid produces 1 mole of bromoacetate ion and 1 mole of hydrogen ion.

Molarity of hydrogen ion = 13.2 % of 0.100 = $\frac{13.2}{100}*0.100=0.0132M$

Molarity of bromoacetate ion = Molarity of hydrogen ion = 0.0132 M

Molarity of bromoacetic acid = Molarity of solution - Molarity of ionized substance

Molarity of bromoacetic acid = 0.100 - 0.0132 = 0.0868 M

Hence, the concentration of hydrogen ion, bromoacetate ion and bromoacetic acid are 0.0132 M , 0.0132 M and 0.0868 M respectively.

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