Answer to Question #217239 in General Chemistry for Haddensen

Question

0.28g of monochloroethanoic acid CH2ClOOH is dissolved in 250cl of distilled water. The pH of the solution obtained is 2.4.

What is the concentration of the monochloroethanoic acid?

What is the value of the equilibrium constant of the reaction of the monochloroethanoic acid with water?

Answer 1

Molar mass of MonoCholoro ethanoic acid =

95.5 g / mole .

Given mass = 0.28 g .

Moles of Chloroacetic acid = 0.28 / 95 .5 = 0.003 mole

Volume = 250 cc = 250 ml = 0.250 L .

1. Concentration =

0.003 / 0.250 mol / L = 0.01185 mol / L .

2. H⁺ = ( C Ka ) ^1/2 .

Given P^H = 2.4

So , H⁺ = 10^-2.4

So, Ka = ( 10^-2.4 )² / C = 10^-4.8 / 0.01185

So, Ka = 1.6 × 10^-5 / 0.12

So, Ka = 160 /12 × 10^-5 .

= 1.33 × 10^-4 . Answer ...........