Question

Calculate the analytical and equilibrium molar concentrations of the solute species in an aqueous solution containing 155 mg of trichloroacetic acid (C12CCOOH, 163.4 g/mol) in 20.0 mL. Trichloroacetic acid becomes 73% ionized in water.

Answer 1

Answer:

The analytical molar concentration is 0,04743M

The equilibrium concentrations are HA = 0,01281M; H⁺ and A⁻ = 0,03462M

Explanation:

The analytical molar concentrations is:

155mg ≡ 0,155g÷163,4g/mol = 9,486x10⁻⁴ moles

9,486x10⁻⁴ moles ÷ 0,020L = 0,04743 M

The trichloroacetic acid [HA] dissociates in water in a 73%, thus:

HA ⇄ H⁺ + A⁻

The equilibrium concentrations are:

HA = 0,04743M ×(100-73)% = 0,01281M

H⁺ and A⁻ = 0,04743M ×73% = 0,03462M

I hope it helps!