Question

1. Consider the following substances: CH₅N, C₂H₅O₂COOH, H₃AsO₄, KClO₂, HC₂H₂ClO₂. For each compound
   1. Decide whether the compound behaves as an acid or base in solution
   2. Write the equation for the dissociation of the compound in water
   3. Write the equilibrium expression for the dissociation in water

2. Calculate the [H⁺], the [OH^-], the pH and the pOH for a solution prepared by diluting 50.0 ml of 0.035 M HNO₃ to 250 ml in a volumetric flask.

3. Fluoroacetic acid, C₂H₃FO₂, occurs naturally in at least 40 plants found in Australia, Brazil and in Africa, notably in “Gifblaar” which is one of the most poisonous plants found in South Africa. The compound is highly toxic and causes convulsions and ventricular fibrillation and has been used to poison rats. If the Ka of this acid is 2.57 x 10^-3 determine the [H⁺], the [OH^-], the pH and the pOH for a 0.05 M solution of this acid.

Answer 1

Answer:

a, b CH₅N + H₂O → CH₅NH⁺ + OH^- Act as base

C₂H₅O₂COOH + H₂O → C₂H₅O₂COO^- + H₃O⁺ Act as acid

H₃AsO₄ + H₂O → H₂AsO₄^- + H₃O⁺ Act as acid

KClO₂ + H₂O → HClO₂ + KOH It's a salt. But HClO₂ is weak acid and KOH is strong base. The solution is little basic.

ClCH₂COOH + H₂O → ClCH₂COO^- + H₃O⁺ Act as acid

c. The initial concentration of HNO₃ = 0.035 M

Final Volume = 250 mL

The final concentrations is = (0.035 x 50)/250 = 7 x 10^-3 M

As HNO₃ is strong acid, the dissociation is ~100 %. So, the concentration of [H⁺] = 7 x 10^-3 M

Hence, pH = -log( 7 x 10^-3) = 2.15

and pOH 14 - 2.15 = 11.85

[OH^-] = 10^-11.85 = 1.41 x 10^-12 M

d. Here, HA is an acid = Fluoroacetic acid [C₂H₃FO₂]

	HA -->	H+ +	A-
Initial	0.05	0	0
Change	-a	+a	+a
Equilibrium	0.05 - a	a	a

K_a = [H⁺][A^-]/[HA] = 2.57 x 10^-3

2.57 x 10^-3 = a.a/(0.05 -a)

or a² + 2.57 x 10^-3a - 1.285 x 10^-4 = 0

[H⁺] = 0.010123 M

pH = 1.99

pOH = 12.01 and [OH^-] = 9.77 x 10^-13 M

Thanks