Titration of a monoprotic weak acid (HA) with a stong base (NaOH). a) What is th

Question

Titration of a monoprotic weak acid (HA) with a stong base (NaOH).

a) What is the pH of 50.0 mL of a solution of the weak acid with an initial concentration of 0.45 M that has a Ka = 3.20 x 10-4 ?

b) What volume of NaOH with a concentration of 2.0 M is necessary to neutralize the weak acid if you initially have 50.0 mL of weak acid with a concentration of 0.45 M (i.e. reach the equivalence point)?

c) What is the pH at the equivalence point if you initially have 50.0 mL of weak acid with a concentration of 0.45 M (hint: you need to use the answer from part b ) ?

d) What is the pH at the 1/2 equivalence point for the acid ( initial concentration is 0.45 M) ?   

Explanation / Answer

Titration

a) let x amount of a weak acid HA has dissociated then,

Ka = [H+][A-]/[HA]

3.20 x 10^-4 = x^2/0.45

x = [H+] = 0.012 M

pH = -log[H+] = 1.921

b) moles of weak acid present = 0.45 M x 0.05 L = 0.0225 mol

1 mole of NaOH would neutralise 1 mole of acid

So,

Volume of NaOH required = 0.0225 mol x 1000/2 M = 11.25 ml

c) at equivalence point all of acid is neutralised to form salt

salt NaA (A- being conjugate base of acid HA) hydrolyses

A- + H2O <==> HA + OH-

molarity of salt produced = 0.0225/61.25 = 3.7 x 10^-4 M

let x amount has hydrolyzed

Kb = Kw/Ka = [HA][OH-]/[A-]

1 x 10^-14/3.2 x 10^-4 = x^2/3.7 x 10^-4

x = [OH-] = 1.07 x 10^-7 M

pOH = -log[OH-] = 6.97

pH = 14 - pOH = 7.03

d) at half-equivalenece

[HA] left = [A-] formed

pH = pKa = -log[Ka]

= -log(3.2 x 10^-4) = 3.50

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