Question

H2A is a diprotic acid with Ka1+1.43 x 10^-3 and Ka2=2.01 x 10^-6. Consider the titration of 50.0mL of 0.300 M Na2A with 0.150 M HCL. Find the pH at the following volumes of HCl (a) 0.00ml (b) 50.0ml (c) 100.0ml (d) 175.0ml (e) 200.0ml (f)230.00ml

Answer 1

A2- + H2O <==> HA- + OH-

Kb2 = Kw/Ka2 = 1 x 10^-14/2.01 x 10^-6 = 5.0 x 10^-5

and,

HA- + H2O <==> H2A + OH-

Kb1 = Kw/Ka1 = 1 x 10^-14/1.43 x 10^-3 = 7.0 x 10^-12

(a) with 0 ml of 0.150 M HCl

we have, only Na2A in solution which hydrolyzes.

let x amount be hydrolyzed, then

Kb2 = 5 x 10^-5 = x^2/0.150-x

x^2 + 5 x 10^-5x - 7.5 x 10^-6 = 0

x = [OH-] = 2.71 x 10^-3 M

pOH = -log[H+] = -log(2.71 x 10^-3) = 2.57

pH = 14-2.57 = 11.43

(b) 50 mL of 0.15 M HCl

moles of HCl = molarity x volume = 0.15 x 0.05 = 0.0075 moles

moles of Na2A = 0.3 x 0.05 = 0.015 moles

So, we will have, remaining moles of Na2A = 0.015-0.0075 = 0.0075 moles

moles of HA- = 0.0075 moles

Total volume = 50 + 50 = 100 mL = 0.1 L

molarity of Na2A = 0.0075/0.1 = 0.075 M

molarity of HA- = 0.0075/0.1 = 0.075 M

pKa2 = -logKa2 = -log(2.01 x 10^-6) = 5.70

pH = 5.7 + log(0.075/0.075) = 5.70

(c) when 100 mL of 0.15 M HCl is added

moles of HCl = 0.15 x 0.1 = 0.015 moles

moles of Na2A = 0.3 x 0.05 = 0.015 moles

total volume = 0.150 L

Molarity of HA- = 0.015/0.15 = 0.1 M

Ka2 = [H+][A2-]/[HA-]

let x be the amount dissciated then,

2.01 x 10^-6 = x^2/0.1

x = [H+] = 4.48 x 10^-4

pH = -log[H+] = -log(4.48 x 10^-4) = 3.35

(d) 175 mL of 0.15 M HCl

moles of HCl = 0.15 x 0.175 = 0.026

moles of Na2A = 0.015 moles

moles of H2A = 0.01125 moles

Total volume = 175 + 50 = 225 ml = 0.225 L

Molarity of H2A = 0.01125/0.225 = 0.05 M

H2A <==> H+ + HA-

let x be the amount of dissociation then,

Ka1 = 1.43 x 10^-3 = x^2/0.05-x

x^2 + 1.43 x 10^-3x - 7.15 x 10^-5 = 0

x = [H+] = 7.771 x 10^-3

pH = -log(7.771 x 10^-3) = 2.11

(e) 200 mL of 0.15 M HCl

moles of HCl = 0.15 x 0.2 = 0.03 moles

moles of Na2A = 0.015 moles

moles of H2A = 0.015 moles

Total volume = 200 + 50 = 250 ml = 0.250 L

molarity of H2A = 0.015/0.25 = 0.06 M

Ka1 = 1.43 x 10^-3 = x^2/0.06-x

x^2 + 1.43 x 10^-3x - 8.58 x 10^-5 = 0

x = [H+] = 8.58 x 10^-3 M

pH = -log(8.58 x 10^-3) = 2.07

(f) 230 mL of 0.15 m HCl

excess HCl = 230 - 200 mL = 30 mL

moles of HCl = 0.15 x 0.03 = 4.5 x 10^-3 moles

total volume = 230 + 50 = 280 mL

molarity of HCl = 4.5 x 10^-3/0.280 = 0.016 M

pH = -log(0.016) = 1.80