Question

Below is the titration data of a 50mL of 0.1M triprotic acid B with 0.1M NaOH. The initial pH and the equivalence points are indicated in the data.

Data: the intial pH is 1.650

In the 1st eqv. point, pH is 4.555 and 50mL of NaOH is added

In the 2nd eqv. point, pH is 9.210 and 100mL of NaOH is added

In the 3rd eqv point, pH is 11.970 and 150mL of NaOH is added

a. Using the above data, calculate the pKa1, pKa2, pKa3 values for the triprotic acid B

b. Hence, estimate the pH when 30ml NaOH were added, respectively

Answer 1

No. of moles = M*V (where M = molarity of the sample and V = volume of the solution)

Therefore, molarity of the NaOH = 0.1M

volume (V) of acid = 50ml

Molarity of acid = 0.1M

Also, for acidic buffer use equation: pH = pK_a + log[moles of salt/moles of acid]

i) pH = 4.555 so, [H⁺] = 2.786*10^-5 M

Hence, moles of acid = (2.786*10^-5M)*(50*10^-3litre) =1.393*10^-6 moles

moles of NaOH = (0.1M)*(50*10^-3litre) = 5*10^-3moles

therefore, moles of salt = moles of NaOH added = 5*10^-3moles

using above equation, 4.555 = pK_a1 + log[5*10^-3moles/1.393*10^-6moles]

on solving we get, pK_a1 = 0.999

ii) pH = 9.21 so, [H⁺] = 6.16*10^-10 M

Hence, moles of acid = (6.16*10^-10 M)*(50*10^-3litre) =3.08*10^-11 moles

moles of NaOH = (0.1M)*(150*10^-3litre) = 15*10^-3moles (total volume =(100+50)ml, 50ml from part (i))

therefore, moles of salt = moles of NaOH added = 15*10^-3moles

using above equation, 9.21 = pK_a2 + log[15*10^-3moles/3.08*10^-11moles]

on solving we get, pK_a2 = 0.522

iii) pH = 11.97 so, [H⁺] = 1.07*10^-12 M

Hence, moles of acid = (1.07*10^-12 M)*(50*10^-3litre) =5.36*10^-14 moles

moles of NaOH = (0.1M)*(300*10^-3litre) = 3*10^-2moles (total volume = (150 + 150 )ml, 150ml from part(ii))

therefore, moles of salt = moles of NaOH added = 3*10^-2moles

using above equation, 11.97 = pK_a3 + log[3*10^-2moles/5.36*10^-14moles]

on solving we get, pK_a3 = 0.222

(b) initially, pH = 1.65

using above equation, pH = pK_a + log[0 moles/moles of acid]

Therefore, 1.65 = pK_a + 0

pK_a = 1.65

Volume (V) of NaOH added =30ml

moles of NaOH = (0.1M)*(30*10^-3litre) = 3*10^-3moles

moles of acid = (0.1M)*(50*10^-3litre) = 5*10^-3moles

Therefore, moles of salt = moles of NaOH added = 3*10^-3moles

So, pH = 1.65 + log[3*10^-3moles/5*10^-3moles]

pH = 1.428