Question

Consider the titration of 20 ml of H2CO3 with 0.1M NaOH. The pKa of H2CO3 is 6.35 and the pKa of HCO3 is 10.33. 15mL of NaOH must be added to reach the first midpoint. What is the initial molarity of the H2CO3 solution? What is the pH of the titration solution at the first midpoint? How much total NaOH must be added to reach the first equivalence point? What is the pH of the titration solution at the first equivalence point?

Answer 1

At the first mid point pH = pKa1

So, at this point pH = 6.35

Volume of reach the first equivalent point = 2 * 15 mL = 30 mL

Suppose the initial molarity of H2CO3 is C molar

From Hinderson hasselbalch equation we get :

pH = pka + log [salt/acid]

or, [salt] = [acid] as pH = pka

[salt ] = amount of base added = 0.1 M * 0.015 L = 0.0015 moles

[acid] = C * 0.02 L - 0.0015 moles

C * 0.02 L - 0.0015 moles = 0.0015

or, C = 2*0.0015 /0.02 = 0.15 M

Initial Molarity of H2CO3 = 0.15 M

First equivalence is reached at 30mL NaOH.

Moles of NaOH added = 0.03 L * 0.1 M = 0.003 moles

At first equivalence point all the H2CO3 will be converted to HCO3^-. Some of it will react with water and hydrolyse according to the following equilibrium :

HCO3^- + H2O <==> H2CO3 + OH-

Kb of this reaction = 1*10^-14 /10^-6.35 = 2.24*10^-8

	HCO3^-	H2CO3	OH-
initial	0.003	0	0
change	-x	+x	+x
equilibrium	0.003-x	x	x

Kb = x^2/0.003-x

2.24*10^-8 = x^2/0.003-x

or, x = 8.18*10^-8 moles

molarity of [OH] = 8.18*10^-8 moles/(total volume in L) = 1.64*10^-6 M

pOH = -log[OH-] = 5.78

pH = 14- 5.78 = 8.22