Question

For the triprotic acid, H+glutamic acid, pK1=2.23, pK2=4.42 and pK3=9.95. If one has 0.10L of a 1.0E-2 M solution and titrates it with 0.255 L of a 1.0E-2 M KOH solution, what is the pH of the resulting solution?

The answer is 10.04. I just have no clue how to get there.

Answer 1

For the triprotic acid, H+glutamic acid -

pKa1 = 2.23 Ka1 = 10 ^{- 2.23} = 5.88x10^-3

pKa2 = 4.42 Ka2 = 10^-4.42 = 3.80x10^-5

pKa3 = 9.95 Ka3 = 10^-9.95 = 1.122x10^-10

Let the acid be represented as H₃A.

[H₃A] = 1.0E-2 = 1x10^-2

H₃A      H₂A^-   + H⁺

initially 0.01 0 0

finally 0.01 - x x x

Ka1 = x*x / 0.01 -x

5.88x10^-3 = x² / 0.01 - x

5.88x10^-5 - x*5.88x10^-3 = x²

x² + x*5.88x10^-3 - 5.88x10^-5 = 0

x = 5.27x10^-3

[H+] = [H₂A-] = 5.27x10^-3 M

H₂A^-      HA^2-   + H⁺

initially   5.27x10^-3 0 0

finally 5.27x10^-3 - x x x

Ka2 = x*x / 5.27x10^-3 - x

3.80x10^-5 = x*x / 5.27x10^-3 - x

2.00x10^-7 - x*3.80x10^-5 = x²

x² + x*3.80x10^-5 - 2.00x10^-7 = 0

x = 4.286x10^-4

[H+] = [H₂A-] = 4.286x10^-4 M

As, pKa3 are very large, so only pKa1 and pKa2 will contribute to the H⁺ ion concentration.

Volume of acid = 0.10 L

moles of H⁺ ions = 0.10*(5.27x10^-3 + 4.286x10^-4 )

= 5.696x10^-4 moles

[KOH] = 1.0E-2 = 0.01 M

volume of KOH = 0.255 L

moles of OH^- = 0.255*0.01

= 2.55x10^-3

moles of OH^- in excess = 2.55x10^-3 - 5.696x10^-4

= 1.98x10^-3 moles

Total volume of solution = 0.10 + 0.255 = 0.355 L

[OH^-] = 1.98x10^-3 / 0.355

= 5.57x10^-3

pOH = - log [OH-]

= - log (5.57x10^-3)

= 2.25

pH = 14 - pOH

= 14 - 2.25

= 11.75