Question

Consider the titration of 100.0 mL of 0.43 M H3A by 0.43 M KOH for the next three questions. The triprotic acid has Ka1 = 1.0 x 10-4, Ka2 = 1.0 x 10-7, and an unknown value for Ka3.

1) Calculate the pH after 100.0 mL of KOH has been added.

2) Calculate the pH after 150.0 mL of KOH has been added.

3) The pH of the solution after 200.0 mL of KOH has been added is 9.00. Determine the value of K_a3 for this triprotic acid. Use scientific notation to enter this answer, e.g., 1.0 x 10^-3 = 1.0E-3.

Answer 1

Equelibrium reaction is as follows:

H3A + KOH --------> H2AK + H2O

1) tripotic acid millimoles = 100 * 0.43 = 43

base millimoles = 100 * 0.43 = 43

1 mol og H3A needed one mole of KOH so,

[H2AK] = mmol H2KA / vt = 43 mmol / (100 + 100)ml = 0.21 M

[H+]E2 = Ka1Ka2[H2AK] + Ka1Kw / Ka1 + [H2AK] = (1E-4)(1E-7)(0.21) + (1E-4)(1E-14) / (1E-4)+0.21 = 3.16E-6 M

pH = -log[H+] = -log(3.16E-6)

pH = 5.50

2)

tripotic acid millimoles = 100 * 0.43 = 43

base millimoles = 150 * 0.43 = 64.5

H3A + KOH ------------> H2AK + H20

43 64.5 0 0

-43 -43 +43   

0 21.5 43

H2AK + KOH ------------->HAK2 + H20

43 21.5 0 0

-21.5 -21.5 +21.5

21.5 0 21

pH = pKa2 + log [HAK2]/[H2AK] = -log (1E-7) + log (21.5mmol/250 ml)/(21.5mmol/250ml) = -log(1E-7) + log 1

pH = 7

3) an addition of 200ml for HAK2

[HAK2] = mmol HAK2 / vt = 43mmol / 300mol = 0.1433 M

pH = 9

[H+] = 10E-pH = 10E-9

[H+]E2 = Ka2Ka3 [HAK2 + Ka2Kw / Ka2 + [H2AK]

10E-9 = (1E-7)Ka3(0.1433) + (1E-7)(1E-14) / (1E-7)(0.1433)

Ka3 = 9.93E-12