Question

Consider the dissociation of triprotic acid, H3Y:

H3Y(aq) + H2O(l) <-----> H2Y-(aq) + H30+(aq)                                  Ka1 = 2.0 x 10-4

H2A-(aq) + H2O(l) <-----> HY2-(aq) + H30+(aq)                                Ka2 = 8.0 x 10-8

HY2-(aq)+ H2O(l) <-----> Y3-(aq) + H30+ (aq)                                    Ka3 = 4.0 x 10-13

2H2O(l) <-----> H30+ + OH-                                                                Kw =[H30+][ OH-] = 1.0 x 10-14

Use the values of Ka1, Ka2, Ka3, and Kw and evaluate the equilibrium constant (Keq) associated with each of the three reactions below. Star with Ka1 and Kw and determine a numerical value for Keq1. Repeat the same calculation and determine a numerical value for Keq2 using the numerical values of Ka2 and Kw. Repeat the same calculation and determine a numerical value for Keq3 using the numerical values of Ka3 and Kw.

H3Y + OH- ↔ H2Y- + H20(l)                                                         Keq1 =

H2A- + OH- ↔ HY2- + H20(l)                                                        Keq2 =

HY2- + OH- ↔ Y3- + H20(l)                                                           Keq3 =

b) Use your answers to Q2:a) to explain why only two equivalence points would be observed in the titration of H₃Y with NaOH

please show work.

Answer 1

For Keq1,

Invert Kw and add Ka1 and this new equation, we get Keq1

So, Keq1 = Ka1/Kw = 2 x 10^-4/1 x 10^-14 = 2 x 10^10

For Keq2,

Invert Kw and add Ka2 and this new equation, we get Keq2

So, Keq2 = Ka2/Kw = 8 x 10^-8/1 x 10^-14 = 8 x 10^6

For Keq3,

Invert Kw and add Ka3 and this new equation, we get Keq3

So, Keq3 = Ka3/Kw = 4 x 10^-13/1 x 10^-14 = 40

b) Only two equivalent points are observed as Keq3 is too large and remains unchanged through out the reaction system.