Question

Using chemical equations, show how the triprotic acid H₃PO₄ ionizes in water. Phases are optional.

Answer 1

The answer to this question is quite simple

H₃PO₄ is triprotic in nature and ionizes in 3 steps in the process as follows

And for this reactions ka1, ka2 and ka3 can be written as

I hope this helps. If you have any query or want more detailed explanation, feel free to ask in the comments section below.

Answer 2

Answer 3

Answer 4

Answer 5

Answer 6

Solution:

H3PO4(Phosphoric Acid) is a Triprotic Acid which has three Protons(H+) in it. It Undergoes Ionisation in Water in Three Steps and which are given by the Following Chemical Equations and also the Rate of Ionisation (Ka) of the three steps is given below:

a)H3PO4 + H2O ----> H2PO4- + H3O+

Rate Ka1 = {[H2PO4-]×[H3O+]} ÷ [H3PO4] = 7.1×10^-3.

b)H2PO4- + H2O ----> HPO4 2- + H3O+

Rate Ka2= {[HPO4 2-]×[H3O+] ÷ [H2PO4-] = 6.3× 10^-8.

c) HPO4 2- + H2O ----> PO4 3- + H3O+

Rate Ka3 = {[PO4 3-]×[H3O+] ÷ [HPO4 2-] = 7.1×10^-13.

The Above given Values are the Rate Constant for the three Ionisation steps from the Rate Equations and are based on Standard Concentration Values of the Acid,Ions and Water.

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Answer 7