What is the Ka value for the acid H3PO4 + H2O -> H2PO4- + H3O+.Fill in the following ICE table to help you.

Answer 1

H3PO4(aq) + H2O(l) ⇌ H2PO4-(aq) + H3O+(aq)

Initial (I): [H3PO4] = 0.10 M, [H2O] = --, [H2PO4-] = 0 M, [H3O+] = 0 M
Change (C): -x, +x, +x, +x
Equilibrium (E): 0.10-x, x, x, x

The equilibrium expression for the reaction is:
Ka = [H2PO4-][H3O+]/[H3PO4]

Substitute the equilibrium concentrations into the expression:

Ka = (x)(x)/(0.10-x)

Since the reaction is in equilibrium, we can assume that x is small compared to 0.10, so we can approximate 0.10-x to be 0.10:

Ka = (x)(x)/0.10

From the ICE table, we see that at equilibrium, [H2PO4-] = [H3O+] = x.

Therefore, Ka = x^2 / 0.10

The Ka value for the acid H3PO4 is equal to x^2 divided by 0.10.