Final answer:
To calculate the [H3O+] of a 0.350 M H3PO4 solution, consider primarily the first dissociation step due to its higher Ka value, setting up an ICE table, and solving the equilibrium expression.
Step-by-step explanation:
Calculating the concentration of hydronium ions ([H3O+]) in a solution of a polyprotic acid such as phosphoric acid (H3PO4) requires an understanding of acid dissociation constants (Ka) and the significant difference between each dissociation step. Since H3PO4 is a triprotic acid with three dissociation constants (Ka1=7.5×10−3, Ka2=6.2×10−8, and Ka3=4.2×10−13), the concentration of H3O+ can be calculated considering mostly the first dissociation step due to the larger Ka value. As a rough approximation, we can ignore the contributions of the second and third ionizations, as their Ka values are significantly smaller by factors of 105 to 106.
For the first ionization of H3PO4:
H3PO4 (aq) ⇌ H+ (aq) + H2PO4− (aq),
we can set up an ICE table and apply the equilibrium constant expression:
Ka1 = [H+][H2PO4−]/[H3PO4].
For a 0.350 M solution of H3PO4, we assume x is the amount of H3PO4 that dissociates. The expression becomes:
Ka1 = (x)(x)/(0.350-x)≈(x)2/0.350, since x is small and can be neglected from the denominator. Solving for x gives us the concentration of H+, after which the concentration of hydronium ions [H3O+] is equal to [H+], since both represent the same entity in solution.