What is the Ka value of phosphoric acid (H3PO4) given a 0.15 M solution with a pH of 1.54?

To determine the Ka value of phosphoric acid (H3PO4) given a 0.15 M solution with a pH of 1.54, we can follow these steps:

1. Calculate the concentration of H3O+ ions:

   The pH of the solution is given as 1.54. The concentration of H3O+ ions can be calculated using the formula:

   [H3O+] = 10^-pH

   [H3O+] = 10^-1.54 ≈ 0.0288 M

2. Write the expression for the acid dissociation constant (Ka):

   For the dissociation of phosphoric acid:

   H3PO4(aq) + H2O(l) ⇌ H2PO4^–(aq) + H3O⁺(aq)

   The Ka expression is:

   Ka = [H2PO4^–][H3O⁺] / [H3PO4]

3. Assume the initial concentration of H3PO4 is 0.15 M and the change in concentration is x:

   At equilibrium:

   [H3PO4] = 0.15 – x ≈ 0.15 M (since x is small compared to 0.15)

   [H2PO4^–] = x

   [H3O⁺] = x ≈ 0.0288 M

4. Substitute the values into the Ka expression:

   Ka = (x)(x) / (0.15 – x) ≈ (0.0288)(0.0288) / 0.15

   Ka ≈ 0.000829 / 0.15 ≈ 0.00553

Therefore, the Ka value for phosphoric acid (H3PO4) is approximately 0.00553.