Ask Question
12 August, 02:35

Given a diprotic acid, H2A, with two ionization constants of Ka1 = 2.4x10^-4 and Ka2 = 4.9x10^-11, calculate the pH for 0.182 M solution of NaHA.

+2
Answers (1)
  1. 12 August, 06:08
    0
    pH = 6.964 which rounds to 7.0 if 2 sig figs

    Explanation:

    Ka₁ * Ka₂ = x

    2.4x10⁻⁴ * 4.9x10⁻¹¹ = 1.176x10⁻¹⁴

    √1.176x10⁻¹⁴ = 1.084435337x10⁻⁷

    pH = - log[H⁺]

    pH = - log[1.084435337x10⁻⁷]

    pH = 6.964796339
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “Given a diprotic acid, H2A, with two ionization constants of Ka1 = 2.4x10^-4 and Ka2 = 4.9x10^-11, calculate the pH for 0.182 M solution of ...” in 📘 Chemistry if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers