Ask Question
3 March, 13:04

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.

+1
Answers (1)
  1. 3 March, 13:30
    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 the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers