Ask Question
30 September, 08:42

Given a diprotic acid, H 2 A, with two ionization constants of K a1 = 3.0 * 10 - 4 and K a2 = 4.0 * 10 - 11, calculate the pH for a 0.117 M solution of NaHA.

+1
Answers (1)
  1. 30 September, 08:54
    0
    The pH for a 0.117 M solution of NaHA is 2.227

    Explanation:

    To solve the question we check the difference in the Ka values thus

    Ka₁ / Ka₂ = 7500000 < 10⁸ so we are required to calculate each value as follows

    We therefore have

    H₂X→ H⁺¹+HX⁻¹ with Ka₁ = 3.0 * 10⁻⁴

    Therefore

    3.0 * 10⁻⁴ = (x²) / (0.117)

    x² = 3.0 * 10⁻⁴ * 0.117 and x = 5.925 * 10⁻³ = [H⁺]

    Similarly

    Ka₂ = 4.0 * 10⁻¹¹

    and

    4.0 * 10⁻¹¹ = (x²) / (0.117)

    x² = 0.117 * 4.0 * 10⁻¹¹

    x = 2.16 * 10⁻⁶

    Total H⁺ = 5.925 * 10⁻³+2.16 * 10⁻⁶ = 5.927 * 10⁻³

    Since pH = - log of hydrogen ion concentration,

    pH = - log 5.927 * 10⁻³ = 2.227
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “Given a diprotic acid, H 2 A, with two ionization constants of K a1 = 3.0 * 10 - 4 and K a2 = 4.0 * 10 - 11, calculate the pH for a 0.117 M ...” 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