Ask Question
20 July, 05:01

Which may be used to prepare a buffer having a ph of 8.8? ka = 7 * 10-3 for h3po4; 8 * 10-8 for h2po4-; 5 * 10-13 for hpo42-?

+1
Answers (1)
  1. 20 July, 07:43
    0
    h3po4 and h2po4 - can be used to prepare buffer of pH = 8.8

    The ph of buffer can be shown as:

    pH = pKa + log [Salt] / [ Acid ]

    [Salt] / [ Acid ] = x

    For h3po4 with ka = 7 * 10-3

    8.8 = - log (7 * 10^-3) + log x

    8.8 = 2.21 + log x

    Thus, the value of log x is coming positive and therefore can be used for preparing buffer.

    For h2po4 - with ka = 8 * 10-8

    8.8 = - log (8 * 10^-8) + log x

    8.8 = 7.14 + log x

    Thus, the value of log x is coming positive and therefore can be used for preparing buffer.

    For hpo42 - with ka = 5 * 10-13

    8.8 = - log (5 * 10-13) + log x

    8.8 = 12.31 + log x

    Thus, the value of log x is coming negative and therefore can not be used for preparing buffer.
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “Which may be used to prepare a buffer having a ph of 8.8? ka = 7 * 10-3 for h3po4; 8 * 10-8 for h2po4-; 5 * 10-13 for hpo42-? ...” 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