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5 July, 15:52

As a technician in a large pharmaceutical research firm, you need to produce 450. mL of 1.00 M potassium phosphate buffer solution of pH = 6.97. The pKa of H2PO4 - is 7.21. You have the following supplies: 2.00 L of 1.00 M KH2PO4 stock solution, 1.50 L of 1.00 M K2HPO4 stock solution, and a carboy of pure distilled H2O. How much 1.00 M KH2PO4 will you need to make this solution? Express your answer to three significant digits with the appropriate units.

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  1. 5 July, 18:34
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    You have to add 285.642 mL of KH2PO4

    Explanation:

    Step 1: What is a buffer

    A buffer is a mixture of a weak acid and its conjugate base or a weak base and its conjugate acid. Buffers work by reacting with any added acid or base to control the pH.

    pH of a buffer = pKa + log [conj base] / [weak acid]

    Step 2: The ph of the buffer

    6.97 = 7.21 = log [K2HPO4] / [KH2PO4]

    -0.24 = log [K2HPO4] / [KH2PO4]

    10 ^ - 0.24 = [K2HPO4] / [KH2PO4] = 0.5754

    0.5754 = (0.450 - x) / x

    0.5754x = 0.450 - x

    1.5754x = 0.450 L

    x = 0.28564 = volume of KH2PO4

    0.450 - 0.28564 = 0.16436 L K2HPO4

    You have to add 285.642 mL of KH2PO4
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