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As a technician in a large pharmaceutical research firm, you need to produce 100.mL of 1.00 M a phosphate buffer solution of pH = 7.45. The pKa of H2PO4− is 7.21. You have 2.00 L of 1.00 M KH2PO4 solution and 1.50 L of 1.00 M K2HPO4 solution, as well as a carboy of pure distilled H2O. How much 1.00 M KH2PO4 will you need to make this solution?

  • Chemistry -

    Use the Henderson-Hasselbalch equation and the total buffer (both equations) and solve simultaneously.
    7.45 = 7.21 + log (base/acid)
    I get base/acid = estimated 1.8 but you need to do it more accurately.
    And base + acid = 1M
    Solve simultaneously. I obtained approximately 0.36M for acid and 0.64M for base. Again, you need to refine those numbers.
    You want 100 mL (0.100L x 0.36M = 0.036 mols acid). Similarly you need 0.064 mols base.
    Since M = mols/L and rearrange to
    L = mols/M = 0.036/1 = 0.036 L or 36 mL of the acid and similarly 64 mL of the base.

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