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CHEMISTRY

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A buffer that is 0.271 M in acid, HA, and 0.150 M in the potassium salt of its conjugate base, KA, has a pH of 2.85. What is the pH of the buffer after 130. mL of 0.155 M LiOH is added to 0.650 L of this buffer? Assume that the volumes are additive.

  • CHEMISTRY -

    First determine pKa.
    pH = pKa + log(A^-)/(HA)
    I have approximately 3 but you need to be more accurate than that.
    650 mL x 0.271M = about 170 mmoles
    650 mL x 0.150M = about 95 mmoles

    mmoles LiOH added = 130 x 0.155M = about 20 mmoles.
    ...........HA + OH^- ==> A^- + H2O
    initial...170....0........95
    added...........20
    change...-20....-20......+20
    equil....150.....0........115
    Substitute the equil values (after refining them) into the HH equation and solve for pH.

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