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9.00mL of 1.10M CH3COOH and 1.00mL of 0.900M CH3COO- what is the pH?

  • buffers - ,

    Use the Henderson-Hasselbalch equation.

  • buffers - ,

    we havn't learnt that equation.

  • buffers - ,

    pH = pKa + log(base/acid)
    pKa is pKa for CH3COOH
    base is (CH3COO^-)
    acid is (CH3COOH)

  • buffers - ,

    thanks. so for this question is the pH 3.70?

  • buffers - ,

    Or you can do it another way.
    Write ionization of CH3COOH.

    CH3COOH ==> H^+ + CH3COO^-

    Now write the Ka expression.
    Ka = (H^+)(CH3COO^-)/(CH3COOH)

    Now plug in (CH3COO^-) from the problem.
    Substitute (CH3COOH) from the problem.
    Substitute Ka for CH3COOH. Try to find Ka in your text or notes but it's very close to 1.8 x 10^-5. Calculate (H^+) and convert that to pH.

  • buffers - ,

    I calculate 3.698 pH which rounds to 3.70. Yes. Very good.

  • buffers - ,

    thanks a lot :D

  • buffers - ,

    You're welcome.

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