If you need to prepare 250.0 mL of a pH 5.00 buffer tha
t has a total buffer concentration of acetic acid +
sodium acetate of 0.050 M, how many moles of each will you need to prepare the solution? Given solutions
of acetic acid and sodium acetate with concentrations of 0.10 M and pKa= 4.76, describe how to pre
pare this

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  1. pH = pKa + log (Ac^-)/(HAc)
    Plug in pH, pKa, and solve for (Ac^-)/(HAc). That is equation 1.

    Equation 2 is (HAc) + (Ac^-) = 0.05

    Solve equations 1 and 2 simultaneously to find (Ac^-) and (HAc).
    Then mol HAc = M x 0.250 = ?
    and mols NaAc = M x 0.25 = ?

    The above gives you the mols you will need which is what you asked for in the first part of the problem. I don't understand how the 0.1M solutions are to be used. Usually problems of this kind ask for mL of the solutions but this one doesn't do that so I ignored that part of the problem.

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  2. How do you solve those 2 equations simultaneously

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  3. That's math, not chemistry. :).
    To make it easy typing let me call the acetic acid, a (for acid) and the sodium acetate, b (for base). So eqn 1 is pH = pKa + log b/a
    eqn 2 is a+b = 0.05

    Substitute into eqn 1 as follows:
    5.00 = 4.76 + log b/a
    5.00-4.76 = log b/a
    0.24 = log b/a
    b/a = 1.74 or
    b = 1.74a

    eqn 2 is a + b = 0.05
    Substitute b from 1 into 2.
    a + 1.74a = 0.05
    2.74a = 0.05
    a = 0.05/2.74 = ?
    Then you substitute a back into eqn 2 of a + b = 0.05. Now you know the 0.05 and a, solve for b and go from there.

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  4. kindly give me hint for this

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