Chemistry help pliz

Explain (using equations) how a solution of 1.0 mol dm^-3 in both CH3COOH and CH3COONa is resistant to changes n pH when we add either small amounts of acid or small amount of base (such solution is called a buffer)

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  1. continued from above calculate the initial pH of the acetic acid-sodium acetate solution above.The pKa of acetic acid is 4.76

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  2. To do the second part from first principles. Start with the acid dissociation equation:

    HAc -> H+ + Ac-

    Ka is then

    [H+][Ac-]/[HAc]

    at the start

    HAc -> H+ + Ac-

    1.0 M ___0__1.0M

    at the end

    HAc -> H+ + Ac-
    1.0-x__x___1.0+x

    so Ka is
    Ka=(x)(1.0+x)/(1.0-x)

    we can treat this in two ways, we can solve for x

    or

    Assume that x is small so that

    Ka=x(1.0)/(1.0)=x

    thus pKa = pH =4.76

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  3. To give another perspective, you can calculate the pH of a buffer using the Henderson-Hasselbalch equation.

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