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CHEMISTRY LAB urgent assistance

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1 (A) This relation gives the dissociation constant of acetic acid (CH3COOH)

Ka = [H3O+][CH3COO-]\[CH3COOH]

Starting from this relationship show that

pH = pKa + log [base]\[acid]

urgently need assistance

  • CHEMISTRY LAB urgent assistance - ,

    Start with the acid dissociation equation:

    HAc -> H+ + Ac-

    Ka is then

    [H+][Ac-]/[HAc] {given in the question}

    at the start

    HAc -> H+ + Ac-

    and we start from A mole/L and B mol/L for the acid and buffer

    A M ___0__ B M

    at the end

    HAc -> H+ + Ac-
    A-x____x___B+x

    so Ka is
    Ka=(x)(B+x)/(A-x)

    Assume that x is small wrt A and wrt B so that the equation reduces to

    Ka=x(B)/(A)

    which we can rearrange to

    x=Ka(A)/(B)

    take logs of each side

    log(x)=log(Ka(A)/(B))

    which is

    log(x)=log(Ka) + log((A)/(B))

    multiply through by -1

    -log(x)=-log(Ka) - log((A)/(B))

    which is

    -log(x)=-log(Ka) + log((B)/(A))

    or

    pH = pKa + log((B)/(A))

    or

    pH = pKa + log [base]\[acid]

  • CHEMISTRY LAB urgent assistance - ,

    As an added note (perhaps picky but I think needed) is that / is used as a divide sign while \ is used for other purposes.

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