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March 27, 2017

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1L of a buffer composed of acetic acid and sodium acetate has a pH of 4.3. Adding 10mL of 2M sodium hydroxide solution to 100mL of this buffer causes the pH to rise to 4.87. what is the total molarity of the original buffer?

  • chemistry - ,

    pH = pka + log b/a
    4.3 = 4.74 + log b/a
    b/a = 0.363
    ----------------
    add 100 mL x 2M = 20 mmols to obtain
    4.87 = 4.74 + log b/a
    b/a = 1.35 and redo to read
    1.349 = (b+20)/(a-20)
    1.349(a-20) = b+20
    1.349a - 27 = b+20
    Substitute 0.363 for b to obtain
    1.349a - 27 = 0.363a + 20
    0.986a = 47
    a = 47.7
    b = 0.363a = 0.363*47.7 = 17.3
    a + b = 47.7 + 17.3 = 65
    --------------------
    Check:
    ..........HAc + OH^- ==> Ac^- + H2O
    I.........47.7...0.......17.3
    add ............20.........
    C.........-20..-20.......+20
    E.........27.7...0.......37.3

    pH = 4.74 + log(37.3/27.7)
    pH = 4.87

  • chemistry - ,

    i got a total concentration of 0.0774M is this correct?

  • chemistry - ,

    Where are you getting that?

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