A volume of 500.0 mL of 0.140 M NaOH is added to 545 mL of 0.200 M weak acid (𝐾a=2.69×10−5). What is the pH of the resulting buffer?

A volume of 500.0 mL of 0.140 M NaOH is added to 545 mL of 0.200 M weak acid (𝐾a=2.69×10−5). What is the pH of the resulting buffer?

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  • david
    Lv 7
    4 weeks ago
    Favorite Answer

    500 X 0.140 = 70 mmol OH-

    545 X 0.2 = 109mmol H+  << excess

      39mmol weak acid (HA)  ///  1045mL total Vol

      =  0.03732 M ... initial HA conc.

    init conc A-  =  70/1045  =  0.066986 M

      Ka = [H+][A-]/[HA]

      2.69x10^-5  =  x(0.066986 + x) / (0.03732 - x)

     = = = =  Round .. solve

       2.69x10^-5 = x(0.066986) / (0.03732)

        x =  1.4489x10^-5 = [H+]

       pH = 4.824

       ...  use sig figs as needed

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