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anonymous

  • one year ago

A student adds 15.0 mL 2.00 M Acetic Acid and 15.0 mL 2.00 M NaOH. Determine the pH of the resulting solution. Ka HC2H3O2 = 1.80E-5

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  1. anonymous
    • one year ago
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    This was my thought process. But it's saying that I have the answer incorrect by more than 10%

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  2. aaronq
    • one year ago
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    you have to use the concentrations as equilibrium, after the reaction (f any) has occured. The NaOH will neutralize the acetic acid, and you'll be left with 0.03 moles of acetate in 30 mL. so can just use the Kb and an eq. expression to find the pH

  3. anonymous
    • one year ago
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    @aaronq Okay, so would the equilibrium expression be Kc= [acetate] [OH-]/ [acetic acid]??

  4. aaronq
    • one year ago
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    nope, \(Acetate+H_2O\rightleftharpoons acetic~ acid+OH^-\) \(K_B=\dfrac{[acetic~ acid][OH^-]}{[acetate]}\)

  5. aaronq
    • one year ago
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    oh, and \(\sf K_B=\dfrac{K_w}{K_A}\)

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