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anonymous

  • one year ago

A buffer solution is prepared by mixing 15.0 mL of 2.00 M Acetic Acid and 10.0 mL of 1.50 M NaC2H3O2. Determine the pH of the solution after the addition of 0.015 moles HCl (assume there is no change in volume when the HCl is added). Ka HC2H3O2 = 1.80E-5

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  1. aaronq
    • one year ago
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    Use the Henderson-Hasselbalch equation

  2. anonymous
    • one year ago
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    @aaronq I used heenderson-Hasselbalch equation and I got 4.745 for my Pka and for the pH I got 5.0298 but it is saying that my answer is incorrect by more than 10%

  3. aaronq
    • one year ago
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    What you did write for the \(log\dfrac{[A^-]}{[HA]}\) part

  4. anonymous
    • one year ago
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    pH= 4.744727495 + log [0.615]/ [1.185] = 5.029843235

  5. anonymous
    • one year ago
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    @aaronq

  6. aaronq
    • one year ago
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    so, the 0.015 moles of HCl neutralizes the 15 mL of 1.5 M of sodium acetate. so in fact you have 0.03 moles (from 15.0 mL of 2.0 M acetic acid) + 0.15 moles 0.045 moles total, so use an equilibrium expression to find \([H^+]\)

  7. aaronq
    • one year ago
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    \(\sf K_A=\dfrac{[H^+][acetate^-]}{[acetic~acid]}=\dfrac{x^2}{(0.045-x)}\)

  8. aaronq
    • one year ago
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    and \(\sf x=[H^+]\)

  9. anonymous
    • one year ago
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    @aaronq Hey, so when i worked it out I got 3.049903011 for the pH and its saying that I am still incorrect by 10%. This is so confusing :(

  10. aaronq
    • one year ago
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    that's really weird man, i dont know what could be wrong

  11. anonymous
    • one year ago
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    @aaronq me either, thank you so much though for your assistance!

  12. aaronq
    • one year ago
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    no problem !

  13. aaronq
    • one year ago
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    do you think it could be 4.04? because 5.4 is 10% away, and 3.04 is also 10%, so the value should be directly between them

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