anonymous
  • anonymous
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
Chemistry
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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aaronq
  • aaronq
Use the Henderson-Hasselbalch equation
anonymous
  • anonymous
@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%
aaronq
  • aaronq
What you did write for the \(log\dfrac{[A^-]}{[HA]}\) part

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anonymous
  • anonymous
pH= 4.744727495 + log [0.615]/ [1.185] = 5.029843235
anonymous
  • anonymous
@aaronq
aaronq
  • aaronq
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 \([H^+]\)
aaronq
  • aaronq
\(\sf K_A=\dfrac{[H^+][acetate^-]}{[acetic~acid]}=\dfrac{x^2}{(0.045-x)}\)
aaronq
  • aaronq
and \(\sf x=[H^+]\)
anonymous
  • anonymous
@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 :(
aaronq
  • aaronq
that's really weird man, i dont know what could be wrong
anonymous
  • anonymous
@aaronq me either, thank you so much though for your assistance!
aaronq
  • aaronq
no problem !
aaronq
  • aaronq
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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