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

Question

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

aaronq · Answer

Use the Henderson-Hasselbalch equation

anonymous · Answer

@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 · Answer

What you did write for the $log\dfrac{[A^-]}{[HA]}$ part

anonymous · Answer

pH= 4.744727495 + log [0.615]/ [1.185] = 5.029843235

anonymous · Answer

@aaronq

aaronq · Answer

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 · Answer

$\sf K_A=\dfrac{[H^+][acetate^-]}{[acetic~acid]}=\dfrac{x^2}{(0.045-x)}$

aaronq · Answer

and $\sf x=[H^+]$

anonymous · Answer

@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 · Answer

that's really weird man, i dont know what could be wrong

anonymous · Answer

@aaronq me either, thank you so much though for your assistance!

aaronq · Answer

no problem !

aaronq · Answer

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