Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

Looking for something else?

Not the answer you are looking for? Search for more explanations.

- 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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

Get your **free** account and access **expert** answers to this and **thousands** of other questions

- anonymous

- katieb

See more answers at brainly.com

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- aaronq

Use the Henderson-Hasselbalch equation

- 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

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

- anonymous

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

- anonymous

- 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

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

- aaronq

and \(\sf x=[H^+]\)

- 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

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

- anonymous

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

- aaronq

no problem !

- 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

Looking for something else?

Not the answer you are looking for? Search for more explanations.