Shit yo

[danv1086]

You have been asked to prepare a buffer of pH 3.20 and have only two reagents available. The first is 500.00 mL of .1 M acetic acid and the other is a jar containing 100 g of KOH (Ka=1.8x10^-5

Comments

[esparami78119]

well, im dropping out and starting a bakery.

[kait807]

i wanna drop out and move to europe... i cant hack it and so much for calc 116 this summer since ill be retaking 115

fucking bitchez they set you up for failure!

at least we're all on the same page

[santiago125]

That's not too bad. Just use the Henderson-Hasselbalch equation.
pH = pKa + log([A-]/[HA])
You've got the desired pH and the pKa (pKa = -log(Ka)), so you can solve for the ratio of [A-]/[HA]. Once you've got that, you have the molarity of [HA] (acetic acid). Using the required molarity of KOH ([A-]) and the volume 200 mL, you can get how many moles of KOH salt you need and subsequently, the mass.

But I think that KOH and acetic acid don't even form a buffer solution. Conventionally, don't you use an acid and its conjugate base in a buffer solution?

[esparami78119]

WOW.
you really are smart.

[greenpants87]

LMAO! Doooo iiiit!

[bellefille]

i'm dropping out and becoming a golddigger.

WHOOP THAT TRICK.