Biochemistry homework (More so G.Chem review)

[Thuronn]

Have some problems on my 1st homework assignment in biochemistry that I'm having trouble with.

The biggest reason I'm having this trouble is because I lack a lot of my G. Chem notes, books etc and that was two plus years now.

In O. Chem my professor didn't really go over, pH's, pka, buffers at all essentially. Especially compared to other classes it seems.

"You prepared a soultion made of 50ml of 2.0M K2HPO4 and 25ml of 2.0M KH2PO4. Then you diluted the solution to a final volume of 200ml. What is the pH of the final solution? (For H3PO4, pka1= 2.2, pka2= 6.9, pka3=12.4)"

I know this involved the titration curve and acid and conjugate bases. I just don't get the relation between K2HPO4, and KH2PO4. It seemed like it would go from H3PO4, H2PO4-, HPO4 2-. The mixture of the two different compounds is blocking me.

"What is the pka of the weak acid HA if the solution containing 0.1M of the protonated form and 0.2M of deprotonated form at pH of 6.5?"

"What is the volume of a solution of 5 M NaOH that must be added to adjust the pH from 4 to 9 in 100mL of a 100mM solution of phosphate buffer?"

EDIT; I know Q#3, is M1V1= M2V2. Solved, done. And with #1 and #2 it's pH= pka + log (A-/HA). Just having trouble I guess with molarity somewhere.

[kuronosan]

To answer your second question, use the Henderson-Hasselbach equation.

[CDF]

"You prepared a soultion made of 50ml of 2.0M K2HPO4 and 25ml of 2.0M KH2PO4. Then you diluted the solution to a final volume of 200ml. What is the pH of the final solution? (For H3PO4, pka1= 2.2, pka2= 6.9, pka3=12.4)"

(Assume the buffer is being diluted with water.) Conceptually, if you keep diluting the solution with "pure" water (pretend carbon dioxide is not being solubilized), which can have a Ka of 1.8 E-16, the pH of the final solution should approach 7.0. The initial buffer is only 75 mL and you add 125 mL of water, so this seems to be the case.

If you were just interested in getting the final pH of the 75 mL buffer:

Monobasic potassium phosphate (KH2PO4) => acid (as a proton donor, with the conjugate base being HPO4 2-)
Dibasic potassium phosphate (K2HPO4) => conjugate base

Then apply Henderson-Hasselbalch.