It is easy but I cannot solve or do buffer questions/slns.. Please help.

Question

Describe the preparation of 500 ml of 1 M phosphate buffer at pH 7.2.

Answer 1

pH = pKa + log(base)/(acid)
pH = 7.2
pKa is pKa2 for H3PO4
base is HPO4^2-
acid is H2PO4^-
and you want 1 M buffer which means
a + b = 1

First you determine the values of a and b. You need two equations. The first one is
7.2 = pKa2 + log b/a
Solve for b/a

The second one is a + b = 1 and I would convert that 1 M to mols. That would be
a + b = 0.5*1

Solve them simultaneously for a and b. a and b will be in mols.
Then mols = grams/molar mass.
You weigh out that many grams of the Na or K salt (molar mass must be the one you want to use) and dissolve that amount of each in water and make up to 500 mL. If you have an example that has you stumped work it to the point where you are stuck then explain what you trouble is and I can help you through it.

Answer 2

I did this:
as you told, this is wrong :(

H3PO4 ↔ H2PO4 + H+ ↔ HPO4 + H+ ↔ PO4 + H+ pH: 7,2

Prepare 500 ml of 1 M NaH2PO4.
500ml = 0,5 liter
0,5 l * 1M = 0,5 mol NaH2PO4
1 mol NaH2PO4 = 120 gr, so; 0,5 mol NaH2PO4 = 60 gr
Solve 60 gr NaH2PO4 in 500 ml distilled water.

Prepare 150 ml of 1 M Na2H2PO4 for HPO4
0,5 l * 1 M = 0,5 mol Na2HPO4
1 mol Na2HPO4 = 142 gr, so; 0,5 mol Na2HPO4 = 71 gr
Solve 71 gr Na2HPO4 in 500 ml distilled water.

Mix prepared NaH2PO4 and Na2HPO4 solutions.

** and I am really bad at buffer solutions, so I have trouble understanding what you tell

Answer 3

No, I disagree. You aren't necessarily bad at solving buffer problems, you are bad about not following instructions. Of course it's wrong. You didn't follow my instructions.

Step 1.
Use the equation 7.2 = pKa2 + log b/a where b is base and a is acid. You will need to look up pKa2 for H3PO4. Solve this equation for the ratio of b/a.

Step 2.
Use your second equation as
a + b = 0.5 mol

Step 3. Solve the equation in step 1 and the equation is step 2 simultaneously for a and b to get a value for acid mols and base mols.

Step 4. Then mols = grams/molar mass for each salt. You know mols of acid and mols base and molar mass of each. Your work looks ok on that part at the end of your second post; i.e., the "wrong" part is that you assumed for some reason that mols was 0.5 for each and worked accordingly. When you find the correct value for mols acid and mols base it should be ok. If you still have problems and you post again please show what pKa2 value you are using.

Answer 4

I did them before you told me, and when I read your steps I understand it is wrong.
I will try again.

I really thank you so much !! :)

Answer 5

Repost if you still have trouble but remember I need what you're using for pKa2 if I'm to come up with the same answer you have.

Answer 6

Ok there is a problem again. My lecturer said take the values showed on lecture.
Values are so: pKa1= 2,14 pKa2= 7,2 pKa3= 12,4

when I take pKa2= 7,2 then log b/a = 0 then b=0 a= 0,5 . Don't we use any base?

7.2 = pKa2 + log b/a
7.2 = 7.20 + log b/a
log b/a = 0 then b/a = 0 then b= 0
a + b = 0.5 mol , then a= 0.5

Answer 7

I saw this : when pKa = pH, there is equal concentration of acid and its conjugate base.

So is what ý did before true ?

Answer 8

OMG!! log 1 = 0 so b/a = 1

I am deeply embarrassed :(
why ý write log 0 = 0
:O

Answer 9

a + b = 0,5
then a=b= 0,25 mol

0.25 mol / 0,5 l = 0,5 M NaH2PO4 (30 gr & 0,5 M)
0,25 / 0,5 = 0,5 M Na2HPO4 (35,5 gr & 0,5 M)

Is it ok now?