[bioiskillingme]

I've been stuck on these for an hour and a half now! *delete me* I will be so grateful

Make up l liter of 0.0l M sodium acetate buffer at pH 4.5. The pKa of acetic acid is 4.7. Calculate the grams of anhydrous sodium acetate (m.w. 82) and number of milliliters of glacial acetic acid (m.w. 60; specific density 1.049 g/mL) required. By the way, what is the molarity of glacial acetic acid?
I know that this one requires the HH equation, just not sure how to use it...

If 10 mL of an acid at 0.09 M gave rise to a solution of pH 5.9 when mixed with 20 mL of 0.15 M potassium salt of itself, what is the pKa of this acid?

[Nobby]

HH-equation

pH = pKa + log ((cA^-)/c(HA))

pH = 4,5
pKa = 4,7

0.01 M buffer means 0,6 g/l Acetic acid and 0,82 g/l sodiumacetate

Calculate with this.

-0,2 = log ((cA^-)/c(HA))

[bioiskillingme]

Okay, I got that far. I figured out that the ratio [Base]/[Acid] = 1.585 using that, but I don't know where to go from there.

[rabolisk]

Okay, I got that far. I figured out that the ratio [Base]/[Acid] = 1.585 using that, but I don't know where to go from there.

Are you sure about that?

[Nobby]

I think  you have to fix one variable, because we have 2 variables but only one equation.  

-0,2 = log ((cA-)/c(HA))
 
0,631 = (cA-)/c(HA)

0,631*c(HA) = (cA-)

[bioiskillingme]

That's all the information I was given! I saw what I did wrong, so now I have the ratio as [Base]/[Acid] = .631

I don't understand how you got this: 0.01 M buffer means 0.6 g/l Acetic acid and 0.82 g/l sodium acetate

[AWK]

).01 M buffer means that combined concentration of both reagents is 0.01 M. Then you have a mathematical problem
of typy x/y = 0.631 and simultaneously x + y = 0.01

[Nobby]

And this is soluoble.

Right 0.01 n is the combination of both acetate and acetic acid.

[bioiskillingme]

Could you check my work really quickly? So for the second one, it would be 5.9 = pKa + log (.15/.09)? I don't understand where the volumes would come in.

And for the first one, it would be 1.049 g/mL divided by 60g/mol? That gives me an answer in mol/mL, which is .017 mol/mL.
Then using the HH Equation: pH = pKa + log ([Base]/[Acid])
4.5 = 4.7 + log ([Base]/[Acid])
[Base]/[Acid] = .631
Where x = base and y = acid, x + y = .01, so y = .01 – x and x/y = .631
Therefore x = [Base] = .00387 M and y = [Acid] = .00613 M
So, 82 g/mol * .00387 mol/L = .317 g/L for sodium acetate AND 60 g/mol * .00613 mol/L = .368 g/L for acetic acid
Thank you so, so much!

[bioiskillingme]

For the second one, would it be this instead:
5.9 = pKa + log [(.15M/20mL) /(.09M/10mL)]

[Nobby]

60 g/mol * .00613 mol/L = .368 g/L

And with the specific gravity of 1,049 g/ml you get 0.35 ml/l

[bioiskillingme]

So it should be 60 g/mol * .00613 mol/L / 1.049g/mL = .35 g/L for acetic acid ?

[Nobby]

No.

Calculate the grams of anhydrous sodium acetate (m.w. 82) and number of milliliters of glacial acetic acid (m.w. 60; specific density 1.049 g/mL) required

It will be 0,35 ml/l what means 350 µl/l

[bioiskillingme]

Okay, I see. Thank you so much!