how can i calculate the volume of 3 M HCl needed to change the ph of 75 ml of undiluted buffer by one ph unit (buffer capacity). i know that pka is 4.75, acetic acid is 1.1009 M for 100 ml, and sodium acetate is 1.0113 M for 100ml.
i have calculated the moles so far for both the acid and acetate by doing concentration(ex 1.0113) x .1 L x .075 L and dividing all this by .2 L. is this correct? what do i do after?
6 answers
One thing you need to learn is patience. We're on this board when we can be. All of us volunteer out time. Give me a few minutes and I'll post a solution. By the way, it would help if you would post the problem clearly.
I don't know what the problem is.
I know you have 75 mL of a buffered HAc/NaAc solution with some unknown pH. I know you want to change the pH of that solution to 3.75. What I don't know follows:
The 100 mL 1.009 M HAc and 1.0113 M NaAc. Where are they? Are these the two solution given to add to the buffer in the problem to change the pH to 3.75? Is the 75 mL given to you made up of these two solutions.
If not, then what is the concn of the buffer now or what are the concentrations of the components now.
I know you have 75 mL of a buffered HAc/NaAc solution with some unknown pH. I know you want to change the pH of that solution to 3.75. What I don't know follows:
The 100 mL 1.009 M HAc and 1.0113 M NaAc. Where are they? Are these the two solution given to add to the buffer in the problem to change the pH to 3.75? Is the 75 mL given to you made up of these two solutions.
If not, then what is the concn of the buffer now or what are the concentrations of the components now.
no worries i am not in a rush, thanks for your help
100 ml of acetic acid and 100ml of sodium acetate were mixed (which is the buffer solution). then hcl (3 M) was added to 75 ml of the buffer solution
what i did was calculate the moles of the acid and acetate:
acetic acid: n=(1.1009 x .1L x .075L) / .2L =0.04
did the same for sodium acetate: n=0.03
then 3.75=4.75+log (0.03-x)/(0.04+x)
x= moles of HCl
then to calculate volume of hcl = moles of hcl/3M
not sure if i did it correctly
acetic acid: n=(1.1009 x .1L x .075L) / .2L =0.04
did the same for sodium acetate: n=0.03
then 3.75=4.75+log (0.03-x)/(0.04+x)
x= moles of HCl
then to calculate volume of hcl = moles of hcl/3M
not sure if i did it correctly
As I understand the problem, it is to change the pH of the buffer by 1 unit from what it is. So the first thing to do and to determine the pH of the buffer and it isn't 4.75 so here is how I would go about it. I work in millimols because that's easier for me. HAc = acetic acid. Ac^- = acetate
When you mix the reagents you have 100 mL 1.1009 M HAc mixed with 1.0113 M Ac^- .
millimols HAc = mL x M = 100 x 1.1009 = 110.09
millimols Ac^- = 100 x 1.0113 = 101.13
You take 75 mL of this solution so in that 75 mL you have
mmols HCl = 110.09 x (75/200) = 41.284
mmols Ac^- = 101.13 x (75/200) = 37.924
So pH of this 75 mL is
pH = 4.75 + log (37.924/41.284) = 4.75 - 0.03867 = 4.75 - 0.04 = 4.71
so you want to add enough 3 M HCl to change that to 3.71.
................Ac^- + HCl ==> HAc + Cl^-
I.............37.924.....0..........41.284................
add.........................x............................
C...............-x...........-x............. +x
E...........37.924-x......0........41.284+x
Now plug the E line into the HH equation.
3.71 = 4.75 + log ((37.92-x)/(41.284+x)
and solve for x = millimols HCl to add to move the pH from 4.61 to 3.61. Then M = millimoles/mL. You know M = 3 and you know miliimols (that's the x), solve for mL: and you have it. You should confirm all of these numbers and the procedure. Let me know if you don't understand anything I did. You may not realize it but for the log base/acid term, the HH equation is for CONCENTRATION BASE/CONCENTGRATION ACID. However, since concn is mols/L that is base is mols/L and that divided by acid in mols/L so the L part cancels and you can write it as log (mols base/mols acid). Technically that isn't correct because it is supposed to be concn and not mols BUT the math works out to be the same so it's easier to work in mols (or millimols) and not go through the extra step(s) to get concentration. Do profs like this. No, well some may. When I taught this I ALWAYS counted off if the student didn't do concn BUT I let them do this. pH = pKa + log (base/acid) = pKa + log (basemols/v/acidmols/v) and since the v is ALWAYS the same, v cancels and you end up working in mols or millimols.
When you mix the reagents you have 100 mL 1.1009 M HAc mixed with 1.0113 M Ac^- .
millimols HAc = mL x M = 100 x 1.1009 = 110.09
millimols Ac^- = 100 x 1.0113 = 101.13
You take 75 mL of this solution so in that 75 mL you have
mmols HCl = 110.09 x (75/200) = 41.284
mmols Ac^- = 101.13 x (75/200) = 37.924
So pH of this 75 mL is
pH = 4.75 + log (37.924/41.284) = 4.75 - 0.03867 = 4.75 - 0.04 = 4.71
so you want to add enough 3 M HCl to change that to 3.71.
................Ac^- + HCl ==> HAc + Cl^-
I.............37.924.....0..........41.284................
add.........................x............................
C...............-x...........-x............. +x
E...........37.924-x......0........41.284+x
Now plug the E line into the HH equation.
3.71 = 4.75 + log ((37.92-x)/(41.284+x)
and solve for x = millimols HCl to add to move the pH from 4.61 to 3.61. Then M = millimoles/mL. You know M = 3 and you know miliimols (that's the x), solve for mL: and you have it. You should confirm all of these numbers and the procedure. Let me know if you don't understand anything I did. You may not realize it but for the log base/acid term, the HH equation is for CONCENTRATION BASE/CONCENTGRATION ACID. However, since concn is mols/L that is base is mols/L and that divided by acid in mols/L so the L part cancels and you can write it as log (mols base/mols acid). Technically that isn't correct because it is supposed to be concn and not mols BUT the math works out to be the same so it's easier to work in mols (or millimols) and not go through the extra step(s) to get concentration. Do profs like this. No, well some may. When I taught this I ALWAYS counted off if the student didn't do concn BUT I let them do this. pH = pKa + log (base/acid) = pKa + log (basemols/v/acidmols/v) and since the v is ALWAYS the same, v cancels and you end up working in mols or millimols.