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?
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.
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
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.
To calculate the volume of 3 M HCl needed to change the pH of the buffer by one pH unit, we need to consider the Henderson-Hasselbalch equation and the buffer capacity.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log ([A-]/[HA])
Where:
pH is the desired pH after the addition of HCl (which is 1 unit lower than the initial pH)
pKa is the dissociation constant of the weak acid (acetic acid in this case)
[A-] is the concentration of the conjugate base (sodium acetate in this case)
[HA] is the concentration of the weak acid (acetic acid)
First, let's calculate the initial pH of the buffer. Since we know the pKa (4.75), we can use the Henderson-Hasselbalch equation using the initial concentrations of acetic acid and sodium acetate.
pH initial = 4.75 + log (1.0113/1.1009)
Next, we need to calculate the concentrations of acetic acid and sodium acetate after the addition of HCl. Since we know the volume of the buffer is 75 ml and the concentrations are given for 100 ml, we can adjust the concentrations accordingly:
Concentration of acetic acid after HCl addition = (1.1009 M * 0.075 L) / 0.1 L
Concentration of sodium acetate after HCl addition = (1.0113 M * 0.075 L) / 0.1 L
Now, we can plug these values into the Henderson-Hasselbalch equation to calculate the new pH:
pH final = 4.75 + log ([sodium acetate after addition]/[acetic acid after addition])
Next, subtract the desired pH (which is 1 unit lower than the initial pH) from the initial pH to get the ph unit change:
pH unit change = pH initial - (pH initial - 1)
Finally, we can calculate the required volume of 3 M HCl to achieve this pH unit change. Assuming that there is complete dissociation of HCl, we can use the equation:
moles of HCl = (volume of HCl needed * concentration of HCl)
Now, rearrange the equation to find the volume of HCl needed:
volume of HCl needed = (moles of HCl) / (concentration of HCl)
Just be sure that all concentrations are converted to the same units (e.g., moles per liter).
Remember to handle appropriate algebraic signs and make necessary conversions to ensure accurate calculations.