How many mL of 0.100 M HCl must be added to 100.0 mL of 0.100 M F- to prepare a buffer solution that has a pH of 3.05?
F- (aq) +H20 (l) <--------> HF (aq) +OH- (aq) Kb= 1.47*10^-11
Sally, TP, etc, please don't change screen names. It matters little which name you choose but using the same one helps us serve you better.
See the sally post above about cautions about using mols and not mol/L.
millimols F^- = M x L = 10
...................F^- + H^+ ==> HF
I................10.......0.............0
add......................x................
C.............10-x....-x..............+x
E.............10-x.....0................x
Plug the E line into the HH equation and solve for x in millimols HCl Convert to mL of 0.100 M to be added.
Post your work if you get stuck.
To solve this problem, we will use the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the concentration of the acid (HCl) and its conjugate base (F-).
The Henderson-Hasselbalch equation is:
pH = pKa + log([A-]/[HA])
Where pH is the target pH, pKa is the negative logarithm of the acid dissociation constant (Ka) of the acid, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid.
In this case, the acid is HCl, and the conjugate base is F-. We know the pH (3.05) and can find pKa from the Kb value of HF using the equation:
pKa + pKb = 14
First, let's calculate pKa:
pKb = -log(Kb) = -log(1.47 * 10^-11) = 10.83
pKa = 14 - pKb = 14 - 10.83 = 3.17
Now we have all the necessary values to use the Henderson-Hasselbalch equation:
3.05 = 3.17 + log([F-]/[HCl])
To calculate the ratio [F-]/[HCl], which is the same as [A-]/[HA], we rearrange the equation:
log([F-]/[HCl]) = 3.05 - 3.17 = -0.12
Now, we need to convert this value to the actual ratio. To do that, we use the antilog function, which is the inverse of the logarithm function:
[F-]/[HCl] = antilog(-0.12)
Raising 10 to the power of -0.12:
[F-]/[HCl] = 10^-0.12
Using the logarithm on the right side of the equation:
[F-]/[HCl] = 0.727
Now, we know that the initial concentration of F- is 0.100 M and want to find the volume of HCl to add. Let's assume it is x mL.
The amount of F- in the final solution will be [F-] * (100 ml + x ml) = 0.100 M * (100 mL + x mL) = 0.100(100 + x) moles
The amount of HF (conjugate acid) will also be 0.100(100 + x) moles since the ratio of [F-]/[HCl] is 0.727.
Using the concentration formula, Concentration (C) = moles/volume (V):
0.100(100 + x) M = 0.100(100 + x) moles / (100 + x) mL
Since the concentration and moles are equal on both sides, we can cancel them out:
1 = 1 / (100 + x)
Now, let's solve for x:
1 = 1 / (100 + x)
Multiplying both sides by (100 + x):
100 + x = 1
Subtracting 100 from both sides:
x = 1 - 100
x = -99
Since the volume cannot be negative, this solution is not physically feasible.
Therefore, it is not possible to prepare a buffer solution with a pH of 3.05 using the given concentrations of HCl and F-.