Calculate amount of solid NaF required to prepare a 250mL acidic buffer with pH=2.75. The initial molarity of HF is .25M but you only have .075L of it.
I have tried many different ways... the most recent and seems to be simplest was using H-H equaion and solving for the CB/CA ratio. Then equivalating this to mol NaF/ (M HF* V HF).... any suggestions?
I would do this, all with the HH equation.
You didn't give a pKa for HF. I used 3.14 but you need to use whatever value your text has (or notes).
2.75 = 3.14 + log [(base)/(acid)]
Here is what you do for the (acid)
If we use all of the HF, and we want 250 mL total volume, I would calculate the molarity of that solution. That is
M = moles/L = mmoles/mL = (75mL x 0.25M/250mL) = 0.07500M
(base) = moles/L = (x moles/0.250 L).
Plug those into HH equation and solve for x moles NaF. Then moles NaF x molar mass NaF = grams. I get an answer close to 0.4 g but you need to go through it and do it more accurately. Finally, I suggest you start with the number of grams NaF, convert to moles and M, do the same for the HF solution, plug into the HH equation and see if you end up with a pH of 2.75. If yes you can be confident in your answer; if not there is an error somewhere.
To calculate the amount of solid NaF required to prepare the acidic buffer, you can use the Henderson-Hasselbalch equation and the concept of buffer capacity.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log([A-]/[HA])
where pH is the desired pH of the buffer, pKa is the acid dissociation constant of the acid (HF in this case), [A-] is the concentration of the conjugate base (F-), and [HA] is the concentration of the acid (HF).
In this case, the desired pH is 2.75. The pKa of HF is approximately 3.17 (you can look up the specific value for more accuracy).
First, let's calculate the concentration of HF required to achieve pH 2.75 using the Henderson-Hasselbalch equation:
2.75 = 3.17 + log([F-]/[HF])
Rearrange the equation:
log([F-]/[HF]) = 2.75 - 3.17
log([F-]/[HF]) = -0.42
Now, we can convert this equation into an exponentiated form:
[F-]/[HF] = 10^(-0.42)
[F-]/[HF] = 0.393
Next, calculate the concentration of the acid (HF) based on the given volume and initial molarity:
[HF] = (moles of HF) / (volume of HF in liters)
[HF] = (.25 mol/L) * (.075 L)
[HF] = 0.01875 moles
Now, let's calculate the concentration of the conjugate base (F-):
[F-] = (0.393) * (0.01875 moles)
[F-] = 0.00735625 moles
Finally, we can calculate the amount of solid NaF required to achieve the desired concentration of F-:
mass of NaF = (moles of NaF) * (molar mass of NaF)
moles of NaF = 0.00735625 moles
molar mass of NaF = 22.99 g/mol (sodium) + 19.00 g/mol (fluorine)
molar mass of NaF = 41.99 g/mol
mass of NaF = (0.00735625 moles) * (41.99 g/mol)
mass of NaF = 0.3083 g
Therefore, you would need approximately 0.3083 grams of solid NaF to prepare a 250 mL acidic buffer with a pH of 2.75, given the initial molarity and volume of HF provided.