How many millilitres of a 0.1 M solution of NaF and of 0.1 M solution of HF
must be mixed to obtain 1L of buffer at pH = 4? Ka = 6.7 × 10-4
Currently I am quite confident with using the henderson and Hasselbalch equation but this question has left me stumped...
Any help with the method of working this out would be greatly appreciated...
To solve this problem, we can use the Henderson-Hasselbalch equation and the principles of buffer solutions.
The Henderson-Hasselbalch equation for an acidic buffer is:
pH = pKa + log([A-]/[HA])
In this case, the acid is HF and the conjugate base is F-. The pKa value is given as 6.7 × 10-4.
First, we need to calculate the ratio of [A-] to [HA] required to achieve pH = 4.
Rearranging the Henderson-Hasselbalch equation:
log([A-]/[HA]) = pH - pKa
log([A-]/[HA]) = 4 - (-log(6.7 × 10-4))
log([A-]/[HA]) = 4 + 3.18
log([A-]/[HA]) = 7.18
Now, we can take the antilog of both sides to obtain the ratio:
[A-]/[HA] = 10^7.18
Next, we can express the ratio in terms of the concentrations:
[C(F-)] / [C(HF)] = 10^7.18
Now, we can substitute the molarities:
[C(F-)] / 0.1 = 10^7.18
Solving for [C(F-)]:
[C(F-)] = (10^7.18) * 0.1
Next, we want to find the total volume of the buffer solution. We are adding 1L of solution, so the final volume will be 1L.
Now, let's assume V1 denotes the volume (in mL) of the 0.1 M NaF solution and V2 denotes the volume (in mL) of the 0.1 M HF solution.
We have V1 + V2 = 1000 mL (1L)
Now, we can calculate the concentrations of F- and HF based on the given volumes:
[C(F-)] = (V1/1000) * 0.1
Then, we can substitute the value of [C(F-)] in the previous equation:
(C(F-)) = (10^7.18) * 0.1
Now, we can substitute the value of [C(F-)] and (C(F-)) in the equation:
(C(F-)) = (10^7.18) * 0.1
(V1/1000) * 0.1 = (10^7.18) * 0.1
(V1/1000) = 10^7.18
Now, solve for V1:
V1 = 1000 * (10^7.18)
V1 ≈ 1.85 × 10^7 mL
Therefore, approximately 1.85 × 10^7 mL of the 0.1 M NaF solution and the rest of the volume (approximately 1000 mL - 1.85 × 10^7 mL) of the 0.1 M HF solution must be mixed to obtain 1L of buffer at pH = 4.
To solve this problem, we can use the Henderson-Hasselbalch equation for calculating the pH of a buffer solution:
pH = pKa + log([A-]/[HA])
Where pH is the desired pH of the buffer, pKa is the negative logarithm of the acid dissociation constant (Ka), [A-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid.
In this case, we want a buffer with a pH of 4, so we can substitute the values into the equation:
4 = -log(Ka) + log([A-]/[HA])
We are given that Ka = 6.7 x 10^(-4), so we can plug that into the equation:
4 = -log(6.7 x 10^(-4)) + log([A-]/[HA])
Simplifying,
4 = 3.18 + log([A-]/[HA])
Subtracting 3.18 from both sides,
0.82 = log([A-]/[HA])
Now, we need to rearrange the equation to solve for the ratio of [A-]/[HA]:
10^0.82 = [A-]/[HA]
Using a calculator,
6.92 = [A-]/[HA]
At this point, we can assign a variable to one of the concentrations, such as [HA], and solve for the other concentration in terms of this variable. Let's choose [HA].
Let's say we have x milliliters of the 0.1 M NaF solution and (1000 - x) milliliters of the 0.1 M HF solution to obtain a total volume of 1L (1000 mL).
The concentration of NaF is 0.1 moles/L, so the moles of NaF that we have are 0.1 * (x/1000) = 0.0001x moles.
Similarly, the concentration of HF is 0.1 moles/L, so the moles of HF that we have are 0.1 * ((1000 - x)/1000) = 0.1 - 0.0001x moles.
Since the ratio of [A-]/[HA] is 6.92, we can set up the equation:
0.0001x / (0.1 - 0.0001x) = 6.92
Simplifying,
0.0001x = 6.92(0.1 - 0.0001x)
0.0001x = 0.692 - 0.000692x
Combining like terms,
0.000792x = 0.692
Solving for x,
x = 0.692 / 0.000792
x ≈ 874.25 mL
Therefore, you would need approximately 874.25 mL of the 0.1 M NaF solution and (1000 - 874.25) mL ≈ 125.75 mL of the 0.1 M HF solution to obtain 1L of buffer at pH 4.
How much buffer do you want? The problem doesn't say; if you can choose a volume, say choose 500 mL. Then let x = mL base and 500-x = mL acid, then in the HH equation it becomes
4.0 = pKa + log (0.1*x)/[(500-x)*0.1] and solve for x = volume of the NaF and 500-x is volume of HF. Should work ok; if 500 is not what you want substitute another number for that.