How many milliliters of each solution of 0.50 M KH2PO4 and 0.50 M K₂HPO4 are needed to make 100.0 mL solution of phosphate buffer with pH = 7.50? What are the final concentrations of K+, H2PO4 and HPO42, in the buffer solution? (for H2PO4, K_{a} = 6.2 * 10 ^ - 8 )
To make a phosphate buffer with pH 7.5, we need to use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
where Ka is the acid dissociation constant for H2PO4, [A-] is the concentration of HPO42-, and [HA] is the concentration of H2PO4. We can rearrange this equation to solve for the ratio of [A-] to [HA]:
[A-]/[HA] = 10^(pH - pKa)
In this case, pKa = -log(Ka) = -log(6.2 * 10^-8) = 7.21, so
[A-]/[HA] = 10^(7.5 - 7.21) = 2.14
We want to make a 100 mL solution of buffer, so we can choose any total concentration we like as long as the ratio of [A-] to [HA] is 2.14. Let's choose a total concentration of 0.1 M. Then we can set up two equations for the two unknowns, the volumes of the two stock solutions (let's call them V1 and V2):
V1 + V2 = 100.0 mL
(0.5 M)(V1) / (V1 + V2) + (0.5 M)(V2) / (V1 + V2)(2.14) = [H2PO4]
where [H2PO4] is the final concentration of H2PO4 in the buffer solution. (Note that we can use the same total volume in both equations because the ratio of [A-] to [HA] is fixed.)
Simplifying the second equation, we get:
(0.5 M)(2.14)(V1) + (0.5 M)(V2) = [H2PO4](V1 + V2)
0.5 M(2.14V1 + V2) = [H2PO4](V1 + V2)
We can do a similar calculation for the final concentration of HPO42-, using the fact that [HPO42-] = 2.14[H2PO4]. The final concentration of K+ is just the sum of the concentrations of K+ from the two stock solutions.
We can use the Henderson-Hasselbalch equation again to check if our solution is indeed a buffer with pH 7.5. Using the final concentrations of H2PO4 and HPO42-, we get:
pH = pKa + log([A-]/[HA])
pH = 7.21 + log(2.14)
pH = 7.5 (as desired)
For the final concentrations, we have:
[H2PO4] = 0.048 M
[HPO42-] = 0.103 M
[K+] = 0.5 M
To solve for the volumes of the two stock solutions, we can rearrange the first equation to get V2 = 100.0 mL - V1, and substitute into the second equation:
0.5 M(2.14V1 + (100.0 mL - V1)) = 0.048 M(V1 + (100.0 mL - V1))
1.07V1 + 53.5 = 4.8 + 0.048V1
1.022V1 = 48.7
V1 = 47.6 mL
V2 = 100.0 mL - V1 = 52.4 mL
Therefore, we need 47.6 mL of the 0.50 M KH2PO4 solution and 52.4 mL of the 0.50 M K2HPO4 solution to make 100.0 mL of a phosphate buffer with pH 7.5.
Hey Bot===a small math error.
[A-]/[HA] = 10^(7.5 - 7.21) = 2.14 should be 1.95 but that won't change the volume much.
Thank you for catching that mistake! You're right, the correct value for the ratio of [A-] to [HA] is 1.95, not 2.14. But as you mentioned, the volume calculation should not be affected much by this correction.
To determine the volumes of each solution needed, we can use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
In this case, A- is the concentration of HPO42-, and HA is the concentration of H2PO4-. Let's go through the calculation step by step.
Step 1: Calculate pKa
The pKa can be calculated using the given Ka value. We can use the formula pKa = -log(Ka).
pKa = -log(6.2 * 10^-8)
≈ 7.21
Step 2: Calculate the ratio of [A-] to [HA]
Using the Henderson-Hasselbalch equation, we have:
7.50 = 7.21 + log([A-]/[HA])
Rearranging the equation gives:
0.29 = log([A-]/[HA])
Converting the logarithmic form to exponential form, we have:
[A-]/[HA] = 10^0.29
≈ 1.995
Step 3: Determine the volumes
Let x be the volume of 0.50 M KH2PO4 (HA) needed, and (100 - x) be the volume of 0.50 M K2HPO4 (A-) needed.
Since the concentrations of both solutions are the same, we can set up the equation:
(0.50 M)(x) / (0.50 M)(100 - x) = 1.995
Simplifying, we have:
x / (100 - x) = 1.995
Solving for x, we get:
x = (1.995 * 100) / (1 + 1.995)
≈ 66.448 mL
Therefore, you would need approximately 66.448 mL of 0.50 M KH2PO4 and approximately 33.552 mL of 0.50 M K2HPO4 to make a 100.0 mL phosphate buffer solution with a pH of 7.50.
Step 4: Calculate the final concentrations
To calculate the final concentrations of K+, H2PO4-, and HPO42-, we need to consider the volumes and concentrations of the solutions used.
For K+:
The concentration of K+ in both solutions is 0.50 M. Therefore, the final concentration of K+ in the buffer solution will also be 0.50 M.
For H2PO4-:
The concentration of H2PO4- in the 0.50 M KH2PO4 solution is 0.50 M. The volume of the KH2PO4 solution used is 66.448 mL. Therefore, the amount of H2PO4- added to the buffer solution is:
(0.50 M)(66.448 mL) = 33.224 mmol
Since the total volume of the buffer solution is 100.0 mL, the final concentration of H2PO4- is:
(33.224 mmol) / (100.0 mL) = 0.332 M
For HPO42-:
The concentration of HPO42- in the 0.50 M K2HPO4 solution is 0.50 M. The volume of the K2HPO4 solution used is (100 - 66.448) mL = 33.552 mL. Therefore, the amount of HPO42- added to the buffer solution is:
(0.50 M)(33.552 mL) = 16.776 mmol
Since the total volume of the buffer solution is 100.0 mL, the final concentration of HPO42- is:
(16.776 mmol) / (100.0 mL) = 0.168 M
To summarize, the final concentrations of the species in the buffer solution are approximately:
K+: 0.50 M
H2PO4-: 0.332 M
HPO42-: 0.168 M