This question pertaining to buffer systems!
How would you prepare 10.0mL of 0.0100M phosphate buffer, pH = 7.4
0, from stock solution of 0.100M KH2PO4 and 0.250M K2HPO4, pKa of KH2PO4 is 7.20.
Please help me in figuring out how I even start this problem...as well as attempting to finish it...
pH = pK2 + log b/a
7.4 = 7.2 + log b/a
log b/a = 1.58. You're right, I typed in the wrong numbers in my calculator. I typed in 7.2= 7.4 + log b/a. It helps to use the right numbers doesn't it.
OK. b/a = 1.58 or b = 1.58a
then a + b = 0.1
a + 1.58a = 0.1 and I suspect you can take from there to find a and b (and those will be in mmoles).
From there you will know mmols a, mmols b, and the stock solutions are
M = mmols/mL. You will know mmols a and M of that stock, substitute and solve for mL.
I like to do these in millimoles. That way I don't get bugged by the zeros in front.
pH = pK2 + log (base)/(acid)
mmols you want = 10 x 0.01 = 0.1.
So equation 1 is a + b = 0.1
Equation 2 is
7.4 = 7.2 + log (b)/(a)
Solve for b/a and rearrange to
b = ?*a (whatever that ? number is). I think that should be close to 5 = b/a or a = 5b.
Solve the two equations simultaneously for a and b.
When you know how many millimols a and mmols b you need, calculate mL you need of each of the stock solutions to give you that man mmols of each. Piece a cake. Post your work if you get stuck.
I was able to do Equation 2 and got 1.58 for b/a ...
I was able to solve it! Thank you again Dr.Bob!
This is the 2nd time you've saved me this semester!
I greatly appreciate it!
:D
That's great. Glad to help. Remember to look to see the METHOD so you will remember it.
To prepare a phosphate buffer with a specific pH, you need to find the appropriate ratio of KH2PO4 (acidic component) and K2HPO4 (basic component) to achieve the desired pH.
Here's how you can start solving this problem:
Step 1: Calculate the ratio of KH2PO4 and K2HPO4 required to prepare a buffer with pH 7.4 using the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
Given:
pH = 7.4
pKa = 7.20
Let's assume the total volume of the buffer solution is V. The concentration of KH2PO4 and K2HPO4 in the final buffer will be [HA] and [A-] respectively.
Since you want to prepare 10.0 mL of buffer solution, let V = 10.0 mL.
Now, let's substitute the given values into the Henderson-Hasselbalch equation:
7.4 = 7.20 + log ([A-]/[HA])
Step 2: Convert the pH to [A-]/[HA] using the logarithmic properties:
[A-]/[HA] = 10^(pH - pKa)
[A-]/[HA] = 10^(7.4 - 7.20)
[A-]/[HA] = 10^0.2
[A-]/[HA] = 1.585
This means that the ratio of [A-] to [HA] should be approximately 1.585.
Step 3: Calculate the concentrations of KH2PO4 and K2HPO4 required to make the desired buffer:
Let's assume the concentration of KH2PO4 in the final buffer solution is x M. This means the concentration of K2HPO4 should be (1.585 * x) M, as per the ratio obtained.
Now, we need to determine the specific values for x (concentration of KH2PO4) and (1.585 * x) (concentration of K2HPO4).
Step 4: Use the dilution formula to calculate the amounts of the stock solutions to mix:
We know that concentration (C1) × volume (V1) = concentration (C2) × volume (V2).
Given:
C1 (concentration of KH2PO4 stock solution) = 0.100 M
V1 (volume of KH2PO4 stock solution) = ? (to be determined)
C2 (concentration of KH2PO4 in the final buffer) = x M
V2 (total volume of the final buffer solution) = 10.0 mL
Using the dilution formula, we can solve for V1:
C1 × V1 = C2 × V2
0.100 M × V1 = x M × 10.0 mL
V1 = (x M × 10.0 mL) / 0.100 M
Similarly, let's calculate the volume of the K2HPO4 stock solution (V3):
Given:
C3 (concentration of K2HPO4 stock solution) = 0.250 M
V3 (volume of K2HPO4 stock solution) = ? (to be determined)
C4 (concentration of K2HPO4 in the final buffer) = (1.585 × x) M
V4 (total volume of the final buffer solution) = 10.0 mL
Using the dilution formula:
C3 × V3 = C4 × V4
0.250 M × V3 = (1.585 × x) M × 10.0 mL
V3 = ((1.585 × x) M × 10.0 mL) / 0.250 M
By solving these equations, you can find the volumes of the stock solutions required to prepare the desired buffer solution.