Asked by Shameca
The pKa of HEPES is 7.55 at 20 degrees C, and its MW is 238.31. Calculate the amounts of HEPES in grams and of 1.0 M NaOH in milliliters that would be needed to make 300 mL of 0.2 M HEPES buffer at pH 7.2.
Answers
Answered by
DrBob222
If we let HA = HEPES.
HA + NaOH ==> NaA + H2O
Use the Henderson-Hasselbalch equation.
pH = pKa + log (B/A)
7.2 = 7.55 = log (B/A)
Calculate (B)/(A)
Then you know (A) + (B) = 0.2
Using those two equations will allow you to calculate (A) and (B).
I assume you can finish. You want moles A and moles B for the equation. M x L = moles and since you know M and volume (you want 300 mL), you can calculate moles of each, plug those in to the equation at the top to see how much of the 1 M NaOH must be added to form the moles you want. Post your work if you get stuck.
HA + NaOH ==> NaA + H2O
Use the Henderson-Hasselbalch equation.
pH = pKa + log (B/A)
7.2 = 7.55 = log (B/A)
Calculate (B)/(A)
Then you know (A) + (B) = 0.2
Using those two equations will allow you to calculate (A) and (B).
I assume you can finish. You want moles A and moles B for the equation. M x L = moles and since you know M and volume (you want 300 mL), you can calculate moles of each, plug those in to the equation at the top to see how much of the 1 M NaOH must be added to form the moles you want. Post your work if you get stuck.
Answered by
Lee
pH =pka + log [A-]/[HA]
7.2 = 7.55 + log [A-]/[0.2M]
[A-] = 0.4477M
So using that concentration 0.4477M x 300mL = 134.32 moles NaOH / 1M NaOH
= 134.32 mL NaOH amount needed right?
7.2 = 7.55 + log [A-]/[0.2M]
[A-] = 0.4477M
So using that concentration 0.4477M x 300mL = 134.32 moles NaOH / 1M NaOH
= 134.32 mL NaOH amount needed right?
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