Using the HH equation.
pH = pKa + log (base/acid) . I chose pH of 7.45 because I want to minimize the amount of Hepes I have to use so making the pH higher will mean less HEPES would have had to be added originally..
plug in pH of 7.45 and pka of 7.55 and solve for the base/acid ratio
Then (acid) = (base)/0.794.
Then HEPES + NaOH ==> NaHEPES + H2O
Since we want the NaHEPES to be 0.01M, using the ratio we get .01/.794 = .01259 M. 0.01259 + .01 = moles of HEPES = .02259 mol. .02559 mol *238.29892 g/mol = 5.38 g HEPES.
Volume of NaOH = .01259 mol / 5.695 mol/L = 2.48 mL.
Can Dr.Bob 222 or anyone confirm if this is what you got? This is how I tried to solve it.
I am sorry to be so late answering this. I hope you see it; it is on page 2 now and will move further from the front as time marches. Post a note on page 1 to that effect so I will know this was not in vain.:-)
Using the HH equation.
pH = pKa + log (base/acid) . I chose pH of 7.45 because I want to minimize the amount of Hepes I have to use so making the pH higher will mean less HEPES would have had to be added originally..
plug in pH of 7.45 and pka of 7.55 and solve for the base/acid ratio
Then (acid) = (base)/0.794.
Then HEPES + NaOH ==> NaHEPES + H2O
Since we want the NaHEPES to be 0.01M, using the ratio we get .01/.794 = .01259 M. (This is in 500 mL of solution; therefore, you want to divide this number by 2. So base is 0.005 mols, acid is 0.005 mols + 1/2 of the number above, NaOH is 0.005 mols. Check it out but you should get pH of 7.45 with these numbers. You can't add molarities; you can add mols.) 0.01259 + .01 = moles of HEPES = .02259 mol. .02559 mol *238.29892 g/mol = 5.38 g HEPES.
Volume of NaOH = .01259 mol / 5.695 mol/L = 2.48 mL.
You will need to redo the calculations but I think this should give you the minimum amount of HEPES necessary for a pH of 7.45. I want to make you aware, however, that I ran into trouble when I tried to add some hydrogen ion, (on paper). I found that if I added as much as 0.001 mols of H+ (that would be 100 mL of the enzyme solution), the pH of this buffer changed to some number outside the limits. I found using 10 mL of the enzyme solution would stay within the desired range. So in theory, this probably is the MINIMUM amount but in practice we might need to edge it up a little. I don't know how much enzyme is used at one time but the capacity of this buffer solution is limited.I hope this helps. I will post a note on page 1 addressed to you in the hopes that it will direct you here.
Based on the given information, here's how you can solve the problem:
1. Start with the Henderson-Hasselbalch equation: pH = pKa + log(base/acid)
2. Plug in the given pH of 7.45 and pKa of 7.55 into the equation.
3. Solve for the base/acid ratio: (base/acid) = 10^(pH - pKa) = 10^(7.45 - 7.55) = 0.794
4. Calculate the acid concentration using the base/acid ratio: (acid) = (base)/0.794
5. The desired concentration of NaHEPES is 0.01 M. Use the ratio to calculate the base concentration: (base) = (acid) * 0.01/0.794
6. Convert the moles of base to grams of HEPES using its molar mass (238.29892 g/mol): grams of HEPES = moles of HEPES * molar mass
7. Calculate the volume of NaOH needed to obtain the desired base concentration: Volume of NaOH = (base) / molarity of NaOH
Note that in step 5, you mentioned using the ratio to calculate the NaHEPES concentration. Make sure to divide the result by 2 since you're working with a 500 mL solution.
Keep in mind that the calculations mentioned here are based on the given information and equations. Please double-check your calculations to ensure accuracy. Additionally, if you are unsure or need confirmation, it is always a good idea to consult with an expert in the field or your mentor.