In: Chemistry
Part 1
Calculate the amount (in millimoles) of protonated and unprotonated HEPES in 25 mL of 50 mM HEPES at pH 7.7 (the pKa of HEPES is 7.47)
Part 2
To adjust the pH of the HEPES solution in problem 7, 1mL of 50mM NaOH was added. Calculate the amount (in millimoles) of protonated and unprotonated HEPES. What is the pH after the addition of NaOH?
Hi, look at the structure of HEPES
Since, we are given pKa 7.47, nitrogen is not protonated at pH values around it. We are talking about the deprotonation of sulphonic acid. Now we have gone through the basics. Consider the protonated form HL and deprotonated form L-. Consider equilibrium:
, we will use the Henderson-Hasselbalch equation to solve it
Now total millimoles are 0.05 M x 25 mL = 1.25 mmol
we can write , solving these two equations will give us
Now coming to the part two, 1 mL of 50 mM NaOH has the following millimoles of NaOH
1 mL x 0.05 M = 0.05 mmol
since the base will react with the protonated form, we will subtract these millimoles from HL and add to the L-.
Now we can put these values in the Henderson-Hasselbalch equation again (Ignore the volume change, since it is the same for both forms so we don't need to include it in the equation)
so we get the pH value 7.78. If you need any other information, please do not hesitate to ask.