[jorgelgalvan93]

YOU ARE PREPARING 1L OF 0.1M HEPES BUFFER AND WOULD LIKE THE FINAL PH TO BE 7.4. YOU BEGIN WITH A 0.5M HEPES SOLUTION AT PH 6.6 AND ACCESS TO SOLUTIONS OF HCL AND NAOH AT 0.5M CONCENTRATIONS EACH. DESCRIBE HOW YOU MAKE THE BUFFER. PKA1 = 3, PKA2 = 7.55

Ok, so my approach to this problem:

Hepes ⁺ Hepes + H⁺ Hepes^- + H⁺

I first find the volume of HEPES in the final buffer solution... M1V1 = M2V2... V2 = M1(V1) / M2... V2 = (0.1M)(1L) / (0.5M) = 0.2L = 200mL of HEPES 0.5M

Now Im unsure here as to how to apply HH equation.

pH = pka + log ( HEPES^- / HEPES )

Since my final pH is 7.4, which is below PKA2, does that mean I have to use PK1? How can I figure out how much 0.5M NaOH I need to add to get the right pH? Do I use "ICE" method? Do I use the ratios of ( HEPES^- / HEPES ) ?

Any guidance is greatly appreciated!

Thank you!

[Borek]

You have to use pKa which is the closest to the target pH.

General approach would call for just finding the ratio of deprotonated HEPES and HEPES in the solution from the HH equation, and solving the system of simultaneous equations. See examples here: http://www.chembuddy.com/?left=buffers&right=composition-calculation

Please note however, that pH of a weak acid solution (and HEPES definitely qualifies) should have acidic pH. Your starting solution is almost neutral, so there is apparently more to the problem.

[Babcock_Hall]

If I simply took HEPES and put it into water, the pH will be lower than the pH given in the problem.  The problem could be describing a situation where someone put HEPES into water and added some NaOH.  The calculation would then be describing how much more NaOH would need to be added.  That makes it a little bit more difficult; it seems to me that you will need to apply the Henderson-Hasselbalch equation to obtain the initial concentrations.