Posted by megan on Thursday, July 26, 2012 at 12:58pm.
The three ionization constants are so close that you can't titrate them separately. Since we have 50 mL of 0.25M, that is 12.5 millimoles citric acid (which I will call H3C). Therefore, it will take 12.5 mmoles to neutralize the first H of H3C, another 12.5 to neutralize the second H of H3C and that leaves 12.5 mmols HC^2- for the buffer. pKa3 = 6.4
Plug into the Henderson-Hasselbalch equation and solve for base.
........HC^= + OH^- ==> C^3- + HOH
initial.12.5....0........0.......0
add.............x.................
change...-x....-x........x......x
equl....12.5-x..0.........x
Solve for x, which is the amount of NaOH that must be added to form the desired pH, then add 25 mmols (from neutralization of the two other H ions) for the total NaOH needed. Convert to mols. I obtained approximately 33 millimoles NaOH.
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