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You are given 50 mL of .25 M citric acid. How many moles of NaOH would you need to add to obtain a soln with a pH of 6.7? Assume volume unchanged

Ka1=7.4E-4
Ka2=1.7E-5
Ka3=4E-7

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    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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