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Calculate the volume, in milliliters, of a 0.220 M KOH solution that should be added to 5.750 g of HEPES (MW = 238.306 g/mol, p𝐾a = 7.56) to give a pH of 7.98.

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  1. I haven't used as many significant figures as I should so you should recalculate all of this from scratch.
    mols HEPES = 5.75 h/238.3 = about 0.024. The rxn is
    mols acid(HEPES) + OH^- = moles base(salt) + H2O

    Solve for base needed.
    ...................HEPES acid + OH^- ==> HEPES base(salt) + H2O
    I....................0.024................0...............0.....................................
    add.........................................x....................................
    C......................-x...................-x................x
    E.................0.024-x.................0...............x
    Plug the E line into the Henderson-Hasselbalch equation like this.
    pH = pKa + log [(base)/(acid)]
    7.98 = 7.56 + log (b/a)
    7.98 = 7.56 + log (x/0.024-x)
    Solve for x = mols base
    Solve for 0.24-x = moles acid
    moles base = M x L. You know M and moles KOH, solve for L and convert to mL.
    Post your work if you get stuck.
    Note: Technically, one the HH equation uses CONCENRATION of acid and base and I've used moles instead. Technically, that is wrong BUT molarity = M = mols/L and when I use mols I can get away with that BECAUSE the volume is the same for the acid and the base because its the same solution. Volume in L cancels and moles is left. Your teacher may tell you to use volume so you can always write
    7.58 = 7.56 + log [(x mols/L)/(0.024-x)/L)] and cancel the L.

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  2. Thank you so much for your help!!

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