Posted by guess who on Saturday, June 2, 2012 at 7:29pm.
I couldn't guess.
How many millimoles acetic acid (HAc) do you have? That's 100 mL x 0.12M = 12.
How much NaOH must be added? That's 0.1 M * x mL = 0.1x
.........HAc + OH^- ==> Ac^- + H2O
initial..12.....0........0.......0
add............0.1x...............
change.-0.1x..-0.1x......0.1x....0.1x
equil..12-0.1x..0.......0.1x.....0.1x
pH = pKa + log(base)/(acid)
4.50 = pKa + log(0.1x)/(12-0.1x)
Solve for x = mL 0.1M NaOH to be added to form buffer of pH = 4.50.
I will leave the last two parts for you; the last one is strictly a strong base problem. The other one follows this same kind of procedure using the Henderson-Hasselbalch equation.
Post your work if you get stuck.
uno for the pKa would it simply be 4.50 as well?
No. It's the pKa for acetic acid. Ka = 1.8E-5. pKa = -log Ka.
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