Posted by katie on Tuesday, July 24, 2012 at 10:40pm.
initial mols HBr = M x L = about 0.014
mols NaOH added = about 0.002
but you need to do them more accurately.
........NaOH + HBr ==> NaBr + H2O
I.........0....0.014....0.......0
added....0.002.....................
C.......-0.002..-0.002..0.002..0.002
E.........0.....0.0120...0.002.0.002
So what you have at 10.0 mL is a solution of HBr which is a strong acid (100% ionized); therefore, (H^+) = (HBr) so pH = -log(HBr)
Note: (HBr) = mols/L = 0.012/(0.010+0.020) = ?
Thank you so much that cleared up a lot!
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