Posted by Sara on Friday, October 28, 2011 at 11:04pm.
11.80 mL x 0.1M NaOH = 1.180 mmoles.
8.00 mL x 0.1M HCl = 0.800 mmoles
8.00 mL x 0.1M HAc = 0.800 mmoles.
............HCl + NaOH ==> NaCl + H2O
initial..0.800...1.180.....0.......0
change..-0.800..-0.800.....0.800.0.800
equi.......0.....0.380.....0.800..0.800
So you have 0.380 mmoles NaOH in 11.8+8.00 mL soln and (OH^-) = (NaOH) and convert to pH.
..........HAc + NaOH ==> NaAc + H2O
initial..0.800..1.180.....0......0
change..-0.800..-0.800..0.800..0.800
equil....0......0.380...0.800..0.800
Ignoring the OH^- that might be added due to the hydrolysis of the NaAc salt (which is negligible) pH is determined by the mmoles NaOH/mL.
The second half is all yours. Post your work if you get stuck.
im having trouble with the samee one....
i got 1.72 for the pH and its still wrong
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