Complete the table below: Note: Make simplifying assumptions, do not use the quadratic formula.

Question

What is the pH of the solution created by combining 11.80 mL of the 0.10 M NaOH(aq) with 8.00 mL of the 0.10 M HCl(aq)?
with 8.00 mL of the 0.10 M HC2H3O2(aq)?

Complete the table below: Note: Make simplifying assumptions, do not use the quadratic formula.
What are the pH values if you take into account that the 8.00 mL of 0.10 M Acid was first diluted with 100 mL of water (like it will be in the experiment you perform in lab)?

Answer 1

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.

Answer 2

im having trouble with the samee one....

i got 1.72 for the pH and its still wrong