Question

What is the pH of the solution created by combining 11.40 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)?



Table 1)

mL NaOH pHw/HCl pH w/HC2H3O2
11.40 ______ _______


Table 2)
mL NaOH pHw/HCl pH w/HC2H3O2
11.40 ______ ________



ive done this problem literally 28 times...please help with details i really appreciate it

Answer 1

11.4 mL x 0.1M NaOH = 1.14 mmoles NaOH.
8.00 mL x 0.1M HCl = 0.800 mmoles HCl.
9.00 mL x 0.1M HAc = 0.800 mmoles HAc

...........NaOH + HCl ==> NaCl + H2O
initial..1.14....0.800.....0......0
change..-0.800....-0.800..0.800..0.800
equil....0.34......0......0.800..0.800

Looking at the equilibrium line, NaCl is not hydrolyzed so you have a solution of NaOH. (OH^-) = (NaOH) = mmoles/total mL. Convert to pOH then to pH.

For the acetic acid part,
...........NaOH + HAc ==> NaAc + H2O
initial....1.14...0.800....0......0
change....-0.800..-0.800..0.800..0.800
equil......0.34...0......0.800..0.800

Here the NaAc salt does hydrolyze; however, the OH^- produced by the hydrolysis of the Ac^- is so small it can be neglected so the (OH^-) = (NaOH) = mmoles NaOH/total mL. Convert to pOH then to pH.

I don't know enough about the problem to answer about the dilution part. I'll leave that for you .