Question

Complete the table below: Note: Make simplifying assumptions, do not use the quadratic formula.
What is the pH of the solution created by combining 1.90 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)?

mL NaOH: 1.90
pH w/HCl: 1.21
pH w/HC2H3O2:



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?
mL NaOH: 1.90
pH w/HCl:
pH w/HC2H3O2:



Okay. I found the pH for the first HCl, but I'm not sure on how to do the rest of them. help?

Answer 1

Okay I got the second pH w/HCl and it is 2.23. but still can't find the pH of both of the ones w/HC2H3O2. Please help?

Answer 2

1.9 mL x 0.1M NaOH = 0.19 millimols NaOH
8.00 mL x 0.1M HAc = 0.8 mmols HAc.

.........HAc + NaOH ==> NaAc + H2O
I........0.....0.19.........0......0
added...0.8........................
C......-0.19...-0.19....0.19....0.19
E......0.61......0.......0.19....0.19

(HAc) = 0.61mmols/9.9 mL = ?M (9.9 is total volume from 8.00+1.9 = 9.9 total)
(NaAc) = (Ac^-) = 0.19 mmols/9.9 mL = ?
Then pH = pKa + log(base)/(acid)

For the diluted part,
(HAc) = 0.61 mmols/101.9 mL = ?
(Ac^-) = 0.19 mmol/101.9 mL = ?