Posted by Jenna on Monday, May 7, 2012 at 1:07pm.
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?
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 = ?
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