# Chemistry - acids & bases

4. Calculate the pH of each of the solutions and the change in pH to 0.01 pH units caused by adding 10.0 mL of 2.37-M HCl to 320. mL of each of the following solutions.

Change is defined as final minus initial, so if the pH drops upon mixing the change is negative.

c) 0.115 M HC2H3O2
pH before mixing =
pH after mixing=
pH change =

My work:

Before Mixing:
HC2H3O2 + H2O --> (C2H3O2)- + (H3O)+
36.8 mmol 0 0
- x + x + x

1.8e-5 = x^2 / (36.8 -x)
x = .02572813 = [H3O+]

-log(.02572813) = 1.59 (this is not correct)

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1. Nope. You used millimols in this and you should use molarity. Ka is 1.8E-5, yes, but that is Ka = (H3O^+)(Ac^-)/(HAc) and you must substitute M and not mmols. If acetic acid is HAc, then
.......HAc+ + H2O ==> Ac^- + H3O^+
I......0.15..........0......0
C......-x.............x......x
E.....0.115-x.........x.......x

This gives you (H3O^+) = ? in mols/L which is concentration and you can use that to calculate pH. .

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2. Thank you so much! That makes sense.

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posted by Ellen
3. I'm sure you picked it up but that's 0.115 M from the problem and not 0.15. My typo.

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