1) Determine the pH before
The acetic acid will dissociate as
CH3COOH --> CH3COO- + H+
Inital 0.5 0 0
change -x x x
Equilibrium 0.5-x x x
So Ka = [H+] [CH3COO-] / [CH3COOH]
1.8 X 10^-5 = x2 / (0.5-x)
As Ka is very small we can ignore x in denominator
1.8 X 10^-5 = x2 / 0.5
0.9 X 10^-5 = x2
x = 3 X 10^-3
[H+] = x = 3 X 10^-3
pH = -log [H+] = 2.52
2) after the addition of 30.00mL OH-
Moles of OH- added = molarity X volume = 0.5 X 30 = 15 millimoles
moles of acid present = molarity X volume = 0.5 X 20 = 10 millimoles
Base will neutralize all of the acid
so moles of base used for neutralization = 10 millmoles
so moles of base left = 5 millimoles
so concentration of base left = millimoles / total volume = 5 / 50 = 0.1 M
we know that
pOH = -log[OH-] = -log [0.1]
pOH = 1
so pH = 14- 1 = 13
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