3. A 30.0 mL sample of 0.165 M propanoic acid (HC3H5O2) is titrated with 0.300 M...

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Answer 1

Answer #1

e)

millimoles of propanoic acid = 30 x 0.165 = 4.95

millimoles of KOH = 25 x 0.300 = 7.5

HC3H5O2 + KOH ----------------> C3H5O2- + H2O

4.95 7.5 0 0

0 2.55 4.95

here strong base KOH remains.

[OH-] = 2.55 / 25 + 30 = 0.0464 M

pOH = -log (0.0464) = 1.33

pH = 12.67

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Answer 2

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