A 100ml sample of 0.250 M propionic acid is titrated with a solution of 0.250 M

Question

A 100ml sample of 0.250 M propionic acid is titrated with a solution of 0.250 M potassium hydroxide. Ka= 1.3*10^-5 for propionic acid. Calculate the ph at the equivalence point (when just enough potassium hydroxide has been added to completely neutralize the propionic acid.) A 100ml sample of 0.250 M propionic acid is titrated with a solution of 0.250 M potassium hydroxide. Ka= 1.3*10^-5 for propionic acid. Calculate the ph at the equivalence point (when just enough potassium hydroxide has been added to completely neutralize the propionic acid.)

Explanation / Answer

propanoic acid millimoles = 100 x 0.250 = 25

pottasium hydroxide millimoles = 0.250 x 100 = 25

volume of KOH at equivalence point = 100 mL

Ka = 1.3 x 10^-5

propanoic acid + KOH ---------------> potassium propanoate + H2O

25 25 0 0

0 0 25 25

at equivalence point salt only remained.

[salt] concentration = millimoles of salt / total volume

= 25 / (100 + 100)

= 0.125 M

it is salt of strong base and weak acid so.

pH = 7 + 1/2 (Pka + log C)

= 7 + 1/2 (4.89 + log 0.125)

= 9.04

pH = 9.04

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