A professor titrated 0.47 g of an unknown monoprotic weak acid with 0.0973 M NaO

Question

A professor titrated 0.47 g of an unknown monoprotic weak acid with 0.0973 M NaOH.

The titration data is below.

a) Calculate # of moles of NaOH required to reach the equivalence point (endpoint).

b) Determine # of moles of acid titrated.

c) Calculate Molar mass of unknown acid.

d) determine the pKa of the unknown acid.

e) Calculate the Ka of the unknown acid.

Volume NaOH (mL) pH 0.00 2.73 2.00 3.30 4.00 3.64 6.00 3.86 8.00 4.03 10.00 4.18 12.00 4.32 14.00 4.46 16.00 4.60 18.00 4.76 20.00 4.94 22.00 5.18 23.00 5.35 23.50 5.47 24.00 5.60 24.50 5.81 25.00 6.18 25.50 10.30 26.00 11.02 27.00 11.44

Explanation / Answer

at equivalence point

there is a sudden jump in pH

so

25 is the equivalence point

a)

we know that

moles = conc x volume (L)

so

moles of NaOH required = 0.0973 x 25 x 10-3 = 2.4325 x 10-3

b)

now

at equivalence point

moles of base added = moles of acid present

so

moles of acid = 2.4325 x 10-3

c)

now

moles = mass / molar mass

so

2.4325 x 10-3 = 0.47 / molar mass

molar mass = 193.2

so

the molar mass of the acid is 193.2

d)

consider 20 ml of NaOH added

moles of NaoH added = 0.0973 x 20 x 10-3 = 1.946 x 10-3

now

the reaction is

HA + NaOH ---> NaA + H20

moles of HA reacted = moles of NaoH added = 1.946 x 10-3

moles of NaA formed = moles of NaoH added = 1.946 x 10-3

so

finally

moles of HA = 2.4325 x 10-3 - 1.946 x 10-3 = 0.4865 x 10-3

moles of NaA = 1.946 x 10-3

now

pH = pKa + log [NaA/ HA]

4.94 = pKa + log [ 1.946 x 10-3 / 0.4865 x 10-3 ]

pKa = 4.338

so

the pKa is 4.338

e)

we know that

pKa = -log Ka

so

4.338 = - log Ka

Ka = 4.59 x 10-5

so

Ka for the acid is 4.59 x 10-5

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