Answer to Question #120534 in Inorganic Chemistry for billy

Question #120534
In an experiment, 13.88 g of potassium propionate, C2H5CO2K, were weighed in a beaker and
dissolved in a small amount of distilled water. The mixture is transferred to a 100 mL volumetric
flask, made up to the mark and shaken well. 50.0 mL of this standard solution is mixed with 50.0
mL of a standard solution of propionic acid, C2H5CO2H in order to prepare a buffer solution of
pH 5.66. The molar mass of potassium propionate is 112 g mol-1
and the pKa of propionic acid is
4.87.
(i) What is the concentration of the propionic acid used to make the buffer?
1
Expert's answer
2020-06-08T15:28:11-0400

pH=pKacid-lg(Cacid/Csolt) - a formula of buffer solution pH.

Then Cacid=Csolt*10^(pKacid-pH)

pKacid=4.87

pH=5.66

Csolt=nsolt/V.

nsolt=msolt/Msolt

msolt=13.88g

Msolt=112g*mol-1

The volume of the buffer will be approximately equal to the sum of the volumes of salt and acid solutions, this is true for dilute solutions.

Therefore V=50ml+50mi=100ml=0.1L

And Csolt=13.88g/(112g*mol-1*0.1L)=1.24mol*L-1

So, Cacid=1.24mol*L-1*10^(4.87-5.66)=2.27mol*L-1


Answer:

The concentration of the propionic acid used to make the buffer is 2.27mol*L-1

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