Answer to Question #120534 in Inorganic Chemistry for billy

Question

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?

Answer 1

pH=pK_acid-lg(C_acid/C_solt) - a formula of buffer solution pH.

Then C_acid=C_solt*10^(pK_acid-pH)

pK_acid=4.87

pH=5.66

C_solt=n_solt/V.

n_solt=m_solt/M_solt

m_solt=13.88g

M_solt=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 C_solt=13.88g/(112g*mol^-1*0.1L)=1.24mol*L^-1

So, C_acid=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