Answer to Question #120517 in Inorganic Chemistry for Aartee

Question #120517
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
(i) What is the concentration of the propionic acid used to make the buffer?
Expert's answer

Concentration of "C_2H_5COOK" in "100" ml "=\\frac{13.88}{112\\times 100}" "mol\/l"

Now "50" ml of this solution is taken and then mixed with equal amount of propionic acid"(" say "HA)"

So,total volume of buffer solution will be "100 \\ ml" .

Concentration of "C_2H_5COOK" or "C_2H_5COO^-" "(" Lets say "A^-)" "=\\frac{13.88}{112\\times 100}\\times \\frac{50}{100}=6.2\\times 10^{-4}\\ mol\/l"

Now ,for buffer solution,

"pH=5.66;pK_a=4.87" and

"pH=pK_a+log_e\\frac{[A^-]}{[HA]}\\implies" "5.66=4.87+log_e\\frac{6.2\\times 10^{-4}}{[HA]}"

"\\implies" "log_e\\frac{6.2\\times 10^{-4}}{[HA]}=0.79\\implies\\frac{6.2\\times 10^{-4}}{[HA]}=e^{0.79}=2.2034"

"\\implies [HA]=\\frac{6.2\\times 10^{-4}}{2.2034}=2.81\\times 10^{-4}\\ mol\/l"

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