Answer to Question #120517 in Inorganic Chemistry for Aartee

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

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

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

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

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