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
4.87.
(i) What is the concentration of the propionic acid used to make the buffer?
1
Expert's answer
2020-06-08T15:31:09-0400

Concentration of C2H5COOKC_2H_5COOK in 100100 ml =13.88112×100=\frac{13.88}{112\times 100} mol/lmol/l

Now 5050 ml of this solution is taken and then mixed with equal amount of propionic acid(( say HA)HA)

So,total volume of buffer solution will be 100 ml100 \ ml .

Concentration of C2H5COOKC_2H_5COOK or C2H5COOC_2H_5COO^- (( Lets say A)A^-) =13.88112×100×50100=6.2×104 mol/l=\frac{13.88}{112\times 100}\times \frac{50}{100}=6.2\times 10^{-4}\ mol/l

Now ,for buffer solution,

pH=5.66;pKa=4.87pH=5.66;pK_a=4.87 and

pH=pKa+loge[A][HA]    pH=pK_a+log_e\frac{[A^-]}{[HA]}\implies 5.66=4.87+loge6.2×104[HA]5.66=4.87+log_e\frac{6.2\times 10^{-4}}{[HA]}

    \implies loge6.2×104[HA]=0.79    6.2×104[HA]=e0.79=2.2034log_e\frac{6.2\times 10^{-4}}{[HA]}=0.79\implies\frac{6.2\times 10^{-4}}{[HA]}=e^{0.79}=2.2034

    [HA]=6.2×1042.2034=2.81×104 mol/l\implies [HA]=\frac{6.2\times 10^{-4}}{2.2034}=2.81\times 10^{-4}\ mol/l



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