Concentration of C2H5COOK in 100 ml =112×10013.88 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 C2H5COOK or C2H5COO− ( Lets say A−) =112×10013.88×10050=6.2×10−4 mol/l
Now ,for buffer solution,
pH=5.66;pKa=4.87 and
pH=pKa+loge[HA][A−]⟹ 5.66=4.87+loge[HA]6.2×10−4
⟹ loge[HA]6.2×10−4=0.79⟹[HA]6.2×10−4=e0.79=2.2034
⟹[HA]=2.20346.2×10−4=2.81×10−4 mol/l
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