A solution of formic acid has a pH of 2.70. Calculate the initial concentration of formic acid in this solution. Ka=1.8×10−4 for formic acid

Answer:- [HC(=O)OH]=0.0241⋅mol⋅L−1INITIALLY

Explanation:

We talk the equilibrium

HC(=O) OH (aq)+H2O(l)⇌HCO−2+H3O+

And at equilibrium…Ka=[HCO−2] [H3O+][HC(=O)OH(aq)]=1.80×10−4

But pH=−log10[H3O+] =2.70…

And so [H3O+] =[HCO−2] =10−2.70⋅mol⋅L−1=

And so, AT EQUILIBRIUM…1.80×10−4={10−2.70}2[HC(=O) OH (aq)]

[HC(=O) OH (aq)]at equilibrium={10−2.70}21.80×10−4

≡0.0221⋅mol⋅L−1…but this is the symmetry value….and [H3O+] =1.995×10−3⋅mol⋅L−1 by definition…

And since the hydronium ion is PRESUMED to derive from the formic acid… [HC(=O)OH]initially=(0.0221+1.995×10−3)⋅mol⋅L−1=0.0241⋅mol⋅L−1…the degree of dissociation was miniscule….