Question

In pure HCOOH liquid, the concentration of $HCO{O^- } = {10^{-3}}M$ at ${27^\circ }C$ . What is the value of self ionization constant at ${27^\circ }C$ given that $k = [HCOO{H_2}^+][HCO{O^-}]$ ?
A) ${10^{-3}}$
B) ${10^3}$
C) ${10^6}$
D) ${10^{-6}}$

Answer 1

Hint:The given compound Formic acid self ionizes to form $HCO{O^-}$ and $HCOO{H_2}^+ $ ions. Write the equation for this reaction and calculate the value of self ionization constant from it. At equilibrium the concentration of both the ions will be equal to ${10^{ - 3}}M$ .


Complete solution:
The process in which one molecule donates a proton to another molecule of the same compound to yield a cation and an anion is called self ionization. So,first we write the equation for self ionization reaction of Formic acid. Two molecules of Formic acid combine to give the following ions.
$2HCOOH \to HCO{O^-} + HCOO{H_2}^ + $
From this equation, we can write the formula for the self ionization constant using the concentrations of the ions formed. It is already given to us that the value of $k = [HCO{O^ - }][HCOO{H_2}^+]$
At equilibrium, the concentrations of both the ions are equal and it is given to us that the concentration of $HCO{O^-}$ is ${10^{-3}}M$ .
Therefore, $[HCO{O^-}] = [HCOO{H_2}^+] = {10^{-3}}M$
By substituting these values in the formula for self ionization constant, we get $k = {10^{-3}} \times {10^{-3}}$
By multiplying, we get $k = {10^{-6}}$
Hence, the value of self ionization constant of the given reaction is ${10^{-6}}$.

Therefore, from the above explanation we can say that the correct option is (D).


Note: It is important to balance the reaction equation before deriving the self ionization constant formula since this constant depends on both concentration and the number of molecules. In this case, two molecules of Formic acid are generating one molecule of $HCO{O^-}$ and one molecule of $HCOO{H_2}^- $ .