Answer:- pH = 3.37
Solution:- The balanced equation for the reaction of KOH with [tex]HNO_2[/tex] is written as:
[tex]HNO_2(aq)+KOH(aq)\rightleftharpoons KNO_2(aq)+H_2O(l)[/tex]
Nitrous acid is a weak acid and KOH is a strong base. So, at half-stoichiometric point half of the acid will be neutralized to form its conjugate base and half of the acid will still be remaining.
It means at half stoichiometric point the solution will have equal moles of weak acid(nitrous acid) and its conjugate base(nitrite ion). It will act as a buffer solution and the pH of the buffer solution is calculated by using Handerson equation:
[tex]pH=pKa+log(\frac{base}{acid}) [/tex]
Since, for the half stoichiometric point, [acid] = [base]
The ratio of their concentrations becomes 1 and the log of 1 is 0.
So, pH = pKa
pKa = - log Ka
[tex]pKa=-log(4.3*10^-^4)[/tex]
pKa = 3.37
So, pH = 3.37
Hence, the pH of the solution at half equivalence point will be 3.37.
The pH at the half-stoichiometric point for the titration of 0.22 M HNO₂(aq) with 0.1 M KOH(aq) is 3.4.
Let's consider the equation for the titration of 0.22 M HNO₂(aq) with 0.1 M KOH(aq).
HNO₂(aq) + KOH(aq) ⇒ KNO₂(aq) + H₂O(l)
At the half-stoichiometric point, we have a buffer system formed by equal concentrations of the weak acid (HNO₂) and its conjugate base (NO₂⁻).
We can calculate the pH using the Henderson-Hasselbach's equation.
[tex]pH = pKa+log(\frac{[NO_2^{-} ]}{[HNO_2]} )= pKa+log1=pKa=-log(4.3\times 10^{-4} )=3.4[/tex]
The pH at the half-stoichiometric point for the titration of 0.22 M HNO₂(aq) with 0.1 M KOH(aq) is 3.4.
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