The equilibrium constant for the reaction
HNO₂(aq)+ H₂O(l) $\rightarrow$ NO₂-(aq)+ H₃O⁺(aq)
is 4.3 * 10^-⁴ at 25^$\omicron$C.
Will nitrous acid spontaneously dissociate when
[HNO₂(aq)] = [NO₂-(aq)] = [H₃O⁺(aq)] = 1.0 M? Show your calculations.

Question

The equilibrium constant for the reaction
HNO₂(aq)+ H₂O(l)

\rightarrow

NO₂-(aq)+ H₃O⁺(aq)
is 4.3 * 10^-⁴ at 25^$\omicron$C.
Will nitrous acid spontaneously dissociate when
[HNO₂(aq)] = [NO₂-(aq)] = [H₃O⁺(aq)] = 1.0 M? Show your calculations.

Quizplus · Accepted Answer

The reaction quotient Q for the reaction, when $[HNO_2(aq)] = [NO_2^-(aq)] = [H_3O^+(aq)] = 1.0 M$, is calculated as $Q = \frac{[NO_2^-][H_3O^+]}{[HNO_2]}$. Given that all concentrations are 1.0 M, Q simplifies to $Q = \frac{1 \times 1}{1} = 1$. Since the equilibrium constant $K = 4.3 \times 10^{4}$ is much larger than Q, the reaction will proceed towards the products to reach equilibrium, indicating that nitrous acid will spontaneously dissociate under these conditions.

The Equilibrium Constant for the Reaction HNO2(aq)+ H2O(l) →\rightarrow→ NO2-(aq)+ H3O+(aq) Is 4

The Equilibrium Constant for the Reaction
HNO₂(aq)+ H₂O(l) $\rightarrow$ NO₂-(aq)+ H₃O⁺(aq)
Is 4