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# One drop of $1~M\mathrm{~OH}^{-}$ion is added to a $1~M$ solution of $\mathrm{HNO}_2$. What will be the effect of this addition on the equilibrium concentration of the following?$\mathrm{HNO}_2(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{NO}_2^{-}(a q)$(a) $\left[\mathrm{OH}^{-}\right]$ (b) $\left[\mathrm{H}^{+}\right]$ (c) $\left[\mathrm{NO}_2^{-}\right]$ (d) $\left[\mathrm{HNO}_2\right]$

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In this task, we need to explain influence of the addition of OH$^-$ on the equilibrium concentrations.

The $\textbf{equilibrium position}$ can be affected by changing the $\textbf{concentration, pressure and temperature}$. If one of the conditions under which the system is in a state of chemical equilibrium changes, then the equilibrium shift will be in the direction of that reaction which tends to oppose that change and to establish the previous conditions. This legality is known as the $\textbf{Le Chatelier's principle}$.

As the $\textbf{concentration}$ of $\textbf{reactants}$ $\textbf{increases}$ or the $\textbf{product}$ is $\textbf{removed }$from the reaction mixture, equilibrium shifts in the direction of $\textbf{product formation}$. An equilibrium shift in the $\textbf{direction of the reactants}$ in a reaction mixture is achieved by the opposite action, ie by $\textbf{reducing the concentration of reactants}$ or by $\textbf{increasing the concentration of the product}$.

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