Question

Predict the products of the following acid-base reactions, and predict whether the equilibrium lies to the left or to the right of the equation: (a) \(\mathrm{O}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\) (b) \(\mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{HS}^{-}(a q) \rightleftharpoons\) (c) \(\mathrm{NO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\)

Short answer

\(a)\): The products are \(OH^-(aq)\) and \(OH^-(aq)\), and the equilibrium lies to the left. \(b)\): The products are \(CH_3COO^-(aq)\) and \(H_2S(aq)\), and the equilibrium lies slightly to the right. \(c)\): The products are \(OH^-(aq)\) and \(HNO_2(aq)\), and the equilibrium lies to the left.

Step-by-step solution

a) Identifying Acid and Base in the equation

In this reaction, we have \(\mathrm{O^{2-}}(a q)\) reacting with \(\mathrm{H_2O}(l)\). \(\mathrm{O^{2-}}\) is a base, as it can accept protons, while \(\mathrm{H_2O}\) acts as an acid.

b) Predicting the products

Based on the given reactants, the products will be formed upon the transfer of a proton from \(\mathrm{H_2O}\) to \(\mathrm{O^{2-}}\). Therefore, the products will be: \(OH^-(aq) + OH^-(aq)\)

c) Determining the equilibrium direction

In this case, the equilibrium constant for this reaction can be represented as: \(K=\frac{[\mathrm{OH^-}][\mathrm{OH^-}]}{[\mathrm{O^{2-}}][\mathrm{H_2O}]}\) Since \(\mathrm{O^{2-}}\) is a weak base, and \(\mathrm{H_2O}\) is a weak acid, the equilibrium will lie to the left, favoring the formation of reactants.

a) Identifying Acid and Base in the equation

In this reaction, we have \(\mathrm{CH_3COOH} (a q)\) reacting with \(\mathrm{HS^-(a q)}\). \(\mathrm{CH_3COOH}\) is an acid, while \(\mathrm{HS^-}\) is a base.

b) Predicting the products

The products will be formed upon the transfer of a proton from \(\mathrm{CH_3COOH}\) to \(\mathrm{HS^-}\). Therefore, the products will be: \(CH_3COO^-(aq) + H_2S(aq)\)

c) Determining the equilibrium direction

In this case, the equilibrium constant for this reaction can be represented as: \(K=\frac{[\mathrm{CH_3COO^-}][\mathrm{H_2S}]}{[\mathrm{CH_3COOH}][\mathrm{HS^-}]}\) Since \(\mathrm{CH_3COOH}\) is a weak acid and \(\mathrm{HS^-}\) is a weak base, the equilibrium direction depends on their relative strengths. In this case, the equilibrium will lie slightly to the right, favoring the formation of products.

a) Identifying Acid and Base in the equation

In this reaction, we have \(\mathrm{NO_2^{-}}(a q)\) reacting with \(\mathrm{H_2O}(l)\). \(\mathrm{NO_2^{-}}\) acts as a base, while \(\mathrm{H_2O}\) will be an acid.

b) Predicting the products

The products will be formed upon the transfer of a proton from \(\mathrm{H_2O}\) to \(\mathrm{NO_2^{-}}\). Therefore, the products will be: \(OH^-(aq) + HNO_2(aq)\)

c) Determining the equilibrium direction

In this case, the equilibrium constant for this reaction can be represented as: \(K=\frac{[\mathrm{OH^-}][\mathrm{HNO_2}]}{[\mathrm{NO_2^-}][\mathrm{H_2O}]}\) \(\mathrm{NO_2^-}\) is a weak base, and \(\mathrm{H_2O}\) is a weak acid, so the equilibrium will lie to the left, favoring the formation of reactants.