Question

Identify the first reactant in each equation as an acid or a base. (a) \(\mathrm{HCN}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{CN}^{-}(a q)\) (b) \(\mathrm{SO}_{4}{ }^{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{HSO}_{4}^{-}(a q)+\mathrm{OH}^{-}(a q)\) (c) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}(a q)+\mathrm{NaOH}(a q) \rightleftharpoons\) \(\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}^{-}(a q)+\mathrm{Na}^{+}(a q)\)

Short answer

In the given equations, the first reactant in (a) HCN is an acid, in (b) SO₄²⁻ is a base, and in (c) C₆H₅OH is an acid.

Step-by-step solution

Analysing Equation (a)

For the first equation, HCN(aq) + H₂O(l) ⇌ H₃O⁺(aq) + CN⁻(aq), we can see that HCN is donating a proton to water, forming H₃O⁺, while itself becoming CN⁻. Thus, HCN is acting as an acid.

Analysing Equation (b)

In the second equation, SO₄²⁻(aq) + H₂O(l) ⇌ HSO₄⁻(aq) + OH⁻(aq), SO₄²⁻ is accepting a proton from water, forming HSO₄⁻, while water turns into OH⁻. Therefore, SO₄²⁻ is acting as a base.

Analysing Equation (c)

For the third equation, C₆H₅OH(aq) + NaOH(aq) ⇌ H₂O(l) + C₆H₅O⁻ (aq) + Na⁺ (aq), C₆H₅OH donates a proton to NaOH, in which process NaOH becomes water (H₂O) and C₆H₅OH becomes C₆H₅O⁻. Here, C₆H₅OH is acting as an acid.

Key concepts

Chemical Equilibrium

Understanding chemical equilibrium is crucial when studying reactions, like the ones presented in the textbook exercise. This concept entails a state in which the rate of the forward reaction equals the rate of the reverse reaction, meaning that the concentrations of the reactants and products remain constant over time, but not necessarily equal.

For example, in equation (a), \(\mathrm{HCN}(aq)+\mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3}\mathrm{O}^{+}(aq)+\mathrm{CN}^{-}(aq)\), the reaction can proceed both ways: HCN can react with water to form hydronium and cyanide ions, and those products can in turn react to reform HCN and water. At equilibrium, the amount of HCN turning into products equals the amount of product reconstituting HCN.

It is important for students to grasp that reaching equilibrium does not mean that the reactants and products stop reacting. Rather, there is a dynamic balance where the reactions continue, but the overall concentrations stay the same, which can be influenced by changes in temperature, pressure, or concentration.

Proton Donation and Acceptance

The transfer of protons is a fundamental process in acid-base chemistry, and it is well illustrated in the reactions from our exercise. Proton donation and acceptance are the actions that define an acid and a base within the Brønsted-Lowry framework.

In equation (b), \(\mathrm{SO}_{4}^{2-}(aq)+\mathrm{H}_{2}\mathrm{O}(l)\rightleftharpoons\mathrm{HSO}_{4}^{-}(aq)+\mathrm{OH}^{-}(aq)\), the sulfate ion \(\mathrm{SO}_{4}^{2-}\) accepts a proton from the water molecule, becoming \(\mathrm{HSO}_{4}^{-}\). Meanwhile, water loses a proton and becomes \(\mathrm{OH}^{-}\), the hydroxide ion. In this context, \(\mathrm{SO}_{4}^{2-}\) is the proton acceptor, classifying it as a base.

This acceptance and donation of protons is how substances display their acidic or basic character and is the basis for understanding their behavior in various chemical reactions.

Bronsted-Lowry Acids and Bases

The Brønsted-Lowry theory is an acid-base reaction theory which defines an acid as a proton donor and a base as a proton acceptor. This theory expands the definitions of acids and bases beyond the substances that contain hydrogen or hydroxide ions.

In equation (c), \(\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{OH}(aq)+\mathrm{NaOH}(aq) \rightleftharpoons \mathrm{H}_{2}\mathrm{O}(l)+\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{O}^{-}(aq)+\mathrm{Na}^{+}(aq)\), phenol (\(\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{OH}\)) donates a proton to sodium hydroxide (NaOH), showcasing its role as a Brønsted-Lowry acid. After donating a proton, phenol becomes the phenoxide ion \(\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{O}^{-}\), while water is formed as one of the products.

This theory helps students to recognize acids and bases in different contexts, making it easier to predict the outcomes of acid-base reactions. It's also essential in understanding the broader implications of acid-base chemistry in biological, environmental, and industrial processes.