Question

Write the chemical equation and the $K_a$ expression for the acid dissociation of each of the following acids in aqueous solution. First show the reaction with ${\mathrm{H}}^{+}(aq)$ as a product and then with the hydronium ion: (a) ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}$, (b) ${\mathrm{HCO}}_3^{-}$

Short answer

(a) For $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}$: 1. With H+ as a product: $${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}(aq)\rightleftharpoons {\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{COO}}^{-}(aq)+{\mathrm{H}}^{+}(aq)$$ 2. With hydronium ion as a product: $${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}(aq)+{\mathrm{H}}_2\mathrm{O}(l)\rightleftharpoons {\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{COO}}^{-}(aq)+{\mathrm{H}}_3{\mathrm{O}}^{+}(aq)$$ 3. Ka expression: $$K_a=\frac{[{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{COO}}^{-}][{\mathrm{H}}^{+}]}{[{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}]}$$ (b) For $\mathrm{HCO}_{3}^{-}$: 1. With H+ as a product: $${\mathrm{HCO}}_3^{-}(aq)\rightleftharpoons {\mathrm{CO}}_3^{2-}(aq)+{\mathrm{H}}^{+}(aq)$$ 2. With hydronium ion as a product: $${\mathrm{HCO}}_3^{-}(aq)+{\mathrm{H}}_2\mathrm{O}(l)\rightleftharpoons {\mathrm{CO}}_3^{2-}(aq)+{\mathrm{H}}_3{\mathrm{O}}^{+}(aq)$$ 3. Ka expression: $$K_a=\frac{[{\mathrm{CO}}_3^{2-}][{\mathrm{H}}^{+}]}{[{\mathrm{HCO}}_3^{-}]}$$

Step-by-step solution

Write the chemical equation with H+ as a product

For C6H5COOH, the chemical equation with H+ as a product is: $${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}(aq)\rightleftharpoons {\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{COO}}^{-}(aq)+{\mathrm{H}}^{+}(aq)$$

Write the chemical equation with hydronium ion as a product

The chemical equation with hydronium ion as a product is: $${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}(aq)+{\mathrm{H}}_2\mathrm{O}(l)\rightleftharpoons {\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{COO}}^{-}(aq)+{\mathrm{H}}_3{\mathrm{O}}^{+}(aq)$$

Write the Ka expression for C6H5COOH

The Ka expression is: $$K_a=\frac{[{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{COO}}^{-}][{\mathrm{H}}^{+}]}{[{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}]}$$ (b) Acid dissociation of HCO3-

Write the chemical equation with H+ as a product

For HCO3-, the chemical equation with H+ as a product is: $${\mathrm{HCO}}_3^{-}(aq)\rightleftharpoons {\mathrm{CO}}_3^{2-}(aq)+{\mathrm{H}}^{+}(aq)$$

Write the chemical equation with hydronium ion as a product

The chemical equation with hydronium ion as a product is: $${\mathrm{HCO}}_3^{-}(aq)+{\mathrm{H}}_2\mathrm{O}(l)\rightleftharpoons {\mathrm{CO}}_3^{2-}(aq)+{\mathrm{H}}_3{\mathrm{O}}^{+}(aq)$$

Write the Ka expression for HCO3-

The Ka expression is: $$K_a=\frac{[{\mathrm{CO}}_3^{2-}][{\mathrm{H}}^{+}]}{[{\mathrm{HCO}}_3^{-}]}$$

Key concepts

Chemical Equation

Understanding chemical equations is essential when studying the behavior of substances in various reactions. A chemical equation represents the transformation of reactants into products using symbols and formulas. For acid dissociation reactions, the acid (HA) in aqueous solution dissociates into its conjugate base (A-) and a hydrogen ion (H+).

In the example of benzoic acid, ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}$, the chemical equation with ${\mathrm{H}}^{+}$ as a product is:
$${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}(aq)\rightleftharpoons {\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{COO}}^{-}(aq)+{\mathrm{H}}^{+}(aq)$$
When we include water in the equation, hydronium ions, ${\mathrm{H}}_3{\mathrm{O}}^{+}$ are formed, indicating the transfer of a proton to water:
$${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}(aq)+{\mathrm{H}}_2\mathrm{O}(l)\rightleftharpoons {\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{COO}}^{-}(aq)+{\mathrm{H}}_3{\mathrm{O}}^{+}(aq)$$
The same logic applies for bicarbonate ${\mathrm{HCO}}_3^{-}$, showing its dissociation in water.

Ka Expression

The acid dissociation constant, $K_a$, is a quantitative measure of the strength of an acid in solution. It is expressed as the equilibrium constant for a chemical reaction known as the dissociation of acid molecules in water. The $K_a$ expression is derived from the concentrations of the products and reactants at equilibrium.

For benzoic acid (${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}$ ), the $K_a$ expression is:
$$K_a=\frac{[{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{COO}}^{-}][{\mathrm{H}}^{+}]}{[{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}]}$$
This expression relates the concentration of the acid with its dissociated form and the free hydrogen ions in solution. For bicarbonate (${\mathrm{HCO}}_3^{-}$), the expression is:
$$K_a=\frac{[{\mathrm{CO}}_3^{2-}][{\mathrm{H}}^{+}]}{[{\mathrm{HCO}}_3^{-}]}$$
The $K_a$ values are essential for predicting the direction of the reaction and the extent of dissociation.

Aqueous Solution Chemistry

Aqueous solution chemistry involves the study of substances dissolved in water. Acids, bases, and salts when dissolved in water can undergo various interactions and reactions.

In the context of acid dissociation, an acid releases hydrogen ions (${\mathrm{H}}^{+}$ ) into the water, which can join with water molecules to form hydronium ions (${\mathrm{H}}_3{\mathrm{O}}^{+}$ ). This conversion is described by the equation:
$${\mathrm{H}}^{+}(aq)+{\mathrm{H}}_2\mathrm{O}(l)\rightleftharpoons {\mathrm{H}}_3{\mathrm{O}}^{+}(aq)$$
The presence of hydronium ions is what typically characterizes an acidic solution. In an alkaline solution, the hydroxide ions (${\mathrm{OH}}^{-}$ ) are more prevalent. Balancing these acid-base reactions in an aqueous solution is crucial for studying pH levels and chemical reactivity.

Hydronium Ion

The hydronium ion is the ion ${\mathrm{H}}_3{\mathrm{O}}^{+}$ and plays a pivotal role in the acid-base chemistry of water. The creation of a hydronium ion occurs when a hydrogen ion, which is essentially just a proton (${\mathrm{H}}^{+}$ ), is donated by an acid to a water molecule (${\mathrm{H}}_2\mathrm{O}$).

These ions are responsible for the acidic properties of solutions and are formed in all acid-water reactions:
$${\mathrm{H}}^{+}(aq)+{\mathrm{H}}_2\mathrm{O}(l)\to {\mathrm{H}}_3{\mathrm{O}}^{+}(aq)$$
Consequently, in any solution where an acid is present, hydronium ions will be formed, influencing the pH and reactivity of the solution. The concentration of hydronium ions is also involved in the calculation of pH, which is the negative logarithm of the hydronium ion concentration.