Question

Water is a very weak electrolyte that undergoes the following ionization (called autoionization): $$ \mathrm{H}_{2} \mathrm{O}(l) \stackrel{k_{1}}{\stackrel{\mathrm{m}_{-1}}} \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ (a) If \(k_{1}=2.4 \times 10^{-5} \mathrm{~s}^{-1}\) and \(k_{-1}=1.3 \times 10^{11} / M \cdot \mathrm{s}\) calculate the equilibrium constant \(K\) where \(K=\left[\mathrm{H}^{+}\right]\) \(\left[\mathrm{OH}^{-}\right] /\left[\mathrm{H}_{2} \mathrm{O}\right] .\) (b) Calculate the product \(\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right],\) \(\left[\mathrm{H}^{+}\right],\) and \(\left[\mathrm{OH}^{-}\right] .\) (Hint : Calculate the concentration of liquid water using its density, \(1.0 \mathrm{~g} / \mathrm{mL}\).)

Short answer

K = 1.846 × 10^{-16} M, [H^+] = [OH^-] = 1.012 × 10^{-7} M.

Step-by-step solution

Understanding the problem

We are given the rate constants for the forward reaction, \(k_1\), and the reverse reaction, \(k_{-1}\), of the autoionization of water. The goal is to find the equilibrium constant \(K\), and then calculate the ion product and concentrations of ions in equilibrium.

Calculate the equilibrium constant (K)

The equilibrium constant \(K\) can be determined by the ratio of the rate constant of the forward reaction to the reverse reaction: \[K = \frac{k_1}{k_{-1}}\]Substituting the values, \[K = \frac{2.4 \times 10^{-5} \text{ s}^{-1}}{1.3 \times 10^{11} \text{ M}^{-1} \cdot \text{s}^{-1}}\]\[K = 1.846 \times 10^{-16} \text{ M}\]

Calculate the concentration of liquid water

The density of water is given as \(1.0 \text{ g/mL}\), which is equivalent to \(1.0 \times 10^3 \text{ g/L}\). The molar mass of water is approximately \(18.015 \text{ g/mol}\), so the concentration of water is:\[\text{Concentration of } H_2O = \frac{1.0 \times 10^3 \text{ g/L}}{18.015 \text{ g/mol}} \]\[= 55.49 \text{ M}\]

Calculate ion concentrations and their product

For a very weak electrolyte such as water, the ion concentrations are extremely small. Since \(K = [H^+][OH^-]/[H_2O]\), solving for \([H^+][OH^-]\): \[[H^+][OH^-] = K \times [H_2O]\]Thus,\[[H^+][OH^-] = (1.846 \times 10^{-16} \text{ M}) \times (55.49 \text{ M})\]\[= 1.024 \times 10^{-14} \text{ M}^2\]From this, we know that \([H^+] = [OH^-]\) and hence:\[[H^+] = [OH^-] = \sqrt{1.024 \times 10^{-14} \text{ M}^2} \]\[= 1.012 \times 10^{-7} \text{ M}\]

Conclusion

The equilibrium constant \(K\) dictates the extent of ionization in water. From the calculated \([H^+][OH^-]\), we find equal concentrations of hydrogen and hydroxide ions due to water's neutral nature.

Key concepts

Equilibrium Constant

The equilibrium constant, denoted as \( K \), is a crucial concept in understanding chemical reactions at equilibrium. It provides a ratio of the concentration of products to reactants. For the autoionization of water, this ratio is expressed as: \[ K = \frac{[\mathrm{H}^+][\mathrm{OH}^-]}{[\mathrm{H}_{2}\mathrm{O}]} \] This formula shows that the equilibrium constant is determined by the concentrations of hydrogen ions \([\mathrm{H}^+]\), hydroxide ions \([\mathrm{OH}^-]\), and water molecules \([\mathrm{H}_{2}\mathrm{O}]\). The constant \( K \) can be calculated using the rates of the forward and reverse reactions. In the case of water, these values are extremely small, indicating that pure water has low concentrations of these ions. Understanding \( K \) helps us comprehend how far a reaction proceeds and allows us to predict the behavior of the system at equilibrium.

Ion Product of Water

The ion product of water, represented by \( K_w \), is the product of the molar concentrations of hydrogen ions and hydroxide ions at equilibrium. This is a specific type of equilibrium constant for water: \[ K_w = [\mathrm{H}^+][\mathrm{OH}^-] \] At 25°C, this constant is typically \( 1.0 \times 10^{-14} \; \text{M}^2 \). This value is derived from the autoionization of water and remains constant regardless of changes in the concentration of water. It is important because it describes the degree of ionization and helps us understand the self-ionizing nature of water. Knowing \( K_w \) is essential for calculating \([\mathrm{H}^+]\) and \([\mathrm{OH}^-]\), especially in slightly acidic or basic solutions, to maintain the balance between hydrogen and hydroxide ions.

Weak Electrolyte

Water is considered a weak electrolyte, meaning it only partially ionizes in solution. Unlike strong electrolytes, which dissociate completely, weak electrolytes dissociate to a small extent. This weak ionization leads to low concentrations of ions in the solution. The equation for water's ionization is: \[ \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{H}^+ + \mathrm{OH}^- \] This reversible reaction illustrates that only a minute fraction of water molecules are ionized. The weak electrolyte nature of water explains why \([\mathrm{H}^+]\) and \([\mathrm{OH}^-]\) are very low in pure water. Because of this partial ionization, water maintains neutrality with equal concentrations of hydrogen and hydroxide ions at equilibrium.

Hydrogen Ion Concentration

Hydrogen ion concentration \([\mathrm{H}^+]\) is a measure of the acidity of a solution. The concentration is often expressed in molarity (M), and it plays a crucial role in determining the pH level of a solution. For pure water at 25°C, the concentration of hydrogen ions is: \[ [\mathrm{H}^+] = 1.0 \times 10^{-7} \; \text{M} \] This low concentration indicates that pure water is neutral. In any given solution, if \([\mathrm{H}^+]\) increases, the solution becomes more acidic. Understanding \([\mathrm{H}^+]\) allows chemists to predict the behavior of solutions and determine the necessary adjustments to reach desired acidity or neutrality conditions.

Hydroxide Ion Concentration

Hydroxide ion concentration \([\mathrm{OH}^-]\) complements the hydrogen ion concentration in maintaining the pH balance of a solution. In pure water, hydroxide ions are also present at a concentration of: \[ [\mathrm{OH}^-] = 1.0 \times 10^{-7} \; \text{M} \] This equality in ion concentrations is characteristic of a neutral pH of 7. Changes in \([\mathrm{OH}^-]\) affect the basicity of the solution. An increase in \([\mathrm{OH}^-]\) makes the solution more basic, whereas a decrease leads to more acidic conditions. The balance between \([\mathrm{H}^+]\) and \([\mathrm{OH}^-]\) is critical in many chemical and biological processes, as it affects the overall stability and reactivity within systems.