Question

The dissociation of water into \(\mathrm{H}_{3} \mathrm{O}^{+}\) and \(\mathrm{OH}^{-}\) ions depends on temperature. At \(0^{\circ} \mathrm{C}(273 \mathrm{~K})\) the \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=3.38 \times 10^{-8}\) \(M\), at \(25^{\circ} \mathrm{C}(298 \mathrm{~K})\) the \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=1.00 \times 10^{-7} M,\) and at \(50^{\circ} \mathrm{C}(323 \mathrm{~K})\) the \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=2.34 \times 10^{-7} \mathrm{M}\) (a) Calculate the pH of water at \(0^{\circ} \mathrm{C}(273 \mathrm{~K})\) and \(50^{\circ} \mathrm{C}\) \((323 \mathrm{~K})\) (b) What is the value of \(K_{\mathrm{w}}\) at \(0{ }^{\circ} \mathrm{C}(273 \mathrm{~K})\) and \(50^{\circ} \mathrm{C}(323 \mathrm{~K})\) ? (c) Is the dissociation of water endothermic or exothermic?

Short answer

(a) pH at 0°C is 7.47; at 50°C is 6.63. (b) \(K_w\) at 0°C is \(1.14 \times 10^{-15}\); at 50°C is \(5.48 \times 10^{-14}\). (c) Dissociation is endothermic.

Step-by-step solution

Calculate pH at 0°C

The pH is calculated using the formula: \( \text{pH} = -\log_{10} [\text{H}_3\text{O}^+] \). For \( 0^{\circ} \text{C} \), \( [\text{H}_3\text{O}^+] = 3.38 \times 10^{-8} \text{ M} \). Calculate:\[ \text{pH}_{0^{\circ}C} = -\log_{10} (3.38 \times 10^{-8}) \approx 7.47 \]

Calculate pH at 50°C

Use the same formula for pH as in Step 1. For \( 50^{\circ} \text{C} \), \( [\text{H}_3\text{O}^+] = 2.34 \times 10^{-7} \text{ M} \).Calculate:\[ \text{pH}_{50^{\circ}C} = -\log_{10} (2.34 \times 10^{-7}) \approx 6.63 \]

Calculate Kw at 0°C

The ion product of water \( K_w \) is calculated as \( K_w = [\text{H}_3\text{O}^+] [\text{OH}^-] \). Since the solution is neutral, \([\text{H}_3\text{O}^+] = [\text{OH}^-]\). Then, \( K_w = ([\text{H}_3\text{O}^+])^2 \).For \( 0^{\circ} \text{C} \):\[ K_w = (3.38 \times 10^{-8})^2 = 1.14 \times 10^{-15} \]

Calculate Kw at 50°C

Use the same approach as in Step 3. For \( 50^{\circ} \text{C} \): \[ K_w = (2.34 \times 10^{-7})^2 = 5.48 \times 10^{-14} \]

Determine if dissociation is endothermic or exothermic

To determine if dissociation is endothermic or exothermic, observe how \( K_w \) changes with temperature. An increase in \( K_w \) with an increase in temperature suggests that the reaction absorbs heat, hence it is endothermic.

Key concepts

pH Calculation

The pH of a solution is a measure of its acidity or basicity. It is calculated using the formula:

• \( \text{pH} = -\log_{10} [\text{H}_3\text{O}^+] \)

The pH scale typically ranges from 0 to 14, where values less than 7 indicate acidic solutions, and values greater than 7 indicate basic solutions. A pH of 7 is neutral, which is typical for pure water at 25°C.
Calculating pH involves taking the negative logarithm of the hydronium ion concentration, \([\text{H}_3\text{O}^+]\). For instance, at 0°C with a hydronium ion concentration of \(3.38 \times 10^{-8} \ \text{M}\), the pH is:

• \[ \text{pH}_{0^{\circ}\text{C}} = -\log_{10} (3.38 \times 10^{-8}) \approx 7.47 \]

At 50°C, with \([\text{H}_3\text{O}^+] = 2.34 \times 10^{-7} \ \text{M}\),

• \[ \text{pH}_{50^{\circ}\text{C}} = -\log_{10} (2.34 \times 10^{-7}) \approx 6.63 \]

This indicates that water is slightly less neutral at higher temperatures.

Ion Product of Water (Kw)

The ion product of water, \(K_w\), is the product of the concentrations of hydronium and hydroxide ions in water. It is expressed as:

• \( K_w = [\text{H}_3\text{O}^+][\text{OH}^-] \)

In a neutral solution, the concentrations of hydroxide and hydronium ions are equal. Therefore, \(K_w\) can also be calculated as the square of the hydronium ion concentration:

• \( K_w = ([\text{H}_3\text{O}^+])^2 \)

At 0°C, the hydronium ion concentration is \(3.38 \times 10^{-8} \ \text{M}\), so:

• \[ K_w = (3.38 \times 10^{-8})^2 = 1.14 \times 10^{-15} \]

At 50°C, with \([\text{H}_3\text{O}^+] = 2.34 \times 10^{-7} \ \text{M}\),

• \[ K_w = (2.34 \times 10^{-7})^2 = 5.48 \times 10^{-14} \]

This increase of \(K_w\) with temperature shows that water dissociates more as it gets warmer.

Endothermic and Exothermic Reactions

Chemical reactions are termed endothermic or exothermic based on whether they absorb or release heat. An endothermic reaction absorbs heat, resulting in an increase in the reaction’s equilibrium constant \(K\) with temperature.
The dissociation of water is determined to be endothermic if the ion product \(K_w\) increases with rising temperatures. As seen from previous calculations:

• At 0°C, \(K_w = 1.14 \times 10^{-15}\)
• At 50°C, \(K_w = 5.48 \times 10^{-14}\)

The notable increase in \(K_w\) suggests that the dissociation of water into ions absorbs heat, confirming its endothermic nature.