Question

At \(50^{\circ} \mathrm{C}\), the ion-product constant for \(\mathrm{H}_{2} \mathrm{O}\) has the value \(K_{w}=5.48 \times 10^{-14}\). (a) What is the pH of pure water at \(50^{\circ} \mathrm{C}\) ? (b) Based on the change in \(K_{w}\) with temperature, predict whether \(\Delta H\) is positive, negative, or zero for the autoionization reaction of water: $$ 2 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q) $$

Short answer

(a) To find the pH of pure water at 50°C, we use the given value of \(K_w = 5.48 \times 10^{-14}\) to find the concentration of \(\mathrm{H}^+\), which is equal to the concentration of \(\mathrm{OH}^-\) in pure water. Calculate \([\mathrm{H}^+] = \sqrt{5.48 \times 10^{-14}}\), and then use the pH formula: \(\mathrm{pH} = -\log_{10}[\mathrm{H}^+]\) to find the pH. (b) To predict the sign of ΔH, we use the Van't Hoff equation and the knowledge that as temperature increases, the pH of pure water decreases slightly, implying that \(K_w\) increases with temperature. If \(K_w\) increases with temperature, then ΔH must be positive. Therefore, the enthalpy change (ΔH) for the autoionization reaction of water is positive, indicating an endothermic process.

Step-by-step solution

(a) Finding the pH of pure water at 50°C

First, recall the definition of the ion-product constant for water, \(K_w\), at a given temperature: $$ K_w = [\mathrm{H}^+][\mathrm{OH}^-] $$ For pure water, \([\mathrm{H}^+] = [\mathrm{OH}^-]\). Thus, we can rewrite the equation as: $$ K_w = [\mathrm{H}^+]^2 $$ From the exercise, we know that at 50°C, \(K_w = 5.48 \times 10^{-14}\). Substitute this value in the equation above, and solve for \([\mathrm{H}^+]\): $$ 5.48 \times 10^{-14} = [\mathrm{H}^+]^2 $$ Now, calculate \([\mathrm{H}^+]\): $$ [\mathrm{H}^+] = \sqrt{5.48 \times 10^{-14}} $$ Then, recall the definition of pH: $$ \mathrm{pH} = -\log_{10}[\mathrm{H}^+] $$ Substitute the value of \([\mathrm{H}^+]\) found above, and calculate the pH of the solution.

(b) Predicting the sign of ΔH for the autoionization of water

We are given the temperature of 50°C and asked to predict the change in enthalpy (ΔH) for the autoionization reaction of water. To do this, we will use the Van't Hoff equation, which relates the coefficient of temperature with the change in enthalpy and the equilibrium constant: $$ \frac{d \ln K_w}{dT} = \frac{\Delta H}{RT^2} $$ Remember that \(R\) is the ideal gas constant, and T is the temperature in Kelvin. For this problem, we are not given a temperature change or the equilibrium constant at another temperature, but we can make a prediction based on the known behavior of \(K_w\) and pH at different temperatures. We know that as temperature increases, the pH of pure water decreases slightly. This means that the ionization constant (\(K_w\)) increases with temperature. If \(K_w\) is increasing with temperature as to make the solution more acidic, the positive sign in the Van't Hoff equation must come from \(\Delta H\) (ΔH) being positive. So, we can predict that the enthalpy change (ΔH) for the autoionization reaction of water is positive, implying that the process is endothermic.

Key concepts

Autoionization of Water

Water can split into ions, a process known as autoionization. This happens when two water molecules interact. One acts as an acid and donates a proton (hydrogen ion) to the other molecule, which acts as a base. This results in the creation of hydronium ions (\[\text{H}_3\text{O}^+\]) and hydroxide ions (\[\text{OH}^-\]).
In pure water, the concentrations of these ions are equal because water splits into equal numbers of them. This equilibrium condition is represented by the ion-product constant, \(K_w\), and is given by the equation:

• \[K_w = [\text{H}^+][\text{OH}^-]\]

For pure water, since \([\text{H}^+] = [\text{OH}^-]\), the equation simplifies to:

• \[K_w = [\text{H}^+]^2\]

Understanding this concept is fundamental in analyzing the acidity or basicity of solutions.

pH Calculation

The pH of a solution measures its acidity or basicity. In simple terms, it tells us how many hydrogen ions are in a solution. A low pH indicates a high concentration of hydrogen ions, thus the solution is acidic. Conversely, a high pH means the solution is basic.
To calculate pH from hydrogen ion concentration, use the equation:

• \[\text{pH} = -\log_{10}[\text{H}^+]\]

This method is used for pure water or any solution where you know the \([\text{H}^+]\) concentration.
For instance, at 50°C, if the ion-product constant \(K_w\) is given as \(5.48 \times 10^{-14}\), first find the \([\text{H}^+]\) using:

• \[[\text{H}^+] = \sqrt{5.48 \times 10^{-14}}\]

Then, you can find the pH by substituting \([\text{H}^+]\) into the pH equation.

Temperature Dependence of Kw

The value of the ion-product constant, \(K_w\), changes with temperature. This constant affects how acidic or basic pure water is with temperature variations. Typically, as the temperature increases, \(K_w\) increases too, making water slightly more acidic and lowering the pH.
To understand why this happens, we consider the enthalpy change (ΔH) of water's autoionization. According to the Van't Hoff equation, the relationship between the change in temperature and \(K_w\) is expressed as:

• \[\frac{d \ln K_w}{dT} = \frac{\Delta H}{RT^2}\]

Here,

• \(R\) is the ideal gas constant
• \(T\) is the temperature in Kelvin

Because \(K_w\) increases with rising temperature, this implies an endothermic process where ΔH is positive. Therefore, the increase in \(K_w\) with temperature is evidence that the autoionization of water absorbs heat.