Question

At \(0^{\circ} \mathrm{C}\), the density of liquid water is \(0.9998 \mathrm{~g} / \mathrm{mL}\) and the value of \(K_{w}\) is \(1.14 \times 10^{-15}\). What fraction of the molecules in liquid water are dissociated at \(0^{\circ} \mathrm{C}\) ? What is the percent dissociation at \(0^{\circ} \mathrm{C}\) ? What is the \(\mathrm{pH}\) of a neutral solution at \(0^{\circ} \mathrm{C}\) ?

Short answer

Fraction dissociation is \(1.922 \times 10^{-10}\), percent is \(1.922 \times 10^{-8}\%\), and pH is \(7.97\).

Step-by-step solution

Calculate Molarity of Water

First, we need to convert the density of water to molarity. Knowing that the molar mass of water (H₂O) is approximately 18.015 g/mol, we calculate the molarity by dividing the density by the molar mass: \[\text{Molarity of water} = \frac{0.9998 \text{ g/mL}}{18.015 \text{ g/mol}} \approx 55.509 \text{ mol/L}\] This is the molarity of water at \(0^{\circ} \mathrm{C}\).

Calculate Dissociation Constant

The self-ionization of water is given by the equilibrium \(2H_2O \rightleftarrows H_3O^+ + OH^-\). The equilibrium constant for this process, \(K_w\), is given as \(1.14 \times 10^{-15}\). Assuming \([H_3O^+] = [OH^-] = x\) at equilibrium, \[ K_w = x^2 \]Solving for \(x\), we find:\[ x = \sqrt{1.14 \times 10^{-15}} = 1.067 \times 10^{-8} \text{ mol/L} \]

Calculate Fraction of Dissociated Water

The fraction of dissociated water molecules can be found by comparing the concentration of dissociated water (\(x\)) to the total concentration of water. It is calculated as follows:\[ \text{Fraction dissociated} = \frac{x}{\text{Molarity of water}} = \frac{1.067 \times 10^{-8} \text{ mol/L}}{55.509 \text{ mol/L}} \approx 1.922 \times 10^{-10} \]

Calculate Percent Dissociation

To convert the fraction of dissociated molecules to a percentage, we multiply by 100:\[ \text{Percent dissociation} \approx 1.922 \times 10^{-10} \times 100 \approx 1.922 \times 10^{-8} \% \]

Calculate pH of Neutral Solution

The \(\text{pH}\) is calculated using the hydronium ion concentration \(x\):\[ \text{pH} = -\log_{10}(1.067 \times 10^{-8}) \approx 7.97 \]

Key concepts

Density of Water

Understanding the density of water is crucial for many chemical calculations. At \(0^{\circ} \mathrm{C}\), water has a density of \(0.9998 \mathrm{~g/mL}\), meaning that each milliliter of water weighs 0.9998 grams. This density is used to calculate the molarity of water, which is a measure of the concentration of water molecules in a liter of solution.

To convert this density to molarity, we use the molar mass of water, approximately 18.015 g/mol. By dividing the density by this molar mass, we find that water's molarity is \(\approx 55.509 \ \mathrm{mol/L}\) at \(0^{\circ} \mathrm{C}\). Knowing this molarity is essential for understanding how many water molecules are available to participate in reactions, such as self-ionization.

Self-Ionization of Water

Water is unique because it can ionize itself, even without any added substances. This process is called self-ionization. In this equilibrium reaction, two water molecules \(H_2O\) react to form a hydronium ion \(H_3O^+\) and a hydroxide ion \(OH^-\):

\[ 2H_2O \rightleftharpoons H_3O^+ + OH^- \]

The equilibrium constant for self-ionization, known as \(K_w\), is the product of the concentrations of these ions. At \(0^{\circ} \mathrm{C}\), \(K_w\) is \(1.14 \times 10^{-15}\).

By setting \([H_3O^+] = [OH^-] = x\) and solving \(K_w = x^2\), we find the concentration of \([H_3O^+]\) and \([OH^-]\) to be \(1.067 \times 10^{-8} \ \mathrm{mol/L}\). This value helps calculate how much water dissociates under these conditions.

pH Calculation

The pH of a solution measures its acidity or basicity. For water at \(0^{\circ} \mathrm{C}\), where \([H_3O^+] = 1.067 \times 10^{-8} \ \mathrm{mol/L}\), we calculate the pH using the formula:

\[ \text{pH} = -\log_{10}(1.067 \times 10^{-8}) \]

Plugging in the values, the pH is approximately 7.97. A pH close to 7 indicates a neutral solution, slightly adjusted due to temperature effects.

Understanding these calculations is crucial, as small changes in \([H_3O^+]\) can significantly affect the pH, and consequently, the behavior of biochemical reactions in water.