Question

At the boiling point of water \(\left(100^{\circ} \mathrm{C}\right), K_{w}=5.6 \times 10^{-13} .\) Calculate \(\left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) for a neutral solution at this temperature.

Short answer

In a neutral solution at the boiling point of water \(\left(100^{\circ} \mathrm{C}\right)\), the concentration of H+ and OH- ions are both approximately \(7.48 \times 10^{-7}\ \mathrm{M}\).

Step-by-step solution

Write down the ion product of water equation

The ion product of water equation relates the concentrations of H+ and OH- ions as follows: \[K_w = [\mathrm{H}^+][\mathrm{OH}^-]\] Given that \(K_w = 5.6 \times 10^{-13}\) for the boiling point of water.

Determine the relationship between H+ and OH- ions for a neutral solution

In a neutral solution, the concentrations of H+ and OH- ions are equal. Thus, we can represent their concentrations using a single variable, x. \[[\mathrm{H}^+] = [\mathrm{OH}^-] = x\]

Substitute the value of x in the ion product of water equation

Now, substitute the values of H+ and OH- ions with x in the ion product of water equation: \[K_w = x^2\] As we know that \(K_w = 5.6 \times 10^{-13}\), substitute this value in the equation: \(5.6 \times 10^{-13} = x^2\)

Solve the equation for x, which represents the concentrations of H+ and OH- ions

To find the value of x, simply take the square root of both sides: \(x = \sqrt{5.6 \times 10^{-13}}\) By solving this equation, we find the value of x: \(x \approx 7.48 \times 10^{-7}\)

Write down the result for concentrations of H+ and OH- ions

Since x represents the concentrations of H+ and OH- ions in the neutral solution at the boiling point of water: \[[\mathrm{H}^+] = [\mathrm{OH}^-] = 7.48 \times 10^{-7} \ \mathrm{M}\] Thus, the concentrations of H+ and OH- ions in a neutral solution at the boiling point of water are both approximately \(7.48 \times 10^{-7}\ \mathrm{M}\).

Key concepts

Neutral Solution

A neutral solution is one where the concentrations of hydrogen ions (H\(^+\)) and hydroxide ions (OH\(^-\)) are equal. This balance occurs in pure water under specific conditions. At 25°C, this is easily understood because the ion product of water \(K_w\) is \(1.0 \times 10^{-14}\). However, the situation changes with temperature.

At 100°C, which is the boiling point of water, the \(K_w\) value increases to \(5.6 \times 10^{-13}\). For neutrality under this condition, we assume \([\mathrm{H}^+] = [\mathrm{OH}^-]\).

This leads us to express both concentrations with a variable \(x\), giving us the equation \([\mathrm{H}^+][\mathrm{OH}^-] = x^2 = 5.6 \times 10^{-13}\). Solving for \(x\) helps us determine that both ion concentrations are \(7.48 \times 10^{-7}\ \mathrm{M}\), maintaining neutrality even as conditions change. This understanding is essential when observing the behavior of solutions at different temperatures.

Concentration of Ions

The concentration of ions in a solution is a crucial factor in determining the properties of that solution. In the case of water, these ions are hydrogen ions (H\(^+\)) and hydroxide ions (OH\(^-\)).

For water to be a neutral solution, the concentrations of these ions must be equal. At room temperature (25°C), both ion concentrations are \(1.0 \times 10^{-7}\ \mathrm{M}\) under pure water conditions. However, as temperature increases to the boiling point (100°C), the ion product of water changes. The \(K_w\) at this point is \(5.6 \times 10^{-13}\), which leads to changing concentrations.

Using the ion product of water relationship \(K_w = [\mathrm{H}^+][\mathrm{OH}^-]\), we substitute \([\mathrm{H}^+] = [\mathrm{OH}^-] = x\) into the equation, solve for \(x\), and find the new concentration to be \(7.48 \times 10^{-7}\ \mathrm{M}\). These values represent the delicate balance required for neutrality and underscore the significance of ion concentration in chemical dynamics.

Boiling Point of Water

The boiling point of water is a critical temperature at which water transitions from liquid to gas. This point is significant not just for cooking or weather, but it also affects the chemical properties of water, such as its ion product \(K_w\).

At standard atmospheric pressure, the boiling point of water is exactly 100°C. While boiling, water's physical interactions alter, affecting the balance of H\(^+\) and OH\(^-\) ions. The \(K_w\) of water at 100°C is \(5.6 \times 10^{-13}\), as opposed to \(1.0 \times 10^{-14}\) at 25°C. This change reflects the increased kinetic energy of water molecules, which consequently affects ionization.

These temperature-dependent shifts in \(K_w\) are vital in practical chemistry applications because they influence how reactions proceed. Knowing the effect of temperature on \(K_w\) allows scientists and chemists to accurately predict and utilize the behavior of aqueous systems in experimental and real-world scenarios.