Question

At the freezing point of water \(\left(0^{\circ} \mathrm{C}\right), K_{w}=1.2 \times 10^{-15}\) Calculate \(\left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) for a neutral solution at this temperature.

Short answer

The concentration of hydrogen ions \([\mathrm{H}^{+}]\) and hydroxide ions \([\mathrm{OH}^{-}]\) in a neutral solution at \(0^{\circ}\mathrm{C}\) is approximately \(1.1 \times 10^{-8}\) M.

Step-by-step solution

Write the equation for the ion product of water

The ion product of water, \(K_w\), is given by the following equation: \[K_w = [\mathrm{H}^{+}][\mathrm{OH}^{-}]\]

Plug in the given value of \(K_w\)

We are given \(K_w = 1.2 \times 10^{-15}\) at \(0^{\circ}\mathrm{C}\). So, the equation becomes: \[1.2 \times 10^{-15} = [\mathrm{H}^{+}][\mathrm{OH}^{-}]\]

Set up the equation for a neutral solution

In a neutral solution, the concentration of hydrogen ions is equal to the concentration of hydroxide ions i.e., \([\mathrm{H}^+] = [\mathrm{OH}^-]\). Let's denote the concentration of hydrogen ions (and hydroxide ions) as \(x\). Now, the equation becomes: \[1.2 \times 10^{-15} = x^2\]

Solve for x

To find the concentration of hydrogen ions (and hydroxide ions) in the neutral solution, we need to solve for x: \[x^2 = 1.2 \times 10^{-15}\] \[x = \sqrt{1.2 \times 10^{-15}}\]

Find the concentration of \(\mathrm{H}^+\) and \(\mathrm{OH}^-\) ions

Since it's a neutral solution, the concentration of hydrogen ions will be equal to the concentration of hydroxide ions. Therefore, \[[\mathrm{H}^{+}] = [\mathrm{OH}^{-}] = x = \sqrt{1.2 \times 10^{-15}}\] Now, calculate the square root: \[[\mathrm{H}^{+}] = [\mathrm{OH}^{-}] \approx 1.1 \times 10^{-8} \mathrm{M}\] So, the concentration of hydrogen ions \([\mathrm{H}^{+}]\) and hydroxide ions \([\mathrm{OH}^{-}]\) in a neutral solution at \(0^{\circ}\mathrm{C}\) is approximately \(1.1 \times 10^{-8}\) M.

Key concepts

pH and pOH Calculations

Understanding pH and pOH is essential for studying the properties of solutions, especially their acidity or basicity. pH is a logarithmic measure of the hydrogen ion concentration \( [H^+] \) in a solution, while pOH is a similar measure for hydroxide ion concentration \( [OH^-] \). The pH and pOH scales range from 0 to 14, with 7 being neutral.

Mathematically, the pH is calculated using the formula \( pH = -\log[H^+] \) and pOH is calculated as \( pOH = -\log[OH^-] \). These calculations require knowledge of the concentrations of hydrogen and hydroxide ions in a given solution. For a neutral solution at \(25^\circ C\), \( [H^+] \) and \( [OH^-] \) are both equal to \( 1 \times 10^{-7} \), resulting in a pH and pOH of 7. However, at other temperatures, these values change, as seen in the given exercise.

Neutral Solution

A neutral solution is one where the concentration of hydrogen ions \( [H^+] \) equals the concentration of hydroxide ions \( [OH^-] \). This implies a state of equilibrium where neither an acidic nor a basic condition prevails. At \(25^\circ C\), the concentration for both ions in a neutral solution is \( 1 \times 10^{-7} \ M (molar)\).

In the context of the exercise, however, we're looking at a neutral solution at the freezing point of water \(0^\circ C\), where the ion product of water \(K_w\) is different, hence the concentrations of \( [H^+] \) and \( [OH^-] \) also differ from the standard values at room temperature.

Equilibrium Constant for Water

The equilibrium constant for water \(K_w\) is an expression of the dissociation of water molecules into hydrogen and hydroxide ions. This constant is typically \(1 \times 10^{-14}\) at \(25^\circ C\), but it varies with temperature. The balance between \( [H^+] \) and \( [OH^-] \) is crucial in determining the pH of the solution.

For a neutral solution, where \( [H^+] = [OH^-] \), \(K_w\) equals the product of the concentrations of these ions. As demonstrated in the original exercise's solution, even at different temperatures, for a neutral solution, we can use the \(K_w\) value to calculate the exact concentrations of \( [H^+] \) and \( [OH^-] \) by setting them equal to each other and solving for their value.

Temperature Dependence of Kw

The ion product of water, \(K_w\), is not a constant value but rather varies with temperature. At \(25^\circ C\), \(K_w\) is \(1 \times 10^{-14}\), but at the freezing point of water \(0^\circ C\), it is \(1.2 \times 10^{-15}\), as indicated in the exercise. This temperature dependence shows that water's propensity to dissociate into hydrogen and hydroxide ions is reduced at lower temperatures.

The fact that \(K_w\) decreases with decreasing temperature also implies that both \( [H^+] \) and \( [OH^-] \) concentrations will be lower in a neutral solution at cooler temperatures. Understanding this concept is important, especially when dealing with pH-sensitive reactions that may be affected by temperature changes.