Problem 80 - Chemistry Matter and Change

Question

What is \(\left[\mathrm{OH}^{-}\right]\) in an aqueous solution at 298 \(\mathrm{K}\) in which \(\left[\mathrm{H}^{+}\right]=5.40 \times 10^{-3} \mathrm{M} ?\)

Step-by-Step Solution

Answer

The concentration of hydroxide ions \([\mathrm{OH}^{-}]\) in the aqueous solution at 298 K is \(1.85 \times 10^{-12} \mathrm{M}\).

Step 1: Recall the Kw constant

At 298 K, the ion product of water (Kw) is given by the following equation: \[K_w = [\mathrm{H}^{+}] \cdot [\mathrm{OH}^{-}]\] Where \(K_w = 1.0 \times 10^{-14}\) is the ion product of water constant at 298 K, \([\mathrm{H}^{+}]\) is the concentration of hydrogen ions, and \([\mathrm{OH}^{-}]\) is the concentration of hydroxide ions in the solution.

Step 2: Use the given concentration of hydrogen ions

We have been given the concentration of hydrogen ions \([\mathrm{H}^{+}]\) as \(5.40 \times 10^{-3} \mathrm{M}\). Now, we can use this value to find the concentration of hydroxide ions \([\mathrm{OH}^{-}]\) using the Kw equation: \[K_w = [\mathrm{H}^{+}] \cdot [\mathrm{OH}^{-}]\]

Step 3: Solve for the concentration of hydroxide ions

Rearrange the equation to solve for \([\mathrm{OH}^{-}]\): \[[\mathrm{OH}^{-}] = \frac{K_w}{[\mathrm{H}^{+}]}\] Now, substitute the given values of \(K_w\) and \([\mathrm{H}^{+}]\) into the equation: \[[\mathrm{OH}^{-}] = \frac{1.0 \times 10^{-14}}{5.40 \times 10^{-3}}\]

Step 4: Calculate the concentration of hydroxide ions

Using a calculator, divide the values: \[[\mathrm{OH}^{-}] = 1.85 \times 10^{-12} \mathrm{M}\]

Step 5: Express the final answer

The concentration of hydroxide ions \([\mathrm{OH}^{-}]\) in the aqueous solution at 298 K is: \[[\mathrm{OH}^{-}] = 1.85 \times 10^{-12} \mathrm{M}\]

Key Concepts

Kw constant

The Kw constant, also known as the ion product of water, is a crucial concept in understanding the chemistry of aqueous solutions. At a temperature of 298 K, which is approximately room temperature, the constant value for water's ion product is given as \(K_w = 1.0 \times 10^{-14}\). This value represents the equilibrium constant for the self-ionization of water.
Water naturally dissociates into a small amount of hydrogen ions (\(\text{H}^+\)) and hydroxide ions (\(\text{OH}^-\)). Thus, the Kw constant is the product of these ion concentrations:

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

This relationship is fundamental in chemistry as it helps determine the concentration of either \(\text{H}^+\) or \(\text{OH}^-\) when one of them is provided. Moreover, because water is neutral, in pure water at 298 K, the concentrations of both ions are equal, each having a concentration of \(1.0 \times 10^{-7}\) M.
Knowing the Kw constant enables chemists to solve for missing ion concentrations in various chemical scenarios.

Hydrogen ion concentration

Hydrogen ion concentration, denoted as \([\text{H}^+]\), refers to the amount of hydrogen ions in a solution. These ions are pivotal in determining the acidity of a solution.
In the provided exercise problem, the hydrogen ion concentration is given as \(5.40 \times 10^{-3} \text{ M}\). Understanding this value allows chemists to determine the level of acidity and can be used further to calculate other properties of the solution.
For instance, in aqueous solutions, if the \([\text{H}^+]\) is known, the corresponding \([\text{OH}^-]\) can be easily calculated using the expression derived from the Kw constant:

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

In this case, with a high concentration of \(\text{H}^+\), the solution is more acidic, which inversely means a lower \([\text{OH}^-]\) concentration, maintaining the balance required by the Kw constant.

Hydroxide ion concentration

Hydroxide ion concentration, represented by \([\text{OH}^-]\), indicates the presence of hydroxide ions in a solution, which relates to its basicity. To find \([\text{OH}^-]\), knowing the hydrogen ion concentration is key due to their inverse relationship governed by the Kw constant:
Given the important relationship:

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

One can rearrange this equation to find the concentration of hydroxide ions, if necessary:

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

So, with an \([\text{H}^+]\) given as \(5.40 \times 10^{-3} \text{ M}\), the calculation of \([\text{OH}^-]\) is straightforward. By plugging the values into the formula, we can confirm that \([\text{OH}^-] = 1.85 \times 10^{-12} \text{ M}\). This calculation highlights how the equilibrium maintains a constant product of \(\text{H}^+\) and \(\text{OH}^-\) concentrations even as individual concentrations vary.