Problem 40 - Chemistry The Central Science

Question

Carbon dioxide in the atmosphere dissolves in raindrops to produce carbonic acid \(\left(\mathrm{H}_{2} \mathrm{CO}_{3}\right),\) causing the \(\mathrm{pH}\) of clean, \(\mathrm{un}^{-}\) polluted rain to range from about 5.2 to \(5.6 .\) What are the ranges of \(\left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) in the raindrops?

Step-by-Step Solution

Answer

The range of \(\left[\mathrm{H}^{+}\right]\) in raindrops is from \(2.51 \times 10^{-6} \ \text{M}\) to \(6.31 \times 10^{-6} \ \text{M}\), and the range of \(\left[\mathrm{OH}^{-}\right]\) in raindrops is from \(1.59 \times 10^{-9} \ \text{M}\) to \(3.98 \times 10^{-9} \ \text{M}\).

Step 1: Calculate \(\left[\mathrm{H}^{+}\right]\) using the definition of pH

The pH is defined as the negative logarithm of the hydrogen ion concentration. Therefore, we can find \(\left[\mathrm{H}^{+}\right]\) by taking the inverse logarithm of the pH values. The given range of pH is from 5.2 to 5.6, so we have: \(\left[\mathrm{H}^{+}\right] = 10^{-\text{pH}}\) For pH = 5.2: \(\left[\mathrm{H}^{+}\right] = 10^{-5.2} \approx 6.31 \times 10^{-6} \ \text{M}\) For pH = 5.6: \(\left[\mathrm{H}^{+}\right] = 10^{-5.6} \approx 2.51 \times 10^{-6} \ \text{M}\) Thus, the range of \(\left[\mathrm{H}^{+}\right]\) in raindrops is from \(2.51 \times 10^{-6} \ \text{M}\) to \(6.31 \times 10^{-6} \ \text{M}\).

Step 2: Calculate \(\left[\mathrm{OH}^{-}\right]\) using the ion-product of water

The ion-product of water, \(K_w\), is given by the product of \(\left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) concentrations: \(K_w = \left[\mathrm{H}^{+}\right] \times \left[\mathrm{OH}^{-}\right]\) At 25°C, \(K_w = 1.0 \times 10^{-14}\). We can now solve for \(\left[\mathrm{OH}^{-}\right]\) using the \(\left[\mathrm{H}^{+}\right]\) values we found in Step 1. For \(\left[\mathrm{H}^{+}\right] = 6.31 \times 10^{-6} \ \text{M}\): \(\left[\mathrm{OH}^{-}\right] = \frac{K_w}{\left[\mathrm{H}^{+}\right]} = \frac{1.0 \times 10^{-14}}{6.31 \times 10^{-6}} \approx 1.59 \times 10^{-9} \ \text{M}\) For \(\left[\mathrm{H}^{+}\right] = 2.51 \times 10^{-6} \ \text{M}\): \(\left[\mathrm{OH}^{-}\right] = \frac{K_w}{\left[\mathrm{H}^{+}\right]} = \frac{1.0 \times 10^{-14}}{2.51 \times 10^{-6}} \approx 3.98 \times 10^{-9} \ \text{M}\) Thus, the range of \(\left[\mathrm{OH}^{-}\right]\) in raindrops is from \(1.59 \times 10^{-9} \ \text{M}\) to \(3.98 \times 10^{-9} \ \text{M}\).

Key Concepts

pH calculation

Understanding the concept of pH is essential in chemistry, as it helps determine the acidity or basicity of a solution. The pH scale, which ranges from 0 to 14, represents how acidic or basic a liquid is. Pure water, which is neutral, has a pH of 7. Solutions with a pH less than 7 are acidic, while those with a pH greater than 7 are basic (also known as alkaline).

The pH calculation is a straightforward process that involves taking the negative logarithm (base 10) of the hydrogen ion concentration (\textbf{[H+]}) in a solution. The formula for this calculation is:

\[\text{pH} = -\log_{10}(\text{[H+]}).\]

In the context of rainwater, which usually contains dissolved carbon dioxide forming carbonic acid, this can lead to slightly acidic pH values between 5.2 and 5.6. To find the hydrogen ion concentration from the pH, we use the inverse process, which is raising ten to the power of the negative pH value:

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

Applying this formula, we can see that with a given pH of 5.2 or 5.6, the corresponding hydrogen ion concentrations can be determined, providing insight into the rainwater's acidity.

Hydrogen ion concentration

The hydrogen ion concentration of a solution plays a critical role in its chemical behavior, particularly its reactivity and properties. It is denoted by the symbol \textbf{[H+]}, and it refers to the molarity of hydrogen ions present in the solution. Molarity, in turn, is a measure of the concentration of a solute in a solution, usually expressed in moles per liter (\textbf{M}).

By calculating the hydrogen ion concentration, we can ascertain the level of acidity in rainwater. As shown in the example with rainwater pH ranging from 5.2 to 5.6:

• For pH = 5.2,\(\text{[H+]} = 10^{-5.2} \approx 6.31 \times 10^{-6}\) M,
• For pH = 5.6,\(\text{[H+]} = 10^{-5.6} \approx 2.51 \times 10^{-6}\) M.

This indicates that the hydrogen ion concentration in clean, unpolluted rain can vary but will typically be in the range of micro-molar levels, signaling a mild level of acidity.

Ion-product of water

An intrinsic property of water is its ability to self-ionize. This results in the formation of hydrogen ions (\textbf{H+}) and hydroxide ions (\textbf{OH-}), which are fundamental to the concept of the ion-product of water (\textbf{Kw}). At a temperature of 25°C, the ion-product of water is always constant at \(1.0 \times 10^{-14}\) M² and is expressed as:

\[K_w = [\text{H+}] \times [\text{OH-}].\]

The ion-product plays a crucial role in finding the concentration of hydroxide ions in a solution when the hydrogen ion concentration is known. By rearranging the formula, we get:

\[\text{[OH-]} = \frac{K_w}{\text{[H+]}}.\]

For the rainwater scenario with calculated

• For \(\text{[H+]} = 6.31 \times 10^{-6} \) M,\(\text{[OH-]} = \frac{1.0 \times 10^{-14}}{6.31 \times 10^{-6}} \approx 1.59 \times 10^{-9}\) M,
• For \(\text{[H+]} = 2.51 \times 10^{-6} \) M,\(\text{[OH-]} = \frac{1.0 \times 10^{-14}}{2.51 \times 10^{-6}} \approx 3.98 \times 10^{-9}\) M.

This is important to consider, as the balance between hydrogen ions and hydroxide ions ultimately determines the pH of a solution, further emphasizing the interconnected nature of these concepts.