Problem 38 - Chemistry The Central Science In Si Units

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, unpolluted 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 for \\[ [H^+] \\] is \([2.51 \times 10^{-6}, 6.31 \times 10^{-6}] \text{ M}\) and for \\[ [OH^-] \\], it is \([1.58 \times 10^{-9}, 3.98 \times 10^{-9}] \text{ M}\).

Step 1: Understand the pH Scale

The pH scale measures how acidic or basic a solution is, ranging from 0 to 14. A pH of 7 is neutral, less than 7 is acidic, and greater than 7 is basic. The formula to find \([H^+]\) concentration is \([H^+] = 10^{-pH}\).

Step 2: Calculate [H⁺] for pH 5.2

Using the formula \[ [H^+] = 10^{-pH} \], substitute pH = 5.2: \[ [H^+] = 10^{-5.2} \]. Calculating this, \[ [H^+] \approx 6.31 \times 10^{-6} \text{ M} \].

Step 3: Calculate [H⁺] for pH 5.6

Similarly, for pH = 5.6, substitute in the formula: \[ [H^+] = 10^{-5.6} \]. Calculating this, \[ [H^+] \approx 2.51 \times 10^{-6} \text{ M} \].

Step 4: Determine [H⁺] Range

The \[ [H^+] \] range for the pH values from 5.2 to 5.6 is approximately \([2.51 \times 10^{-6}, 6.31 \times 10^{-6}] \text{ M}\).

Step 5: Calculate [OH⁻] using the Ion Product of Water

The ion product of water, \[ Kw = [H^+][OH^-] = 1.0 \times 10^{-14} \], is used to find \[ [OH^-] \]. Rearranging gives \[ [OH^-] = \frac{Kw}{[H^+]} \].

Step 6: Calculate [OH⁻] for pH 5.2

Using \[ [H^+] \approx 6.31 \times 10^{-6} \text{ M} \], \[ [OH^-] = \frac{1.0 \times 10^{-14}}{6.31 \times 10^{-6}} \approx 1.58 \times 10^{-9} \text{ M} \].

Step 7: Calculate [OH⁻] for pH 5.6

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

Step 8: Determine [OH⁻] Range

The \[ [OH^-] \] range for the pH values from 5.2 to 5.6 is approximately \([1.58 \times 10^{-9}, 3.98 \times 10^{-9}] \text{ M}\).

Key Concepts

pH Scale

The pH scale is a tool used to measure the acidity or basicity of a solution. It ranges from 0 to 14, where 7 is considered neutral. Solutions with a pH less than 7 are acidic, while those with a pH greater than 7 are basic (alkaline). The scale is logarithmic, which means each whole number step represents a tenfold difference in acidity.

• Zero to 6: Acidic solutions; the lower the number, the more acidic.
• Seven: Neutral, like pure water.
• Eight to 14: Basic solutions; the higher the number, the more basic.

Understanding the pH scale is essential because it helps us determine how solutions will behave in various chemical reactions and everyday situations. For instance, in the given exercise, rainwater has a pH that typically ranges between 5.2 and 5.6, indicating it is slightly acidic due to dissolved carbon dioxide forming carbonic acid. This is a common condition for unpolluted rainwater.

Hydrogen Ion Concentration

The concentration of hydrogen ions (02H^+02) in a solution is directly tied to its pH value. By knowing the pH, the 02H^+02 concentration can be calculated using the formula \([H^+] = 10^{-pH}\). This relationship shows that as pH decreases, the hydrogen ion concentration increases, making the solution more acidic.
Let's explore how we calculate this for the exercise:

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

Therefore, the hydrogen ion concentration ranges from about \([2.51 \times 10^{-6}, 6.31 \times 10^{-6}]\text{ M}\) for the given pH values. This decrease in 02H^+02 concentration from pH 5.2 to 5.6 reflects a decrease in acidity as the pH moves closer to neutral.

Ion Product of Water

The ion product of water, symbolized as 02K_w02, is a crucial constant in chemistry. It is defined as the product of the concentrations of hydrogen ions (02H^+02) and hydroxide ions (02OH^−02) in water. Mathematically, this is expressed as \([K_w = [H^+][OH^-] = 1.0 \times 10^{-14}]\). This product remains constant at a given temperature (usually 25°C).
How does this work in practice?

• If you know 02H^+02 concentration, you can find 02OH^−02 by rearranging the formula: \([OH^-] =\frac{K_w}{[H^+]}\).

Applying this to the initial problem:

• For pH 5.2 with \([H^+] \approx 6.31 \times 10^{-6}\text{ M}\): \([OH^-] \approx \frac{1.0 \times 10^{-14}}{6.31 \times 10^{-6}} \approx 1.58 \times 10^{-9}\text{ M}\).
• For pH 5.6 with \([H^+] \approx 2.51 \times 10^{-6}\text{ M}\): \([OH^-] \approx \frac{1.0 \times 10^{-14}}{2.51 \times 10^{-6}} \approx 3.98 \times 10^{-9}\text{ M}\).

Thus, the 02OH^−02 concentration ranges from \([1.58 \times 10^{-9}, 3.98 \times 10^{-9}]\text{ M}\), showcasing how even slight changes in 02H^+02 concentration and pH can affect the balance of ions in water.