Problem 50 - Chemistry and Chemical Reactivity

Question

A \(0.015 \mathrm{M}\) solution of a base has a pH of 10.09 (a) What are the hydronium and hydroxide ion concentrations of this solution? (b) Is the base a strong base, a moderately weak base \(\left(K_{\mathrm{b}} \text { of about } 10^{-5}\right)\) Or a very weak base \(\left(K_{\mathrm{b}} \text { of about } 10^{-10}\right) ?\)

Step-by-Step Solution

Answer

(a) \([\text{H}^+] \approx 8.13 \times 10^{-11} \mathrm{M}\), \([\text{OH}^-] \approx 1.23 \times 10^{-4} \mathrm{M}\); (b) Moderately weak base.

Step 1: Convert pH to pOH

To find the pOH, use the formula: \( \text{pH} + \text{pOH} = 14 \). Therefore, \( \text{pOH} = 14 - 10.09 = 3.91 \).

Step 2: Calculate Hydroxide Ion Concentration

Use the formula \( [\text{OH}^-] = 10^{-\text{pOH}} \). Substitute \( \text{pOH} = 3.91 \) to get \( [\text{OH}^-] \approx 10^{-3.91} \approx 1.23 \times 10^{-4} \mathrm{M} \).

Step 3: Calculate Hydronium Ion Concentration

Use the relationship between the concentrations of hydronium and hydroxide ions: \( [\text{H}^+] \times [\text{OH}^-] = 1 \times 10^{-14} \). Substitute \( [\text{OH}^-] = 1.23 \times 10^{-4} \mathrm{M} \) to find \( [\text{H}^+] = \frac{1 \times 10^{-14}}{1.23 \times 10^{-4}} \approx 8.13 \times 10^{-11} \mathrm{M} \).

Step 4: Determine the Strength of the Base

Since \( [\text{OH}^-] \) came from a \( 0.015 \mathrm{M} \) solution of the base, we use \( K_b = \frac{[\text{OH}^-]^2}{[\text{Base}] - [\text{OH}^-]} \). For very small \( [\text{OH}^-] \), \( K_b \approx \frac{[\text{OH}^-]^2}{[\text{Base}]} \). Substitute \( [\text{OH}^-] = 1.23 \times 10^{-4} \mathrm{M} \) and \( [\text{Base}] = 0.015 \mathrm{M} \), and calculate \( K_b \approx \frac{(1.23 \times 10^{-4})^2}{0.015} \approx 1.01 \times 10^{-7} \).

Step 5: Classify Base Strength

Since the calculated \( K_b \approx 1.01 \times 10^{-7} \), this corresponds to a moderately weak base. It is in between a strong base (high \( K_b \)) and a very weak base (\( K_b \approx 10^{-10} \)).

Key Concepts

pH and pOH relationship

Understanding the relationship between pH and pOH is crucial in solving acid-base chemistry problems. In a neutral solution at 25°C, the sum of pH and pOH is always equal to 14.
This relationship can be expressed by the formula:

• \[ \text{pH} + \text{pOH} = 14 \]

This allows us to calculate one value if the other is known. For instance, if a solution has a pH of 10.09, its pOH can be found by subtracting the pH from 14, resulting in a pOH of 3.91.
This relationship helps in determining the acidity or basicity of a solution by providing insight into the concentration of hydroxide ions \([\text{OH}^-]\) from the given pH.

hydronium and hydroxide ions

The concentration of hydronium ions \([\text{H}^+]\) and hydroxide ions \([\text{OH}^-]\) in a solution provides essential information about the solution's nature. These ions are central in water's self-ionization process, expressed as:

• \[ \text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{OH}^- \]

The product of these concentrations is a constant at a given temperature, known as the ionic product of water \(K_w\), given by:

• \[ [\text{H}^+] \, [\text{OH}^-] = 1 \times 10^{-14} \]

Given the concentration of hydroxide ions from pOH (in this case, \([\text{OH}^-] = 1.23 \times 10^{-4} \, \text{M}\)), we can calculate \([\text{H}^+]\) using the relationship:

• \[ [\text{H}^+] = \frac{1 \times 10^{-14}}{1.23 \times 10^{-4}} \]

This results in a hydronium ion concentration \([\text{H}^+] \approx 8.13 \times 10^{-11} \, \text{M}\). By understanding both concentrations, we can better understand a solution's acidic or basic nature.

base ionization constant (Kb)

The base ionization constant, \(K_b\), is a crucial parameter in assessing the strength of a base. It quantifies the extent to which a base can dissociate in water, indicating how productive it is in forming hydroxide ions.
The formula for calculating \(K_b\) is:

• \[ K_b = \frac{[\text{OH}^-]^2}{[\text{Base}] - [\text{OH}^-]} \]

In cases with low hydroxide concentrations compared to the base, this formula can simplify to:

• \[ K_b \approx \frac{[\text{OH}^-]^2}{[\text{Base}]} \]

For our solution, with \([\text{OH}^-] = 1.23 \times 10^{-4} \, \text{M}\) and base concentration \([\text{Base}] = 0.015 \, \text{M}\), the calculated \(K_b\) is approximately \(1.01 \times 10^{-7}\). This result helps to understand whether a base is strong, moderately weak, or very weak.

acid-base strength classification

Classifying the strength of acids and bases helps predict their behavior in chemical reactions. The strength of a base is determined by its base ionization constant, \(K_b\).
Strong bases have high \(K_b\) values, indicating a greater tendency to donate electrons and form hydroxide ions. In contrast, weak bases have significantly lower \(K_b\) values.
They ionize less in water.
Our calculated \(K_b\) of the example base, \(1.01 \times 10^{-7}\), is indicative of a moderately weak base.
This particular value situates the base's strength between a strong base, with a higher \(K_b\), and a very weak base, with \(K_b\) near \(10^{-10}\).
Understanding this classification allows us to foresee how bases will behave in water and how they will interact with other chemical species.