Problem 46 - Chemistry and Chemical Reactivity

Question

A 0.015 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 \((K_{\mathrm{b}} \text { of about } 10^{-5}),\) or a very weak base \((K_{\mathrm{b}}\) of about \(10^{-10}) ?\)

Step-by-Step Solution

Answer

The hydronium concentration is \(8.13 \times 10^{-11}\;\text{M}\), and hydroxide is \(1.23 \times 10^{-4}\;\text{M}\). It is a very weak base.

Step 1: Calculating pOH from pH

To find the pOH, we use the relation between pH and pOH at 25°C: \( ext{pH} + ext{pOH} = 14\). Given \(\text{pH} = 10.09\), we calculate \(\text{pOH} = 14 - 10.09 = 3.91\).

Step 2: Finding Hydroxide Ion Concentration

The hydroxide ion concentration \([\text{OH}^-]\) can be found using \([\text{OH}^-] = 10^{-\text{pOH}}\). Substitute \(\text{pOH} = 3.91\): \( [\text{OH}^-] = 10^{-3.91} = 1.23 \times 10^{-4} \ \text{M}\).

Step 3: Calculating Hydronium Ion Concentration

The hydronium ion concentration \([\text{H}_3\text{O}^+]\) is related to \([\text{OH}^-]\) by the ion product of water \(K_w = 1.0 \times 10^{-14}\) at 25°C: \([\text{H}_3\text{O}^+] = \frac{K_w}{[\text{OH}^-]}\).Using \([\text{OH}^-] = 1.23 \times 10^{-4} \ \text{M}\), we calculate \([\text{H}_3\text{O}^+] = \frac{1.0 \times 10^{-14}}{1.23 \times 10^{-4}} = 8.13 \times 10^{-11} \ \text{M}\).

Step 4: Determining the Nature of the Base

To classify the base, we first calculate \(K_b\). From the base ionization reaction, \(B + H_2O \rightleftharpoons BH^+ + OH^-\), if \([B]_{initial} = 0.015 \ \text{M}\) and \([\text{OH}^-] = 1.23 \times 10^{-4} \ \text{M}\), then \(x = [\text{OH}^-] = 1.23 \times 10^{-4}\).\(K_b = \frac{[BH^+][OH^-]}{[B]} \approx \frac{(1.23 \times 10^{-4})(1.23 \times 10^{-4})}{0.015}\). Hence, \(K_b = 1.01 \times 10^{-7}\). This value corresponds to a very weak base \((K_b \approx 10^{-10})\).

Key Concepts

pH and pOH relationship

Understanding the relationship between pH and pOH is crucial when working with aqueous solutions. At 25°C, the sum of pH and pOH is always equal to 14.0. This relationship can be represented by the equation:

• \( ext{pH} + ext{pOH} = 14 \)

This equation helps us determine the pOH if we know the pH, and vice versa. For example, if a solution has a pH of 10.09, you can find the pOH by simply subtracting the pH from 14. Thus, \( ext{pOH} = 14 - 10.09 = 3.91 \). This calculation is essential in determining the hydroxide ion concentration, which is important for characterizing the nature of the solution.

hydronium ion concentration

Hydronium ion concentration \([ ext{H}_3 ext{O}^+]\) indicates how acidic a solution is. In water, hydronium ions form when water molecules combine with hydrogen ions. To find \([ ext{H}_3 ext{O}^+]\), we often use the relationship with hydroxide ion concentration due to the ion product of water. Using the ion product formula:

• \( [ ext{H}_3 ext{O}^+] = \frac{K_w}{[ ext{OH}^-]} \)

Where \( K_w = 1.0 \times 10^{-14} \) at 25°C. For example, if the hydroxide ion concentration \([ ext{OH}^-] = 1.23 \times 10^{-4} \, ext{M}\), then \([ ext{H}_3 ext{O}^+] = \frac{1.0 \times 10^{-14}}{1.23 \times 10^{-4}} = 8.13 \times 10^{-11} \, ext{M}\). This very low concentration of hydronium ions indicates that the solution is basic.

hydroxide ion concentration

The concentration of hydroxide ions \([ ext{OH}^-]\) in a solution is a key indicator of its basicity. To find \([ ext{OH}^-]\), we can use the calculated pOH:

• \( [ ext{OH}^-] = 10^{- ext{pOH}} \)

For instance, given a \( ext{pOH} = 3.91\), the hydroxide ion concentration is \([ ext{OH}^-] = 10^{-3.91} = 1.23 \times 10^{-4} \, ext{M}\). Knowing the hydroxide ion concentration allows us to find the hydronium ion concentration using the water ion product. Thus, understanding \([ ext{OH}^-]\) is essential for calculating other related concentrations and determining the strength of the base in the solution.

ion product of water

The ion product of water \(K_w\) is fundamental in understanding aqueous chemistry. It's the product of the concentrations of hydrogen ions \([ ext{H}^+]\) and hydroxide ions \([ ext{OH}^-]\) in water:

• \( K_w = [ ext{H}_3 ext{O}^+][ ext{OH}^-] \)

At 25°C, \(K_w\) is always \(1.0 \times 10^{-14}\), which is central to all aqueous solution calculations. This means that if you have the concentration of one ion, you can find the other by rearranging the formula: \([ ext{H}_3 ext{O}^+] = \frac{K_w}{[ ext{OH}^-]}\). This relationship is particularly useful in solutions where you need to balance acidity and basicity, such as in the given exercise.

base ionization constant (Kb)

The base ionization constant \(K_b\) measures the strength of a base in solution. It is derived from the equilibrium of the reaction involving the base's ionization in water:

• \( B + H_2O \rightleftharpoons BH^+ + OH^- \)

\(K_b\) is calculated as:

• \( K_b = \frac{[BH^+][OH^-]}{[B]} \)

In the given problem, with a hydroxide ion concentration \([ ext{OH}^-] = 1.23 \times 10^{-4} \, ext{M}\) and an initial base concentration \([B] = 0.015 \, ext{M}\), we find that \(K_b = 1.01 \times 10^{-7}\). This indicates that the base is very weak compared to bases with higher \(K_b\) values, which would ionize more fully in water.