Problem 41 - Chemistry and Chemical Reactivity

Question

A 0.015 M solution of hydrogen cyanate, HOCN, has a pH of 2.67 (a) What is the hydronium ion concentration in the solution? (b) What is the ionization constant, \(K_{\alpha}\), for the acid?

Step-by-Step Solution

Answer

(a) The hydronium ion concentration is approximately \(2.14 \times 10^{-3} \, M\). (b) The ionization constant \(K_a\) is approximately \(3.56 \times 10^{-4}\).

Step 1: Calculate Hydronium Ion Concentration

First, we need to find the hydronium ion concentration from the given pH value. The formula to find the hydronium ion concentration \( [H_3O^+] \) from pH is \[ [H_3O^+] = 10^{-pH} \]. Substitute the given pH value: \[ [H_3O^+] = 10^{-2.67} \].Using a calculator, \[ [H_3O^+] \approx 2.14 \times 10^{-3} \, M \].

Step 2: Use Acid Dissociation Equation

The dissociation of HOCN in the solution is given by the equation:\[ \text{HOCN (aq)} \rightleftharpoons \text{H}^+ (\text{aq}) + \text{OCN}^- (\text{aq}) \]At equilibrium, \([H^+] = [OCN^-] = x \), where \( x \) is the concentration of ions dissociated.

Step 3: Calculate Ionization Constant \(K_{a}\)

The acid dissociation constant \( K_a \) is calculated using the expression:\[ K_a = \frac{[H^+][OCN^-]}{[HOCN]} \]Since \( [H^+] = [OCN^-] = 2.14 \times 10^{-3} \) M and the initial concentration of HOCN is 0.015 M, the expression becomes:\[ K_a = \frac{(2.14 \times 10^{-3})^2}{0.015 - 2.14 \times 10^{-3}} \]Now calculate:\[ K_a = \frac{4.58 \times 10^{-6}}{0.01286} \approx 3.56 \times 10^{-4} \].

Key Concepts

Hydronium Ion Concentration

To understand the hydronium ion concentration in a solution, it is crucial to first know what a hydronium ion, \( [H_3O^+] \), is. This ion is formed when a proton \( (H^+) \) combines with a water molecule. In the context of acids, hydronium ions are essential as they essentially define the acidity of a solution. Calculating the hydronium ion concentration from the pH is straightforward. The pH is a measure of how acidic or basic a solution is, and it's calculated as the negative logarithm (base 10) of the hydronium ion concentration. Therefore, to find \( [H_3O^+] \) from pH, you'll use:- The formula: \( [H_3O^+] = 10^{-pH} \).- For a pH of 2.67, the calculation is \( [H_3O^+] = 10^{-2.67} \).- This typically requires a calculator, resulting in \( [H_3O^+] \approx 2.14 \times 10^{-3} \, M \).This simple calculation not only helps in identifying how acidic a solution is, but it also sets the foundation for further calculations like determining the ionization constant.

Ionization Constant

The ionization constant, especially for acids denoted as \( K_a \), is a crucial concept for understanding acid strength. It measures the extent to which an acid dissociates in water. A larger \( K_a \) value indicates a stronger acid, meaning it dissociates more completely.For hydrogen cyanate, HOCN, the dissociation can be written as:- \[\text{HOCN (aq)} \rightleftharpoons \text{H}^+ (\text{aq}) + \text{OCN}^- (\text{aq})\]- At equilibrium, the concentration of \( \text{H}^+ \) ions is equal to that of \( \text{OCN}^- \) ions, denoted by \( x \).The ionization constant \( K_a \) is calculated by:- \( K_a = \frac{[H^+][OCN^-]}{[HOCN]} \)- Given \( [H^+] = [OCN^-] = 2.14 \times 10^{-3} \) M and the initial \( [HOCN] = 0.015 \, M \), the calculation becomes: \( K_a = \frac{(2.14 \times 10^{-3})^2}{0.015 - 2.14 \times 10^{-3}} \).Finally, the result is approximately \( K_a \approx 3.56 \times 10^{-4} \), showing the degree to which HOCN dissociates into ions.

pH Calculation

Calculating the pH of a solution is a fundamental aspect of understanding its acidity or basicity. The term pH stands for "potential of hydrogen" and it represents the negative logarithm of the hydronium ion concentration.Here's how you calculate pH:

• Use the formula: \( \text{pH} = -\log_{10} [H_3O^+] \).
• In our given problem, if \( [H_3O^+] \approx 2.14 \times 10^{-3} \, M \), you’d calculate the pH by substituting this value into the formula.

To obtain the pH:- Use a scientific calculator or logarithm tables to find \( \text{pH} = -\log_{10} (2.14 \times 10^{-3}) \).- The result will confirm the pH provided in the problem, matching closely to 2.67.Understanding this calculation helps in a wide range of chemistry applications, from industrial synthesis to the biochemical processes within the human body.