Problem 45 - Chemistry and Chemical Reactivity

Question

A \(0.015 \mathrm{M}\) 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, \(K_{a \prime}\) for the acid?

Step-by-Step Solution

Answer

(a) \([\text{H}_3\text{O}^+] = 2.14 \times 10^{-3} \text{ M}\). (b) \(K_a' \approx 3.56 \times 10^{-4}\).

Step 1: Calculate Hydronium Ion Concentration

The pH of a solution is calculated using the formula: \(\text{pH} = -\log[\text{H}^+]\). We know the pH is 2.67. To find the hydronium ion concentration \([\text{H}_3\text{O}^+])\), we rearrange to: \([\text{H}_3\text{O}^+] = 10^{-\text{pH}}\). Thus, \([\text{H}_3\text{O}^+] = 10^{-2.67}\).

Step 2: Convert to Concentration Value

Plug the pH value into the equation: \([\text{H}_3\text{O}^+] = 10^{-2.67} \approx 2.14 \times 10^{-3} \text{ M}\). So, the hydronium ion concentration is approximately \(2.14 \times 10^{-3} \text{ M}\).

Step 3: Use ICE Table Approach

To find the ionization constant \(K_a'\), we set up an ICE (Initial, Change, Equilibrium) table. Initial concentrations: \([\text{HOCN}]=0.015\) M, \([\text{H}_3\text{O}^+]=0\) M, \([\text{OCN}^-]=0\) M. Change: \([-x, +x, +x]\). At equilibrium: \([\text{HOCN}]=0.015-x\) M, \([\text{H}_3\text{O}^+]=x\) M, \([\text{OCN}^-]=x\) M. Since we found \([\text{H}_3\text{O}^+]=2.14 \times 10^{-3}\), let \(x = 2.14 \times 10^{-3}\).

Step 4: Solve for Ionization Constant

The expression for \(K_a'\) is \(K_a' = \frac{[\text{H}_3\text{O}^+][\text{OCN}^-]}{[\text{HOCN}]}\). Substitute the values: \(K_a' = \frac{(2.14 \times 10^{-3})(2.14 \times 10^{-3})}{0.015 - 2.14 \times 10^{-3}}\).

Step 5: Calculate Ionization Constant

Now compute \(K_a'\): \(K_a' = \frac{(2.14 \times 10^{-3})^2}{0.015 - 2.14 \times 10^{-3}} \approx \frac{4.58 \times 10^{-6}}{0.01286} \approx 3.56 \times 10^{-4}\). The ionization constant \(K_a'\) is approximately \(3.56 \times 10^{-4}\).

Key Concepts

pH Calculation

Understanding the calculation of pH is crucial in acid-base chemistry. The pH scale is a concise way to express the acidity or alkalinity of a solution. It ranges from 0 to 14, with 7 being neutral. A pH less than 7 indicates an acidic solution, while greater than 7 denotes a basic medium. The formula to calculate pH is:

• \( \text{pH} = -\log[\text{H}^+] \)

Here, \([\text{H}^+]\) represents the concentration of hydrogen ions in moles per liter (M).
To find the pH, you need to know the hydronium ion concentration, using the inverse is possible too. In cases where the pH is given, as in the exercise, you can find the hydronium ion concentration

• \([\text{H}_3\text{O}^+] = 10^{-\text{pH}} \)

So, with a pH of 2.67, the hydronium ion concentration is approximately \(2.14 \times 10^{-3} \text{ M}\). This helps us understand the solution’s acidity level effectively.

Ionization Constant

The ionization constant, \(K_a\), is an essential parameter in determining the strength of an acid. It indicates how well an acid donates protons to the solution. A higher \(K_a\) value means a stronger acid. For weak acids, which do not disassociate completely, \(K_a\) helps in calculating equilibrium concentrations.
To find \(K_a\), consider the reaction:

• \( \text{HOCN} \leftrightarrow \text{H}^+ + \text{OCN}^- \)
• \(K_a = \frac{[\text{H}^+][\text{OCN}^-]}{[\text{HOCN}]}\)

Using the initial and equilibrium concentrations from our ICE table, we substitute the known values. Given \([\text{H}_3\text{O}^+]\) is \(2.14 \times 10^{-3} \text{ M}\), and \([\text{OCN}^-] = 2.14 \times 10^{-3} \text{ M}\). Substitute into the formula:
\(K_a' = \frac{(2.14 \times 10^{-3})(2.14 \times 10^{-3})}{0.015 - 2.14 \times 10^{-3}} \approx 3.56 \times 10^{-4}\). This calculated \(K_a'\) value shows that hydrogen cyanate is a weak acid as the \(K_a\) magnitude confirms its partial ionization.

Hydronium Ion Concentration

Determining the hydronium ion concentration is fundamental to understanding the acid character of solutions. Hydronium ions \( [\text{H}_3\text{O}^+] \) result from the disassociation of acids in water, contributing to a solution's acidity. Here, it's derived from the pH value using the relationship:

• \([\text{H}_3\text{O}^+] = 10^{-\text{pH}}\)

For our exercise, substituting 2.67 for the pH yields \([\text{H}_3\text{O}^+] = 2.14 \times 10^{-3} \text{ M}\).
This shows a relatively low concentration, emphasizing a characteristic trait of weak acids:

• They do not fully dissociate in solution.
• The equilibrium shifts towards the left in their ionization reaction.

This moderate hydronium concentration helps in analyzing equilibrium properties of acid solutions further.