Question

The \(\mathrm{pH}\) of a \(1.00 \times 10^{-2} \mathrm{M}\) solution of cyanic acid (HOCN) is \(2.77\) at \(25^{\circ} \mathrm{C}\). Calculate \(K_{\mathrm{a}}\) for \(\mathrm{HOCN}\) from this result.

Short answer

Given the pH of $2.77$, we can calculate the [H+]: \([H+] = 10^{-2.77} = 1.68 \times 10^{-3} \mathrm{M}\). We find the equilibrium concentrations: [HOCN] = \(8.32 \times 10^{-3} \mathrm{M}\), [H+] = \(1.68 \times 10^{-3} \mathrm{M}\), and [OCN-] = \(1.68 \times 10^{-3} \mathrm{M}\). Using the equilibrium expression: \(K_a = [H+][OCN^-] / [HOCN] = 3.37 \times 10^{-4}\).

Step-by-step solution

Calculate the concentration of H+ ions

Given the pH of the solution, we can determine the concentration of H+ ions using the formula: pH = -log[H+] Rewrite the formula for [H+]: [H+] = 10^(-pH) Now we will plug in the given pH value: [H+] = 10^(-2.77) = \(1.68 \times 10^{-3} \mathrm{M}\) Step 2: Find equilibrium concentrations of all species

Calculate equilibrium concentrations of HOCN, H+, and OCN-

Since the initial concentration of HOCN is 1.00 × 10^-2 M and the change in the concentration of the H+ ions is equal to that of HOCN, we can deduce the equilibrium concentrations. [HOCN] = initial concentration - change in concentration = 1.00 × 10^-2 M - 1.68 × 10^-3 M = \(8.32 \times 10^{-3} \mathrm{M}\) [H+] = 1.68 × 10^-3 M (as calculated earlier) [OCN-] = 1.68 × 10^-3 M (since it is proportional to H+) Step 3: Calculate Ka

Use the equilibrium expression to calculate Ka

Now that we have the equilibrium concentrations of all species, we can use the equilibrium expression to find Ka: Ka = [H+][OCN-] / [HOCN] Ka = (1.68 × 10^-3)(1.68 × 10^-3) / (8.32 × 10^(-3)) Ka = \(3.37 \times 10^{-4}\) Thus, the Ka for HOCN is \(3.37 \times 10^{-4}\).

Key concepts

Cyanic Acid

Cyanic acid, represented chemically as HOCN, is a relatively weak acid that plays a significant role in acid-base equilibrium studies. It dissociates in water to produce hydrogen ions, \([H^{+}]\), and cyanate ions, \([OCN^{-}]\). This dissociation is not complete, which means only a small percentage of the cyanic acid molecules donate protons to the water. Understanding the level of this dissociation is fundamental to understanding cyanic acid’s behavior in solutions.
Cyanic acid is interesting because of its role in organic chemistry and its applications in synthesizing other chemicals. Its dissociation in water allows chemists to assess its strength by calculating its acid dissociation constant, \(K_{a}\). This constant is crucial for predicting the extent to which cyanic acid will ionize under different conditions.

pH Calculation

To understand the acidity or alkalinity of a solution, we use the pH scale, which ranges from 0 to 14. The pH value is the negative logarithm of the hydrogen ion concentration: \(\mathrm{pH} = -\log[H^{+}]\). Lower pH values indicate higher concentrations of hydrogen ions and, therefore, stronger acidity.
For any solution of cyanic acid, knowing the pH is the first step in calculating the concentration of hydrogen ions (\([H^{+}]\)) in the solution. By using the pH provided (like 2.77 in the original exercise), you can determine that the hydrogen ion concentration is \(1.68 \times 10^{-3} \ \mathrm{M}\). This information serves as a crucial building block for further calculations of equilibrium concentrations and acid dissociation constant.

Equilibrium Concentration

When cyanic acid is dissolved in water, it reaches a state of equilibrium where the rate of dissociation into \(H^{+}\) and \(OCN^{-}\) ions is equal to the rate of recombination of these ions into HOCN. At equilibrium, the concentrations of these species are stable, which allows us to calculate them precisely.
In the example given, the equilibrium concentration of \(H^{+}\) can be derived from the pH, and it's found to be \(1.68 \times 10^{-3} \, \mathrm{M}\). Since the dissociation reaction is balanced, the concentration of \(OCN^{-}\) is the same at equilibrium. Meanwhile, the remaining concentration of undissociated cyanic acid can be determined by subtracting the concentration of \(H^{+}\) ions from the initial concentration of cyanic acid, leading to \(8.32 \times 10^{-3} \, \mathrm{M}\) for HOCN.

Acid-Base Equilibrium

The concept of acid-base equilibrium is central to understanding the behavior of cyanic acid in solution. In the dissociation reaction of HOCN, there is an establishment of a dynamic balance between the reactants (HOCN) and products (H\(^{+}\) and OCN\(^{-}\)).
In any weak acid like cyanic acid, the equilibrium lies closer to the reactants because only a small fraction of the acid ionizes. The equilibrium expression for this process is given by: \K_{a} = \frac{[H^{+}][OCN^{-}]}{[HOCN]}\. The acid dissociation constant, \(K_{a}\), quantifies the extent of dissociation and, hence, the strength of the acid. For cyanic acid, the calculated \(K_{a}?????????????????$\) of \[3.37 \times 10^{-4}\], signifies its status as a weak acid, which is crucial for predicting its behavior in chemical reactions and solutions.