Question

Isocyanic acid \((\mathrm{HNCO})\) can be prepared by heating sodium cyanate in the presence of solid oxalic acid according to the equation $$ 2 \mathrm{NaOCN}(s)+\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(s) \longrightarrow 2 \mathrm{HNCO}(l)+\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(s) $$ Upon isolating pure HNCO \((l),\) an aqueous solution of HNCO can be prepared by dissolving the liquid HNCO in water. What is the pH of a 100 -mL solution of HNCO prepared from the reaction of 10.0 g each of NaOCN and \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4},\) assuming all of the HNCO produced is dissolved in solution? \(\left(K_{\mathrm{a}} \text { of HNCO }\right.\) \(=1.2 \times 10^{-4} . )\)

Short answer

The pH of the 100 mL HNCO solution prepared from the reaction of 10.0 g each of NaOCN and H₂C₂O₄ is approximately 1.79.

Step-by-step solution

Determine the limiting reactant

To determine the limiting reactant, first, we need to calculate the moles of NaOCN and H2C2O4 given the masses. Moles of NaOCN = \(\frac{\text{mass NaOCN}}{\text{molar mass NaOCN}} = \frac{10.0 \text{g}}{82.01 \text{g/mol}} = 0.1219 \text{mol}\) Moles of H₂C₂O₄ = \(\frac{\text{mass H₂C₂O₄}}{\text{molar mass H₂C₂O₄}} = \frac{10.0 \text{g}}{90.03 \text{g/mol}} = 0.1111 \text{mol}\) Next, compare the mole ratio between the reactants and the stoichiometry of the balanced chemical equation, which shows that 2 moles of NaOCN react with 1 mole of H₂C₂O₄: Mole ratio = \(\frac{0.1219 \text{mol}\;\text{NaOCN}}{0.1111 \text{mol}\;\text{H₂C₂O₄}} = 1.097\) Since the mole ratio (1.097) is greater than the required ratio (2:1 = 2), H₂C₂O₄ is the limiting reactant.

Calculate the moles of HNCO produced

Using the stoichiometry of the balanced chemical equation, we can calculate the moles of HNCO produced from the limiting reactant: Moles of HNCO = 2 × moles of H₂C₂O₄ = 2 × 0.1111 mol = 0.2222 mol

Calculate the concentration of HNCO in a 100 mL solution

To compute the concentration of HNCO, we will divide the moles of HNCO by the volume of the solution in liters: \[HNCO\: concentration = \frac{0.2222\: mol}{0.100\: L} = 2.222\: M\]

Determine the pH of the HNCO solution using its Ka value

As HNCO is a weak acid and its dissociation is not complete, it will establish the following equilibrium in water: \[ HNCO(aq) + H2O(l) \rightleftharpoons H3O^{+}(aq) + NCO^-(aq) \] We will use the Ka expression for the equilibrium: \[K_a = \frac{[H_3O^{+}][NCO^-]}{[HNCO]}\] Since the initial concentration of HNCO is significantly greater than K_a, we can assume that x moles of HNCO dissociate, resulting in a negligible change: \[ K_a = \frac{x^2}{(2.222 - x)} \] Plug in the given value for K_a and solve for x: \(1.2 \times 10^{-4} = \frac{x^2}{(2.222 - x)}\) Since K_a is small, we can further simplify by assuming x is negligible compared to 2.222: \(x^2 ≈ 1.2 \times 10^{-4} \times 2.222\) \(x ≈ \sqrt{2.666 \times 10^{-4}}\) \(x ≈ 1.63 \times 10^{-2}\) Since x represents the concentration of H³O⁺: \[[H_3O^{+}] \approx 1.63 \times 10^{-2} M\] Now we can calculate the pH of the solution: \[pH = -\log_{10} [H_3O^{+}] = -\log_{10} (1.63 \times 10^{-2}) \approx 1.79\] The pH of the 100 mL HNCO solution is approximately 1.79.

Key concepts

Limiting Reactant

In chemical reactions, the limiting reactant is the substance that is entirely consumed first and thus determines the amount of product that can be formed. To identify the limiting reactant, you need to compare the mole ratio from the balanced chemical equation to the mole ratio from your given quantities.
For the reaction involving sodium cyanate (\(\mathrm{NaOCN}\)) and oxalic acid (\(\mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4}\)), the balanced equation shows that 2 moles of \(\mathrm{NaOCN}\) reacts with 1 mole of \(\mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4}\).

• First, calculate the moles of each reactant using their given masses and molar masses.
• Then, determine the actual mole ratio and compare it to the stoichiometric ratio required by the balanced equation.

In this problem, the moles of \(\mathrm{NaOCN}\) divided by the moles of \(\mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4}\) yields a ratio greater than 2, indicating that \(\mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4}\) is the limiting reactant. This means that all of the \(\mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4}\) will be used up when the reaction reaches completion.

Mole Calculations

Moles are a fundamental unit in chemistry that measures the amount of a substance. Calculating moles accurately allows chemists to predict how much product will form in a reaction.
To determine the moles of a substance, use the formula:
\[\text{Moles} = \frac{\text{Mass of substance}}{\text{Molar mass of substance}}\]
In this exercise, you start by calculating the moles of sodium cyanate and oxalic acid based on their given masses.

• Sodium cyanate has a molar mass of 82.01 g/mol, and by dividing the provided mass (10.0 g) by its molar mass, you find 0.1219 mol of sodium cyanate.
• Oxalic acid has a molar mass of 90.03 g/mol, resulting in 0.1111 mol when the mass is divided by its molar mass.

You can see how these calculations are integral for determining which reactant is limiting, as dictated by the mole ratio provided by the balanced chemical equation.

Weak Acid Dissociation

Weak acids, such as isocyanic acid (HNCO), do not completely ionize in solution. Understanding their dissociation helps in determining the hydrogen ion (\([H_3O^+]\)) concentration, which is a key factor in calculating pH.
For HNCO, the dissociation can be represented as:
\[HNCO(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + NCO^-(aq)\]
Since HNCO is a weak acid, its dissociation is incomplete, establishing an equilibrium state. Using the given equilibrium constant, \(K_a = 1.2 \times 10^{-4}\), you can use the expression:
\[K_a = \frac{[H_3O^+][NCO^-]}{[HNCO]}\]
From here, you estimate dissociation by simplifying the equation to solve for the concentration of \([H_3O^+]\). Recognizing that HNCO's initial concentration (2.222 M) is much larger than \(K_a\), we assume that the change in concentration due to dissociation (\(x\)) is negligible, making the math straightforward.

pH Calculation

pH is a measure of the acidity or basicity of a solution. To find the pH of a solution involves determining the concentration of hydrogen ions and then applying the formula:
\[pH = -\log_{10}[H_3O^+]\]
The earlier steps involving the weak acid dissociation provide the concentration of \([H_3O^+]\), which you've calculated as approximately \(1.63 \times 10^{-2}\, M\). Using this value in the pH formula results in:
\[pH = -\log_{10}(1.63 \times 10^{-2}) \approx 1.79\]
This low pH value aligns with the expectation for a solution of a weak acid with relatively high concentration, indicating its acidic nature. Understanding this calculation is crucial for comprehending many chemical processes where pH impacts reactivity and stability.