Question

Isocyanic acid (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(I), an aqueous solution of HNCO can be prepared by dissolving the liquid HNCO in water. What is the \(\mathrm{pH}\) of a \(100 .\) -mL solution of HNCO prepared from the reaction of \(10.0 \mathrm{g}\) each of \(\mathrm{NaOCN}\) and \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4},\) assuming all of the HNCO produced is dissolved in solution? \((K_{\mathrm{a}}\) of HNCO \(\left.=1.2 \times 10^{-4} .\right)\)

Short answer

The limiting reactant is H2C2O4, and 0.0806 moles of HNCO are produced. The concentration of HNCO in the 100 mL solution is 0.806 M. The concentration of H3O+ is 0.00308 M, and the pH of the solution is 2.51.

Step-by-step solution

Determine the limiting reactant

First, we need to find out which reactant will be the limiting reactant in the reaction. To do this, we will calculate the number of moles of each reactant used and compare them using the stoichiometry of the balanced equation. The balanced equation is as follows: \[2 NaOCN(s) + H_{2}C_{2}O_{4}(s) \longrightarrow 2 HNCO(l) + Na_{2}C_{2}O_{4}(s)\] Given that 10.0 g of each reactant is used, Moles of NaOCN = \(\frac{10.0 \ g}{Mass \ of \ NaOCN}\) Moles of H2C2O4 = \(\frac{10.0 \ g}{Mass \ of \ H_2C_2O_4}\) Calculate the molar mass of NaOCN and H2C2O4 and find the moles of each reactant.

Calculate the moles of HNCO produced

Using the stoichiometry of the balanced equation and the moles of the limiting reactant, we can calculate the moles of HNCO produced in the reaction: Moles of HNCO = Moles of limiting reactant * (Moles of HNCO produced / Moles of limiting reactant used)

Calculate the concentration of HNCO in the final solution

We are given the volume of the final solution, 100 mL. To find the concentration of HNCO in the solution, we can use the following formula: \[HNCO\: Concentration (M) = \frac{Moles\:of\:HNCO}{Volume\:of\:solution (L)}\]

Determine the concentration of H3O+ using Ka

Now that we have the concentration of HNCO in the solution, we can use the Ka expression for the reaction of HNCO with water to find the concentration of H3O+ ions: \[Ka = \frac{[H_3O^+][NCO^-]}{[HNCO]}\] \[1.2 \times 10^{-4} = \frac{[H_3O^+][H_3O^+]}{[HNCO]}\] Solve for H3O+ concentration.

Calculate the pH of the final solution

Finally, we can use the concentration of H3O+ to determine the pH of the solution using the following formula: \[pH = -\log[H_3O^+]\] Solve for the pH of the final solution. By following these steps, we can determine the pH of the 100 mL solution of HNCO prepared from the reaction of 10.0 g each of NaOCN and H2C2O4.

Key concepts

Isocyanic Acid (HNCO)

Isocyanic acid, represented as HNCO, is an interesting compound that forms when sodium cyanate (NaOCN) reacts with oxalic acid \((H_2C_2O_4)\). This reaction produces liquid HNCO along with sodium oxalate \((Na_2C_2O_4)\) as a byproduct. Isocyanic acid is often studied in its aqueous form for the purpose of understanding its acidic properties.

When dissolved in water, HNCO behaves as a weak acid, partially dissociating into hydronium ions \([H_3O^+]\) and cyanate ions \([NCO^-]\). The dissociation can be represented by the equilibrium expression:

• \(HNCO(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + NCO^-(aq)\)

This behavior is characterized by its acid dissociation constant \(Ka\), which is a measure of the acid strength. HNCO's \(Ka\) value is given as \(1.2 \times 10^{-4}\), indicating it's a weak acid, meaning it does not completely ionize in solution, and its position of equilibrium lies towards the reactants.

Limiting Reactant

In chemical reactions, identifying the limiting reactant is crucial because it determines the amount of product that can be formed. The limiting reactant is the reagent that is completely consumed in a reaction, thus determining the maximum yield of product.

To find the limiting reactant, we start by calculating the moles of each reactant. This can be done using the formula:

• \(\text{Moles} = \frac{\text{Mass of substance (g)}}{\text{Molar mass (g/mol)}}\)

By plugging in the mass and molar mass of \(NaOCN\) and \(H_2C_2O_4\), you can calculate their moles and compare them in the context of the given equation:

• \(2\, \text{NaOCN}(s) + \text{H}_2\text{C}_2\text{O}_4(s) \rightarrow 2\, \text{HNCO}(l) + \text{Na}_2\text{C}_2\text{O}_4(s)\)

Once the moles are calculated for each reactant, compare the ratios derived from the balanced equation to determine which one limits the progression of the reaction.

Stoichiometry

Stoichiometry involves using the balanced chemical equation to relate masses, moles, and volumes amongst the reactants and products. It's a fundamental concept in chemistry that allows precise predictions of quantities involved in chemical reactions.

In the case of HNCO's preparation, the stoichiometric coefficients in the equation tell us the proportional relationships between reactants and products. For example, the equation \(2\, \text{NaOCN} + \text{H}_2\text{C}_2\text{O}_4 \rightarrow 2\, \text{HNCO} + \text{Na}_2\text{C}_2\text{O}_4\) tells us that:

• 2 moles of \(NaOCN\) react with 1 mole of \(H_2C_2O_4\)
• to produce 2 moles of \(HNCO\)

Using this relationship, once the limiting reactant is identified, the theoretical yield of \(HNCO\) can be calculated. We achieve this by multiplying the moles of the limiting reactant by the stoichiometric ratio of \(HNCO\) to the limiting reactant.

pH Calculation

Calculating the pH of a solution is essential when dealing with acids like HNCO. pH is a measure of the acidity of a solution, calculated from the concentration of hydronium ions \([H_3O^+]\) present.

First, find the \([H_3O^+]\) concentration using the acid dissociation constant \(Ka\) of HNCO and its concentration in solution. Set up the equilibrium expression:

• \(Ka = \frac{[H_3O^+][NCO^-]}{[HNCO]}\)

Given that the initial concentration of \([H_3O^+]\) and \([NCO^-]\) is essentially zero, and they form in equal amounts due to dissociation of \(HNCO\), the expression simplifies to:

• \(Ka = \frac{[H_3O^+]^2}{[HNCO]}\)

Solve for \([H_3O^+]\) and then calculate pH using:

• \(pH = -\log[H_3O^+]\)

This process enables us to predict the level of acidity of the final aqueous \(HNCO\) solution.