Problem 88 - Chemistry The Molecular Science

Question

A \(5.00-\mathrm{g}\) sample of white phosphorus is burned in excess oxygen and the product is dissolved in sufficient water to form \(250 . \mathrm{mL}\) of solution. (a) Write the balanced chemical equation for the burning of phosphorus in excess oxygen. (b) Calculate the \(\mathrm{pH}\) of the resulting solution. (c) An excess of aqueous calcium nitrate is added to the solution causing a white precipitate to form. Write a balanced chemical equation for this reaction and calculate the mass of precipitate formed. (d) An excess of zinc is added to the remaining solution. The reaction generates a colorless gas. Identify the gas and calculate its volume at STP.

Step-by-Step Solution

Answer

(a) \(4\text{P} + 5\text{O}_2 \rightarrow \text{P}_4\text{O}_{10}\); (b) \(\text{pH} \approx 0.19\); (c) \(\text{Ca}_3(\text{PO}_4)_2\), mass = 24.98 g; (d) \(\text{H}_2\), volume = 3.61 L.

Step 1: Write the Balanced Chemical Equation for Burning Phosphorus

Phosphorus (P) reacts with excess oxygen (O₂) to form diphosphorus pentoxide (P₄O₁₀). The balanced equation is:\[ \text{4P} + \text{5O}_2 \rightarrow \text{P}_4\text{O}_{10} \]

Step 2: Calculate the Moles of Phosphorus

Given the mass of the phosphorus sample as 5.00 g and its molar mass (P is 30.974 g/mol), we calculate the moles:\[\text{Moles of P} = \frac{5.00 \, \text{g}}{30.974 \, \text{g/mol}} \approx 0.161 \, \text{mol} \]

Step 3: Determine the Moles of P4O10 Formed

From the balanced equation, 4 moles of phosphorus produce 1 mole of P₄O₁₀. Thus, the moles of P₄O₁₀ are:\[\text{Moles of P}_4\text{O}_{10} = \frac{0.161 \, \text{mol of P}}{4} \approx 0.04025 \, \text{mol} \]

Step 4: Calculate the Hydrogen Ion Concentration

P₄O₁₀ reacts with water to form phosphoric acid (H₃PO₄):\[ \text{P}_4\text{O}_{10} + 6\text{H}_2\text{O} \rightarrow 4\text{H}_3\text{PO}_4 \]Each mole of P₄O₁₀ forms 4 moles of H₃PO₄, so:\[\text{Moles of } \text{H}_3\text{PO}_4 = 4 \times 0.04025 = 0.161 \text{ mol}\]The concentration of H₃PO₄ in 0.250 L is:\[[\text{H}_3\text{PO}_4] = \frac{0.161}{0.250} = 0.644 \, \text{M} \]Using the assumption that H₃PO₄ fully dissociates, the [H⁺] is also 0.644 M.

Step 5: Calculate the pH of the Solution

The pH is calculated as follows:\[\text{pH} = -\log_{10}(0.644) \approx 0.19 \]

Step 6: Write the Reaction with Calcium Nitrate

When calcium nitrate is added, phosphoric acid reacts to form calcium phosphate :\[ 2\text{H}_3\text{PO}_4 + 3\text{Ca(NO}_3\text{)}_2 \rightarrow \text{Ca}_3(\text{PO}_4)_2\underline{\phantom{xxx}}(s) + 6\text{HNO}_3 \]

Step 7: Calculate the Mass of Calcium Phosphate Precipitate

The molar mass of Ca₃(PO₄)₂ is approximately 310.18 g/mol. Based on stoichiometry, 1 mole of Ca₃(PO₄)₂ forms from 2 moles of H₃PO₄:\[\text{Moles of } \text{Ca}_3(\text{PO}_4)_2 = \frac{0.161}{2} = 0.0805 \text{ mol}\]Therefore, mass of the precipitate:\[\text{Mass} = 0.0805 \, \text{mol} \times 310.18 \, \text{g/mol} \approx 24.98 \, \text{g} \]

Step 8: Identify the Gas from Zinc Reaction

Zinc reacts with H₃PO₄ to produce hydrogen gas (H₂):\[ \text{Zn} + 2\text{H}_3\text{PO}_4 \rightarrow 3\text{Zn}_3\text{(PO}_4)_2 + \text{H}_2 \]

Step 9: Calculate the Volume of Hydrogen Gas at STP

At STP, 1 mole of any gas occupies 22.4 liters. The number of moles of H₂ formed is equal to half the moles of H₃PO₄, due to stoichiometry:\[\text{Moles of } \text{H}_2 = 0.161 \, \text{mol} \]\[\text{Volume} = 0.161 \, \text{mol} \times 22.4 \, \text{L/mol} \approx 3.61 \, \text{L} \]

Key Concepts

Phosphorus Combustion

Phosphorus is a nonmetal element that can react with oxygen in a chemical process known as combustion. When white phosphorus is subjected to excess oxygen, it undergoes a transformation into diphosphorus pentoxide, denoted as \( \text{P}_4\text{O}_{10} \). This is represented in the balanced chemical equation as:

• \( \text{4P} + \text{5O}_2 \rightarrow \text{P}_4\text{O}_{10} \)

The equation shows that four molecules of phosphorus react with five molecules of oxygen to yield one molecule of diphosphorus pentoxide. Understanding balanced chemical equations is essential in stoichiometry as it ensures that the number of atoms on each side of the equation is equal, showcasing the law of conservation of mass.

pH Calculation

One of the products of phosphorus combustion with water is phosphoric acid (\( \text{H}_3\text{PO}_4 \)). When water is involved, it reacts with diphosphorus pentoxide to form phosphoric acid:

• \( \text{P}_4\text{O}_{10} + 6\text{H}_2\text{O} \rightarrow 4\text{H}_3\text{PO}_4 \)

In an acidic solution, hydrogen ions \( (\text{H}^+) \) influence the pH level. Since \( \text{H}_3\text{PO}_4 \) can dissociate completely, the concentration of hydrogen ions is approximately equal to the concentration of \( \text{H}_3\text{PO}_4 \). pH is determined by the formula:

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

For this solution, with \( [\text{H}^+] = 0.644 \text{ M} \), the pH calculation results in a very low pH of approximately 0.19, indicating a highly acidic medium.

Precipitate Formation

When a solution contains phosphoric acid and excess calcium nitrate is introduced, a solid known as a precipitate forms. The ions in the solution interact leading to the formation of calcium phosphate, a white solid that precipitates out:

• \( 2\text{H}_3\text{PO}_4 + 3\text{Ca(NO}_3\text{)}_2 \rightarrow \text{Ca}_3(\text{PO}_4)_2(s) + 6\text{HNO}_3 \)

This process is a double displacement reaction where calcium ions \( (\text{Ca}^{2+}) \) swap partners with hydrogen ions \( (\text{H}^+) \). The formation of calcium phosphate confirms the presence of a precipitate by demonstrating how product solubility influences chemical reactions in solutions.

Gas Evolution

Gas evolution reactions occur when a gas is released as a product of a chemical reaction. In this exercise, zinc metal reacts with phosphoric acid producing a colorless gas:

• \( \text{Zn} + 2\text{H}_3\text{PO}_4 \rightarrow 3\text{Zn}_3\text{(PO}_4)_2 + \text{H}_2 \)

The gas evolved in this scenario is hydrogen gas \( (\text{H}_2) \), which forms as zinc reacts with the acidic solution. The reaction between a metal and an acid releasing hydrogen gas is a classic example of a gas evolution reaction.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions. It involves using balanced equations to determine the quantities of substances consumed and produced in a reaction. In this scenario:

• 4 moles of phosphorus produce 1 mole of \( \text{P}_4\text{O}_{10} \),
• Each mole of \( \text{P}_4\text{O}_{10} \) results in 4 moles of \( \text{H}_3\text{PO}_4 \).

Also, we calculate that 0.161 mol of \( \text{H}_3\text{PO}_4 \) will lead to evolving 0.161 mol of hydrogen gas. Stoichiometry helps in approximating the theoretical yield and understanding the relationships between different chemicals in a reaction.

Chemical Equations

Chemical equations describe transformations in a succinct form using chemical symbols and formulas to represent the reactants and products. Each equation balances the types and numbers of atoms, ensuring the conservation of mass and charge. In the exercise, we see the transformations:

• Phosphorus burns to form diphosphorus pentoxide,
• Phosphoric acid reacts with calcium nitrate to form a precipitate,
• Zinc and phosphoric acid produce hydrogen gas.

Balanced equations not only provide information about the reactants and products but also reveal the proportionate quantities involved, which is essential for calculations in stoichiometry and subsequent analysis.