Question

A \(10.0-\mathrm{g}\) sample of white phosphorus was burned in an excess of oxygen. The product was dissolved in enough water to make \(500.0 \mathrm{~mL}\) of solution. Calculate the \(\mathrm{pH}\) of the solution at \(25^{\circ} \mathrm{C}\).

Short answer

The pH is approximately 0.19, considering only the first ionization of phosphoric acid.

Step-by-step solution

Determine the reaction equation

The reaction of white phosphorus (\(P_4\)) with oxygen (\(O_2\)) produces phosphorus pentoxide (\(P_4O_{10}\)). The balanced chemical equation is \(P_4 + 5O_2 \rightarrow P_4O_{10}\). Next, \(P_4O_{10}\) will react with water to form phosphoric acid (\(H_3PO_4\)), with the equation: \(P_4O_{10} + 6H_2O \rightarrow 4H_3PO_4\).

Calculate moles of white phosphorus

First, calculate the number of moles of white phosphorus (\(P_4\)) in the 10.0 g sample. The molar mass of white phosphorus is approximately 123.88 g/mol. Moles of \(P_4\) = \(\frac{10.0 \text{ g}}{123.88 \text{ g/mol}} \approx 0.0807\).

Calculate moles of phosphoric acid produced

From Step 1, we see that 1 mole of \(P_4\) produces 4 moles of \(H_3PO_4\). Using this stoichiometry ratio: Moles of \(H_3PO_4\) = \(0.0807 \times 4 = 0.3228\).

Determine the concentration of phosphoric acid solution

The \(H_3PO_4\) is dissolved to make 500.0 mL (0.500 L) of solution. Hence, the concentration \([H_3PO_4]\) is \(\frac{0.3228 \text{ moles}}{0.500 \text{ L}} = 0.6456 \text{ mol/L}\).

Determine the ionization and pH calculation

Phosphoric acid (\(H_3PO_4\)) is a triprotic acid with multiple ionization steps. Considering the first ionization, \(H_3PO_4 \rightarrow H^+ + H_2PO_4^-\), and assuming only the first ionization significantly affects the pH: \[ [H^+] \approx \text{concentration of } H_3PO_4 = 0.6456 \text{ mol/L} \]. \[ \text{pH} = -\log([H^+]) = -\log(0.6456) \approx 0.19 \].

Verify and conclude

Given assumptions like solving only for the first dissociation step can underestimate the pH; the triprotic nature is complex. In reality, the pH may be slightly higher. However, for this calculation's purposes focusing on the first step, the \(\text{pH}\) is estimated around 0.19.

Key concepts

Chemical Reactions

Chemical reactions involve the transformation of substances through the breaking and forming of chemical bonds. In the problem of burning white phosphorus, we start with the element phosphorus in its molecular form, represented as \(P_4\). This is then burned in the presence of oxygen \(O_2\) to form a compound called phosphorus pentoxide \(P_4O_{10}\).

The reaction equation illustrating this transformation is \(P_4 + 5O_2 \rightarrow P_4O_{10}\). This equation is crucial as it provides a balanced picture showing how many molecules of each substance are involved. It helps us grasp the stoichiometry of the reaction, which is the quantitative relationship between reactants and products.

Phosphorus pentoxide then reacts with water \(H_2O\) to produce phosphoric acid \(H_3PO_4\) as shown in the equation \(P_4O_{10} + 6H_2O \rightarrow 4H_3PO_4\). This second reaction is important as it forms our central compound of interest, phosphoric acid, which impacts the overall chemical equilibrium and the ultimate pH of the solution.

Acid-Base Chemistry

At the heart of acid-base chemistry lies the concept of acids and bases and how they interact in solutions. Phosphoric acid \(H_3PO_4\) is a prime example of a triprotic acid, meaning it can donate three protons \(H^+\) in a step-wise manner. In the context of this problem, we consider primarily the first ionization step where \(H_3PO_4\) donates a single proton:

\(H_3PO_4 \rightarrow H^+ + H_2PO_4^-\).

This is significant because the first dissociation step largely determines the initial acidity of the solution. However, phosphoric acid can dissociate further:

- Second step: \(H_2PO_4^- \rightarrow H^+ + HPO_4^{2-}\)
- Third step: \(HPO_4^{2-} \rightarrow H^+ + PO_4^{3-}\)

Each step contributes to the overall ion balance, but the effect on pH decreases with each step. Understanding these dissociation steps is essential in predicting the behavior of phosphoric acid in a solution and how it influences pH.

Molarity

Molarity is a core concept in chemistry to express the concentration of a solution. It is defined as the number of moles of a solute per liter of solution. This problem involves calculating the molarity of phosphoric acid \(H_3PO_4\) after dissolving the phosphorus combustion product into water.

After determining the amount of \(H_3PO_4\) produced, we find it to be approximately 0.3228 moles. Since this is dissolved in 500.0 mL of water, which is equivalent to 0.500 liters, we calculate the molarity as:

\[ [H_3PO_4] = \frac{0.3228 \text{ moles}}{0.500 \text{ L}} = 0.6456 \text{ mol/L} \]

This concentration is crucial as it represents the initial condition used in the next step of pH calculation. The molarity provides us with a quantitative measure that is indispensable for understanding the solution's strength and the extent of the acid's dissociation.

pH Calculation

The pH of a solution is a measure of its acidity or basicity, with lower pH values indicating higher acidity. In the case of phosphoric acid, we calculate the pH based on its ionization in water. Phosphoric acid, being a triprotic acid, dissociates in steps, but often the first ionization step has the most significant impact on the pH calculation.

Given the concentration of \([H_3PO_4]\) as 0.6456 mol/L and simplifying by considering only the first ionization step, we estimate:

- \([H^+] \approx 0.6456 \text{ mol/L}\)
- \(\text{pH} = -\log([H^+]) \approx -\log(0.6456) \approx 0.19\)

This estimate uses the assumption that subsequent ionization contributions are minimal compared to the first. However, in practice, considering all ionizations would adjust the pH slightly higher due to cumulative effects of the additional \(H^+\) ions from further dissociations. It's important for students to recognize the assumptions and approximations made in such calculations and to understand that more precise methods could yield a different pH value.