A concentrated solution of aqueous ammonia is 28.0% w/w NH3 and has a density of 0.899 g/mL. What is the molar concentration of NH3 in this solution?
What is pNa for a solution of 1.76 × 10–3 M Na3PO4?
The Correct Answer and Explanation is:
1. Molar concentration of NH₃ in 28.0% w/w aqueous ammonia
Given:
- 28.0% w/w NH₃ → 28.0 g NH₃ in 100 g solution
- Density of solution = 0.899 g/mL
- Molar mass of NH₃ = 17.03 g/mol
Step 1: Convert total mass of solution to volume
Mass of solution = 100 g
Volume = mass / density = 100 g / 0.899 g/mL ≈ 111.23 mL = 0.11123 L
Step 2: Calculate moles of NH₃
Moles NH₃ = 28.0 g / 17.03 g/mol ≈ 1.644 mol
Step 3: Calculate molar concentration (mol/L)
Molarity = moles / volume (L) = 1.644 mol / 0.11123 L ≈ 14.78 M
✅ Answer: 14.78 M NH₃
2. pNa of 1.76 × 10⁻³ M Na₃PO₄ solution
Step 1: Understand dissociation
Na₃PO₄ dissociates completely in water:
Na₃PO₄ → 3Na⁺ + PO₄³⁻
So, for every 1 mole of Na₃PO₄, we get 3 moles of Na⁺.
Step 2: Calculate [Na⁺]
[Na₃PO₄] = 1.76 × 10⁻³ M
⇒ [Na⁺] = 3 × (1.76 × 10⁻³) = 5.28 × 10⁻³ M
Step 3: Calculate pNa
pNa = –log₁₀[Na⁺] = –log₁₀(5.28 × 10⁻³) ≈ 2.28
✅ Answer: pNa = 2.28
Explanation
The molar concentration (or molarity) of a solute in a solution refers to the number of moles of solute per liter of solution. For aqueous ammonia, a 28.0% w/w solution indicates 28.0 g of NH₃ is present in every 100 g of the total solution. Given the density of the solution is 0.899 g/mL, we use this to convert the mass of the solution to volume, which allows us to express molarity in mol/L. By dividing the mass of NH₃ by its molar mass (17.03 g/mol), we obtain the number of moles of ammonia. Dividing this by the calculated volume in liters yields a concentration of 14.78 M, which is highly concentrated, as expected for industrial-grade aqueous ammonia.
On the other hand, pNa is a logarithmic scale similar to pH or pOH, but it refers to the sodium ion concentration. In a solution of sodium phosphate (Na₃PO₄), the salt dissociates completely into 3 sodium ions and one phosphate ion. This stoichiometry is crucial because it triples the effective sodium ion concentration compared to the initial Na₃PO₄ concentration. Multiplying 1.76 × 10⁻³ M by 3 gives the sodium ion concentration of 5.28 × 10⁻³ M. Applying the negative logarithm base 10, the pNa is calculated to be approximately 2.28. Low pNa values indicate high concentrations of Na⁺, which can be important in contexts like electrochemistry, biochemistry, or ion-selective electrode measurements.
