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:
Part 1: Molar concentration of NH₃
We are given:
- 28.0% w/w NH₃ → 28.0 g NH₃ per 100 g of solution
- Density = 0.899 g/mL
Step 1: Find the mass of the solution per liter
Since density = mass/volume: Mass of 1 L solution=0.899 g/mL×1000 mL=899 g\text{Mass of 1 L solution} = 0.899 \, \text{g/mL} \times 1000 \, \text{mL} = 899 \, \text{g}
Step 2: Find mass of NH₃ in 1 L solution
28.0% of 899 g is NH₃: Mass of NH₃=0.28×899=251.72 g\text{Mass of NH₃} = 0.28 \times 899 = 251.72 \, \text{g}
Step 3: Convert NH₃ mass to moles
Molar mass of NH₃ = 17.03 g/mol Moles of NH₃=251.7217.03=14.78 mol\text{Moles of NH₃} = \frac{251.72}{17.03} = 14.78 \, \text{mol}
Answer: Molarity=14.78 mol1.00 L=14.78 M\text{Molarity} = \frac{14.78 \, \text{mol}}{1.00 \, \text{L}} = \boxed{14.78 \, \text{M}}
Part 2: pNa of 1.76 × 10⁻³ M Na₃PO₄
Step 1: Dissociation of Na₃PO₄ Na3PO4→3Na++PO43−\text{Na}_3\text{PO}_4 \rightarrow 3\text{Na}^+ + \text{PO}_4^{3-}
For every 1 mol of Na₃PO₄, 3 mol of Na⁺ are produced.
Step 2: Calculate [Na⁺] [Na+]=3×(1.76×10−3)=5.28×10−3 M[\text{Na}^+] = 3 \times (1.76 \times 10^{-3}) = 5.28 \times 10^{-3} \, \text{M}
Step 3: Calculate pNa
pNa is defined as: pNa=−log10[Na+]\text{pNa} = -\log_{10}[\text{Na}^+] pNa=−log10(5.28×10−3)≈2.28\text{pNa} = -\log_{10}(5.28 \times 10^{-3}) \approx \boxed{2.28}
Explanation
In chemistry, concentration calculations and logarithmic expressions are crucial for understanding solution properties. For the ammonia solution, “28.0% w/w” means that 28 grams of ammonia are present per 100 grams of solution. To determine the molarity (mol/L), we must first find the total mass of one liter of solution using its density. With a density of 0.899 g/mL, one liter (1000 mL) weighs 899 grams. Using the 28.0% ratio, the actual mass of ammonia in that volume is 251.72 grams. Dividing this mass by the molar mass of ammonia (17.03 g/mol) yields the number of moles: 14.78 mol. Since this is in one liter, the molarity is 14.78 M. This value indicates a highly concentrated basic solution, typical for industrial-strength ammonia.
For pNa, we use the concept similar to pH. Sodium phosphate (Na₃PO₄) dissociates in water to give 3 Na⁺ ions for every formula unit. Therefore, a 1.76 × 10⁻³ M solution of Na₃PO₄ releases three times as many Na⁺ ions: 5.28 × 10⁻³ M. The pNa is calculated as the negative logarithm of the sodium ion concentration: −log₁₀[Na⁺]. This gives a pNa of about 2.28, meaning the concentration of sodium ions is relatively high. This calculation is important in fields like electrochemistry or biochemistry, where specific ion concentrations affect cell behavior or reaction mechanisms. Understanding how to go from percentage concentrations and stoichiometric relationships to molarity and p-functions (like pH, pKa, or pNa) is fundamental in both lab and theoretical chemistry.
