Question

A typical commercial grade aqueous phosphoric acid is \(75 \% \mathrm{H}_{3} \mathrm{PO}_{4}\) by mass and has a density of \(1.57 \mathrm{g} / \mathrm{mL}\) What is the molarity of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) in this solution?

Short answer

The molarity of H3PO4 in the solution is 11.95 M.

Step-by-step solution

Calculation of Mass of the Solute

Extract 100g of the solution since it's more practical to calculate with base 100% when dealing with percentages. Given this, the solution consists of 75g H3PO4 (from the 75% w/w information)

Calculate the moles of H3PO4

To convert grams to moles, the molar mass of H3PO4 is required. It can be calculated as \(3 \times 1.01g/mol (H) + 15.99 g/mol (P) + 4 \times 16.00g/mol (O) = 97.99g/mol\). Thus, the moles of H3PO4 in 75g is \( \frac{75 g}{97.99 g/mol} = 0.765 mol\)

Calculate the volume of the solution in liters

With the given density of 1.57 g/mL, convert this density to g/L to obtain \(1.57 g/mL \times 1000 = 1570 g/L \). Then, find the volume occupied by 100g of the solution by using the formula 'mass = volume \times density' rearranged to 'volume = mass / density' giving \( \frac{100g}{1570 g/L} = 0.064 L \)

Compute the Molarity

Finally, molarity can be calculated by substituting the obtained values into the molarity formula: \( Molarity = \frac{Amount of Solute (moles)}{Volume of solution (L)} = \frac{0.765mol}{0.064L} \)= 11.95 M.

Key concepts

Density

Density refers to how much mass is contained within a certain volume. In this context, it is the mass of the phosphoric acid solution per unit volume. Density is an essential concept because it helps us connect mass with volume, two critical factors when dealing with solutions.
The density given is 1.57 g/mL, which means that each milliliter of the solution contains 1.57 grams of mass. This information allows us to convert between different units of measurement, such as grams and liters, which is crucial when calculating molarity.
In practical terms, you can transform this value to g/L by multiplying by 1000, making it easier to use in subsequent calculations. Thus, 1.57 g/mL becomes 1570 g/L, facilitating volume adjustments during molarity determinations.

Molar mass calculation

Molar mass is a fundamental concept in chemistry that allows us to convert between grams and moles, effectively bridging the gap between macroscopic measurements and molecular scale quantities. The molar mass is the mass of one mole of a substance, which is equivalent to the atomic mass of all the atoms in a molecule expressed in g/mol.
For phosphoric acid (H₃PO₄), calculating the molar mass involves summing the atomic masses of its constituent elements:

• Hydrogen (H) has an atomic mass of 1.01 g/mol and there are 3 H atoms, so: 3 × 1.01 = 3.03 g/mol.
• Phosphorus (P) has an atomic mass of 30.97 g/mol, contributing 30.97 g/mol.
• Oxygen (O) has an atomic mass of 16.00 g/mol and there are 4 O atoms, so: 4 × 16.00 = 64.00 g/mol.

Adding these masses together gives 97.99 g/mol for H₃PO₄. This value is fundamental when converting grams of a solute to moles, which is a crucial step in calculating molarity.

Aqueous solution

An aqueous solution is one in which the solvent is water. The term "aqueous" is derived from "aqua," meaning water. In this chemical environment, substances like acids, bases, salts, or other solutes dissolve and distribute uniformly throughout the solvent, forming a homogeneous mixture.
In this exercise, we are dealing with an aqueous solution of phosphoric acid (H₃PO₄). The solution is described as 75% phosphoric acid by mass. This mass percentage indicates that out of every 100 grams of the solution, 75 grams are the acid itself, while the remaining 25 grams are likely water.
Understanding the nature of an aqueous solution helps us calculate specific concentrations such as molarity, as it dictates how solutes are dispersed within the solvent. Proper comprehension of this concept is vital for determining solution properties and behavior in various chemical reactions.