Osmolarity and Ion Calculations in Solution

Osmolarity and Ion Calculations in Solution

Osmolarity: Definition and Formula

Osmolarity (ionic molarity) represents the number of moles of ions per liter of solution. It is a measure of the total concentration of all ions present in a solution and is important in understanding colligative properties and physiological effects.

• Osmolarity Formula:

$\text{Osmolarity} = \frac{\text{meter ions}}{\text{liter solution}}$

Method 1: Direct Calculation of Osmolarity

In this method, use the moles of ions and liters of solution to calculate osmolarity directly.

• Identify the number of ions produced per formula unit of solute.

• Calculate the total moles of ions in the given volume.

Example: Calculate the molarity of chloride ions when dissolving 58.1 g AgCl in enough water to make 500 mL of solution.

• Mole calculation: $\text{M} = \frac{\text{mol solute}}{\text{L solution}}$

• 1 mol AgCl = 1 mol Ag+ + 1 mol Cl-

• Molar mass AgCl = 143.32 g/mol

• Number of moles: $\frac{58.1\ \text{g}}{143.32\ \text{g/mol}} = 0.4055\ \text{mol}$

• Volume: 500 mL = 0.500 L

• Chloride ion molarity: $\frac{0.4055\ \text{mol}}{0.500\ \text{L}} = 0.811\ \text{M}$

Method 2: Osmolarity from Molarity

If the molarity of a compound is known, the osmolarity for each of its ions can be determined by multiplying the molarity by the number of ions produced per formula unit.

• Osmolarity Formula:

$\text{Osmolarity} = \#\ \text{of ions} \times \text{M of Compound}$

Example: What is the concentration of hydroxide ions in a 0.350 M solution of gallium hydroxide, Ga(OH)3?

• Ga(OH)3 dissociates to give 3 OH- ions per formula unit.

• OH- concentration: $3 \times 0.350\ \text{M} = 1.05\ \text{M}$

Method 3: Number of Ions from Molarity

Problems involving number of ions and molarity can use given amount and conversion factors to isolate an end amount. This is useful for calculating the total number of ions or mass of ions in a given volume of solution.

• Use molarity, volume, and Avogadro's number for calculations.

Example: How many moles of Ca2+ ions are in 0.120 L of 0.450 M Ca(ClO4)2 solution?

• Given: 0.120 L, 0.450 M

• Each formula unit produces 1 Ca2+ ion.

• Moles Ca2+: $0.120\ \text{L} \times 0.450\ \text{mol/L} = 0.054\ \text{mol Ca}^{2+}$

Practice Problems and Solutions

Practice problems help reinforce the concepts of osmolarity and ion concentration calculations.

• Which of the following solutions will have the highest concentration of bromide ions?

  • a) 0.10 M NaBr

  • b) 0.10 M CaBr2 (correct answer)

  • c) 0.10 M AlBr3

  • d) 0.05 M MnBr4

• How many milligrams of nitride ions are required to prepare 820 mL of 0.330 M Ba3N2 solution?

  • Answer: 270.6 mg

• How many bromide ions are present in 65.5 mL of 0.210 M CaBr2 solution?

  • Answer: 13.755 (likely in mmol or another unit; context needed)

Summary Table: Osmolarity Calculation Methods

Method	Formula	Application
Direct Calculation	$\text{Osmolarity} = \frac{\text{moles of ions}}{\text{liters of solution}}$	When moles of ions and volume are known
From Molarity	$\text{Osmolarity} = \#\ \text{of ions} \times \text{M of compound}$	When molarity and ion count per formula unit are known
Number of Ions	Use molarity, volume, and Avogadro's number	To find total number or mass of ions in solution

Additional info: The notes provide foundational concepts for GOB Chemistry students, focusing on solution chemistry and ionic calculations, which are essential for understanding physiological and laboratory processes.