University:
University of North Carolina at Charlotte
Course:
CHEM 1251 | General Chemistry I
Academic year:

2024
Views:

256

Pages:

4
Author:

Kamila P.

Image Analysis: Textual Content: CHAPTER 9 CHEMICAL REACTIONS IN AQUEOUS SOLUTIONS SOLUTIONS: is an homogenous mixture of 2 substances A solute dissolves in a solvent to form a solution. NaCl (solute) H₂O (solvent) HCl (aq) solution SOLUTION CONCENTRATION: DILUTE SOLUTIONS: have a small amount of solute compared to solvent. CONCENTRATED SOLUTIONS: have a large amount of solute compared to a solvent. We can quantify the concentration of a solution using MOLARITY (M) Molarity (M) = amount of solute (in mol) / volume of solution (in L) Practice: Calculate the molarity of a solution by adding 45.4g of sodium nitrate to a flask and dissolving it with water to create a total volume of 2.50 L. 45.4g NaNO₃ x 1 mol NaNO₃ / 85.00g NaNO₃ = 0.5341 mol NaNO₃ Molarity (M) = 0.5341 mol NaNO₃ / 2.50 L solution = 0.214 M USING MOLARITY AS A CONVERSION FACTOR: The molarity of a solution can be used as a conversion factor between moles of solute and liters (volume) of solution. Ex: A 0.500 M NaCl solution contains 0.500 mol NaCl for every 1 L of solution. 0.500 mol NaCl / L solution How many moles of NaCl are in 15.0 L of a 0.500 M NaCl solution? 0.500 mol NaCl / L solution x 15.0 L solution = 7.50 moles NaCl What is the volume (in mL) of a 0.500 M NaCl solution that contains 1.25 moles of NaCl? 1.25 mol NaCl x 1 L solution / 0.500 mol NaCl = 2.5 L solution The amount of particles doesn't change. The volume does or can change. SOLUTION DILUTION: MAKING A SOLUTION from ANOTHER SOLUTION A dilution is when a new solution is prepared from a stock solution (more concentrated solution). To make solutions of lower concentrations from these stock solutions, more solvent is added. - The amount of solute doesn't change! - Only the volume of solution changes! mol solute in sol'n 1 = mol solute in sol'n 2 The concentrations and volumes of the stock and new solutions are inversely proportional. mol solute x L solution = moles solute PRACTICE: M₁ = 0.500 M M₂ = V₁ = 0.100 L V₂ = 0.500 L M₂ = M₁V₁ / V₂ = (0.500 M)(0.100 L) / 0.500 L = 0.100 M