Solutions and Molarity

Solutions The solute is the substance dissolved in the solution The solvent is the substance in which the solute is dissolved Aqueous Solution is a solution in which the solvent is water

Heterogeneous Mixture: A mixture in which the composition is not uniform throughout Sand and water A salad Oil and water Homogeneous Mixture: A mixture in which the composition is uniform throughout May be solids, liquids and/or gases Also called a solution Salt and water Ice tea mix and water

Solutions What is an ionic substance? A bond between a metal and a non-metal The metal gives up one or more electrons to the non-metal and becomes positively charged The non-metal accepts one or more electrons and becomes negatively charged. The two ions are held together by electrostatic attraction [Na+][Cl-]

Solutions Ionic substances readily dissociate (come apart) in water. The positive and negative ions become attracted to individual water molecules Salt dissolving Ionic equation for salt dissolving in water: NaCl -> Na+ + Cl-

Solutions An electrolyte is a substance, when dissolved in water, conducts electricity A strong electrolyte is a solute that produces many ions in water (sodium chloride) A weak electrolyte is a solute that produces only a few ions in water A non-electrolyte does not dissociate into ions in water and does not conduct electricity (sucrose)

Conversions 1000 ml = 1 L (liter) To convert from ml to L, divide by 1000 To convert from L to ml, multiply by 1000 350 ml = ? L 350 ml = .350 L 1.7 L = ? ml 1.7 L = 1700 ml

Molarity Molarity is an expression of solution concentration Molarity = moles of solute/liter of solution Steps for calculating molarity Convert volume from ml to L (if necessary) Determine the molarity by dividing the number of moles of solute by the solution volume in liters

Volumetric flasks

Molarity 100.5 ml/1000 = .1005 L Molar mass of glucose = 180.16 g/mol 5.10 g/180.16 g/mol = .03 mol glucose .03 mol/.1005 L = .29 mols per liter .29 M

Molarity Practice Problem #1: .15 M Practice Problem #2: .13 M

Preparation of Solutions Convert ml to L (if necessary) Determine molar mass of solute Set up a proportion Solve to get number of moles of solute needed Multiply that number by the molar mass of the solute.

Preparation of Solutions Example 1: 0.1 M means .1 mol/1 Liter Convert to L -> 100.0 ml/1000 = 0.10 L Molar mass NaOH = 40 g/mol Proportion -> .1mol/1 L * x mol/.1 L X = .01 mol of NaOH .01 * 40.0 g/mol = 0.4 g

Preparation of Solutions Alternate Convert ml to L (if necessary) Determine molar mass of solute Multiply all the numbers: Liters x mols/liter x molar mass Example 1: 0.10 L x .1 mol/1 Liter x 40.0 g/mol = 0.4 g

Preparation of Solutions Practice Problems 11.1 g CaCl2 80.0 g NaOH 30.0 g NaOH

Dilutions Solutions you work with in the lab (dropper bottles) are made by diluting premixed concentrated stock solutions. By adding additional solvent to concentrated solutions decreases the ratio of solute to solvent particles. Use the following relationship to calculate the volume of solvent you may need M1V1 = M2V2 M1V1 represent the molarity and volume of the stock solution M2V2 represent the molarity and volume of the dilute solution Remember to convert ml to L !

Dilutions Example problem 2.00 M * V1 = 0.300 M * 0.50 L Solve for V1 = 0.150/2.00 V1 = .075 L or 75.0 ml

Dilutions Practice Problems V1 = 0.125 L or 125 ml M2 = .7 M