Molarity and all that is Molarity (well not all, but a lot) Solution Chemistry Molarity and all that is Molarity (well not all, but a lot)

SOLUTION CHEMISTRY Concentration: amount of solute in a given amount of solvent (can be determined quantitatively) Dilute: a solution with a small amount of solute per solvent amount (relative term) Concentrated: a solution with a large amount of solute per solvent amount (relative term) BOTH DILUTE AND CONCENTRATED ARE QUALITATIVE

Concentration of Solute The amount of solute in a solution is given by its concentration. Molarity ( M ) = moles solute liters of solution The concentration of a solution is said to be its molarity. Ex. 1 M CuSO4 “1 molar copper II sulfate

MOLARITY M = mol solute L sol’n M = 4.67 mol Li2SO3 2.04 L Li2SO3 A measurement of the concentration of a solution Molarity (M) is equal to the moles of solute per liter of solution M = mol / L Determine the Molarity (M) of the following solution. You have 4.67 moles of Li2SO3 and you want to make 2.04 L of the solution. M = mol solute L sol’n M = 4.67 mol Li2SO3 2.04 L Li2SO3

MOLARITY M = mol of solute / L sol’n Calculate the moles of solute: A measurement of the concentration of a solution Molarity (M) is equal to the moles of solute per liter of solution M = mol of solute / L sol’n Calculate the molarity of a solution prepared by mixing 1.5 g of NaCl in .50 L of water. Calculate the moles of solute: 1.5 g NaCl X 1 mol NaCl = 58.45 g NaCl Plug the appropriate values into the correct variables in the equation: M = mole solute/L sol’n = 0.0257 moles / 0.500 L = 0.051 mol/L 0.0257 mol NaCl

MOLARITY M = mol / L How many grams of LiOH is needed to prepare 250.0 mL of a 1.25 M solution? First calculate the moles of solute needed: M = mole solute  rearrange to solve for mole solute: L sol’n Change mL to L of solution Plug in the variables in the formula mole solute = (1.25 mol / L) (0.2500 L) = 0.3125 moles of solute needed Next calculate the molar mass of LiOH: 23.95 g/mol Last, use diminsional analysis to solve for mass: 0.3125 moles x 23.95 g LiOH / 1 mol LiOH) = 7.48 g of LiOH

MOLARITY M = n / V = mol / L What is the molarity of hydroiodic acid if the solution is 47.0% HI by mass and has a density of 1.50 g/mL? First calculate the mass of solute in the 47.0% solution using the density. The 1.50 g/mL is the density of the solution but only 47.0% of the solution is the solute 47.0% of 1.50 g/mL = (0.470) (1.50 g/mL) = 0.705 g/mL density of solute Since molarity is given in moles per liter and not grams we must convert the g/mL to mol/mL using the molar mass. 0.705 g/mL (1 mole/ 128 g) = 0.00551 mol/mL Convert mL to L: 0.00551 mol/mL (1000 mL/ 1L) = 5.51 mol/L = 5.51 M

MOLARITY & DILUTION M1V1 = M2V2 The act of diluting a solution is to simply add more water (the solvent) thus leaving the amount of solute unchanged. Since the amount or moles of solute before dilution (n1) and the moles of solute after the dilution (n2) are the same: n1 = n2 And the moles for any solution can be calculated by n=MV A relationship can be established such that M1V1 = n1 = n2 = M2V2 Or simply : M1V1 = M2V2

M1 V1 = M2 = (0.05 mol/L) (25.0 mL) = 0.0167 M of KI MOLARITY & DILUTION Calculate the molarity of a solution prepared by diluting 25.0 mL of 0.05 M potassium iodide with 50.0 mL of water (the densities are similar). M1 = 0.05 mol/L M2 = ? V1 = 25.0 mL V2 = 50.0 + 25.0 = 75.0 mL M1V1 = M2V2 M1 V1 = M2 = (0.05 mol/L) (25.0 mL) = 0.0167 M of KI V2 75.0 mL

M2 V2 = V1 = (0.150 mol/L) (250.0 mL) = 6.25 mL of 6 M HCl MOLARITY & DILUTION Given a 6.00 M HCl solution, how would you prepare 250.0 mL of 0.150 M HCl? M1 = 6.00 mol/L M2 = 0.150 V1 = ? mL V2 = 250.0 mL M1V1 = M2V2 M2 V2 = V1 = (0.150 mol/L) (250.0 mL) = 6.25 mL of 6 M HCl M1 6.00 mol/L You would need 6.25 mL of the 6.00 M HCl reagent which would be added to a 250.0 mL graduated cylinder then more water would be added to the mixture until the bottom of the menicus is at 250.0 mL. Mix well.