Quantities and Concentrations

SI Base Units Physical quantityName of UnitsAbbreviation Masskilogramkg Lengthmeterm Timeseconds TemperaturekelvinK Amount of substancemolemol Electric currentampereA Luminous intensitycandelacd

SI Prefixes especially useful in this course giga G 10 9 megaM10 6 kilok10 3 centic10 -2 millim10 -3 micro  nanon10 -9 picop10 -12

Quantities Mole Molar mass Mass Weight What is the difference between mass and weight?

Useful Algebraic Relationships wt A (g) mol A = fw A (g/mol) mol A = V (L) x M (mol A/L soln) or wt A (mg) mmol A = fw A (g/mol) mmol A = V (mL) x M (mmol A/mL soln)

Solution Terminology solute solvent aqueous solution liter atomic weight molecular weight

Molar concentrtion-Molarity no. moles A Molarity => M = no. liters solution or no. millimoles A Molarity => M = no. milliliters solution

Analytical Molarity Equilibrium molarity What are analytical molarity and equilibrium molarity? What is the difference between them?

Percent Composition wt of a solute w - w% = × 100% wt of solution vol of a solute v - v% = × 100% vol of solution wt of a solute w - v% = × 100% vol of solution

p-Functions pX = - log 10 [X] examples: pH pOH pCl pAg

Parts per Million / Billion wt of a solute c ppm = × 10 6 wt of solution wt of a solute c ppb = × 10 9 wt of solution

Empirical Formulas Molecular Formulas Structure Formulas

Chemical Stoichiometry

Preparing Solutions EXAMPLE: Describe the preparation of 1.00 L of M NaOH solution (f.w ) from reagent grade solid.

(1.00 L soln) # g NaOH =

EXAMPLE: Describe the preparation of 1.00 L of M NaOH solution (f.w ) from reagent grade solid. (1.00 L soln)(0.100 mol NaOH) # g NaOH = (1 L soln)

EXAMPLE: Describe the preparation of 1.00 L of M NaOH solution (f.w ) from reagent grade solid. (1.00 L soln)(0.100 mol NaOH) # g NaOH = (1 L soln)

EXAMPLE: Describe the preparation of 1.00 L of M NaOH solution (f.w ) from reagent grade solid. (1.00)(0.100 mol NaOH) # g NaOH = (1)

EXAMPLE: Describe the preparation of 1.00 L of M NaOH solution (f.w ) from reagent grade solid. (1.00)(0.100 mol)(40.00g NaOH) # g NaOH = (1) (1 mol)

EXAMPLE: Describe the preparation of 1.00 L of M NaOH solution (f.w ) from reagent grade solid. (1.00)(0.100 mol)(40.00g NaOH) # g NaOH = (1) (1 mol)

EXAMPLE: Describe the preparation of 1.00 L of M NaOH solution (f.w ) from reagent grade solid. (1.00)(0.100)(40.00g NaOH) # g NaOH = (1) (1)

EXAMPLE: Describe the preparation of 1.00 L of M NaOH solution (f.w ) from reagent grade solid. (1.00)(0.100)(40.00g NaOH) # g NaOH = (1) (1) = 4.00 g NaOH

EXAMPLE: Describe the preparation of 1.00 L of M NaOH solution (f.w ) from reagent grade solid. (1.00)(0.100)(40.00g NaOH) # g NaOH = (1) (1) = 4.00 g NaOH Weigh 4.00 g of NaOH, transfer to a 1.00 L volumetric flask, and dilute to the line.

Dilution #moles solute in conc. soln equals #moles solut in dil. soln therefore M conc V conc = M dil V dil