Solution stoichiometry Volumetric calculations Acid-base titrations

Learning objectives  Calculate molarity and dilution factors  Use molarity in solution stoichiometry problems  Apply solution stoichiometry to acid-base titrations

Solution stoichiometry  In solids, moles are obtained by dividing mass by the molar mass  In liquids, it is necessary to convert volume into moles using molarity

Molarity (M) Molarity (M) = Moles of solute/Liters of solution  Stoichiometric calculations are facile  Amounts of solution required are volumetric  Concentration varies with T  Amount of solvent requires knowledge of density

Example  What is molarity of 50 ml solution containing g H 2 SO 4 ? Molar mass H 2 SO 4 = 98.1 g/mol Molar mass H 2 SO 4 = 98.1 g/mol Moles H 2 SO 4 = mol Moles H 2 SO 4 = mol Volume of solution = L Volume of solution = L Concentration = moles/volume Concentration = moles/volume = M = M

What is concentration of solution containing 60 g NaOH in 1.5 L

Dilution  More dilute solutions are prepared from concentrated ones by addition of solvent Moles before = moles after: M 1 V 1 = M 2 V 2 Molarity of new solution M 2 = M 1 V 1 /V 2 Molarity of new solution M 2 = M 1 V 1 /V 2 To dilute by factor of ten, increase volume by factor of ten  Do molarity exercises

What is concentration if 2 L of 6 M HCl is diluted to 12 L?

How much water must be added to make a 2 M solution from 100 mL of 6M solution?

Solution stoichiometry  How much volume of one solution to react with another solution Given volume of A with molarity M A Given volume of A with molarity M A Determine moles A Determine moles A Determine moles B Determine moles B Find target volume of B with molarity M B Find target volume of B with molarity M B

Titration  Use a solution of known concentration to determine concentration of an unknown  Must be able to identify endpoint of titration to know stoichiometry  Most common applications with acids and bases

Example  How much M NaHCO 3 is required to neutralize 18.0 mL of M HCl?