Unit 10 Solution stoichiometry Learning objectives Calculate molarity

Unit 10: Solution stoichiometry

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

Solution stoichiometry q q 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) = Moles of solute/Liters of solution q Stoichiometric calculations are facile q Amounts of solution required are volumetric q Concentration varies with T q Amount of solvent requires knowledge of density

Example q What is molarity of 50 ml solution containing 2. 355 g H 2 SO 4? n n Molar mass H 2 SO 4 = 98. 1 g/mol Moles H 2 SO 4 = 0. 0240 mol Volume of solution = 0. 050 L Concentration = moles/volume = 0. 480 M

What is concentration of solution containing 60 g Na. OH in 1. 5 L

Dilution q 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 To dilute by factor of ten, increase volume by factor of ten q 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 m. L of 6 M solution?

Solution stoichiometry q How much volume of one solution to react with another solution n n Given volume of A with molarity MA Determine moles B Find target volume of B with molarity MB

Titration q q q 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 q How much 0. 125 M Na. HCO 3 is required to neutralize 18. 0 m. L of 0. 100 M HCl?