Molarity calculations

( ) And the final answer is? What is the final answer? Calculate the molarity concentration for 80 g of NaOH in 4.00 L of solution. And the final answer is? What is the final answer? How many liters of solution? Is this the only conversion factor we need? How many moles of NaOH? Now consider the definition of molarity? What number should we write first? Next comes a conversion factor What number goes in the numerator? What number goes in the denominator? Calculate the number of moles of NaOH Let’s do it stepwise first. ( ) 1 mol NaOH 80. g NaOH 2.0 moles NaOH = 40. g NaOH molarity = moles solute L solution 2.0 mols NaOH = .50 M NaOH 4.00 L solution

( ) Is this all we need? Put this on one side of the equation What volume of 1.0 M HF contains 20. g of HF? Is this all we need? Put this on one side of the equation What is the unknown variable? What is the final answer? Substitute in the values What goes in the numerator? And the final answer is? What is the first number we write? How many moles are in 20 g of HF? What goes in the numerator? Next comes a conversion factor? What goes in the denominator? Next consider the definition of molarity Let’s do this stepwise again. ( ) 1 mole HF 20. g HF = 1.0 mole HF 20. g HF molarity = moles solute liters of solvent L of solvent moles solute L of solvent = 1.0 mole HF = 1.0 L soln molarity 1.0 M HF

The Concept of Titration NaOH NaOH What happens with this volume of NaOH is added? What volume of NaOH solution is needed to react with all of the HF’s? How many NaOH’s are needed to react with all of the HF’s? NaOH NaOH NaOH 5 NaOH’s are needed. This volume is needed. NaOH NaOH NaOH Na+ F- H2O Na+ F- H2O Na+ F- H2O HF Na+ F- H2O Na+ F- H2O

AgNO3 + MgCl2  AgCl(s)+ Mg(NO3)2 2 Given that 100 mL of 2.0 M AgNO3(aq) reacts completely with 1.00 M MgCl2(aq), what volume of MgCl2(aq) is needed for complete precipitation? 2 AgNO3 + MgCl2  AgCl(s)+ Mg(NO3)2 2 Let’s look at this problem graphically. AgNO3 MgCl2 100 mL Mg2+ 100 mL Ag+

AgNO3 + MgCl2  AgCl(s)+ Mg(NO3)2 2 Given that 100 mL of 2.0 M AgNO3(aq) reacts completely with 1.00 M MgCl2(aq), what volume of MgCl2(aq) is needed for complete precipitation? 2 AgNO3 + MgCl2  AgCl(s)+ Mg(NO3)2 2 What is the denominator of the second conversion factor? What is the numerator of the second conversion factor? What is the numerator of the third conversion factor? What is the denominator of the first conversion factor? What is the denominator of the third conversion factor? What is the numerator of the first conversion factor? What is the second step? What is the third step? And the final answer…? What is the first step? Write the units of the desired quantity at the right. It must be liters MgCl2 Again, it must be moles MgCl2 Again, it must be moles of AgNO3. It must be moles MgCl2 Write the given quantity at the left. It must be L AgNO3. It must be moles of AgNO3. Comes from doing the calculation. Balance the chemical equation. 1 mol MgCl2 1 L MgCl2 2.0 mol AgNO3 .100 L AgNO3 = 1 L AgNO3 2 mol AgNO3 1.00 mol MgCl2 .100 L MgCl2

AlCl3 + KOH  Al(OH)3(s) + KCl 3 3 Given that 50. mL of .20 M AlCl3(aq) reacts completely with .40 M KOH, what volume of KOH is required for complete precipitation? AlCl3 + KOH  Al(OH)3(s) + KCl 3 3 What is in the numerator of the first conversion factor? What is in the denominator of the first conversion factor? What is in the numerator of the second conversion factor? What is in the denominator of the third conversion factor? What is in the numerator of the third conversion factor? What is in the denominator of the second conversion factor? And the final answer comes from…? What is the second step? What is the third step? What is the first step? Balance the equation. It must be mol KOH. Again, it must be mol KOH. It must be mol AlCl3. Write the given quantity on the left. It must be L AlCl3(aq). It must be L KOH. Doing the calculation. Write the units of the desired quantity on the right. .20 mol AlCl3 3 mol KOH 1.0 L KOH .050 L AlCl3 1 L AlCl3 1 mol AlCl3 .40 mol KOH = .075 L KOH