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Ch. 5 part 2

# Ch. 5 part 2 - of HNO3 M1V1 = M2V2 1.0 M x V1 =.2M x 100 mL...

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Definition The best way express concentration is “molarity”. The Molarity of a solute in a solution is the number of moles of the solute divided by the volume of the solution in liters. The units are the mol/L, and are abbreviated M. Problem statement How many grams of Sodium Hydroxide do you need to weight out to get 14mL of a .65M solution of Sodium Hydroxide. 14mL (1L/1000mL)(.65mol/1L)(39.9950g/mol) = 3.6x10^-1g NaOH If you have a solution of known concentration, you can prepare a solution of a lower concentration by “dilution”. The concentration and volume of the new solution are related to that of the old by M1V1=M2V2 Problem Statement How do we prepare 100 mL of a 0.2 M solution of HNO3 given a 1.0M solution
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Unformatted text preview: of HNO3? M1V1 = M2V2 1.0 M x V1 = .2M x 100 mL V1 = 1.0M V1 = 20mL And conclude we need to take 20 mL of the 1.0 M and dilute it to 100mL Most calculation for solution chemistry are carried out using molarity. The basic procedure is the same as all the Grams -> moles -> moles -> grams calculations we’ve been doing, except it’s Molarity -> moles-> moles -> molarity Grams->-> grams When working with Molarities of ionic solutions it’s often useful to work with the concentrations of the ions. For example, a 0.1 M solution of Fe2(SO4)3 has a Fe3+ concentration of .2 M and a .3M concentration of SO2-,4. Once we have ion concentrations, it’s usually simple to work in terms of net ionic reactions....
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