SOLUTIONS Concentration Measurement: Molarity Mr. Shields Regents Chemistry U12 L02

Molarity MOLARITY Let’s start our discussion of solution concentration Measurement with MOLARITY Molarity: the most common solution concentration measurement is defined as follows: M = moles of solute LITERS of solution Which means Molarity = # of moles per liter of solution Capital M

Molarity Let’s Consider the following Problem: What is the molarity of a solution in which 50g of CuSO4 is dissolved in water to make a 2 liter solution. Step 1. What’s the formula for MOLARITY? M = moles of solute liters of solution

Molarity Step 2. We need to determine how many moles we have in 50g of CuSO4 . How do we do that? Remember the mole Hole? 1. Determine the molar mass of CuSO4 (159.5g/mol) 2. Calculate the # of moles of CuSO4 Since we’re going into the mole hole we need to DIVIDE # of grams by molar mass (50/159.5) = 0.31 moles

Molarity Step 3. Calculate the Molarity of the solution using the formula for MOLARITY M = moles of Solute liters of solution How many moles do we have? How many liters do we have? So M = 0.31 moles 2 liters 0.31 mol 2 L M = 0.155 mol/liter

Making a 1M NaCl Solution Weigh out and add 1 mole of NaCl to a volumetric Flask Add a small amount of solvent to dissolve the salt 3. Once the solute is dissolved add additional solvent up to the 1 liter mark

Problem Calculate how many moles of ammonia there are What is the molarity of a solution made by dissolving 2.1 liters of NH3 (g) in sufficient water to make 2 liters of solution? What do we need to begin calculating the Molarity of the solution? Calculate how many moles of ammonia there are # of moles = volume = 2.1 liters = 0.094 mol molar vol 22.4 l/mol

Problem Step 2. Calculate the molarity of the solution We now know the number of moles of NH3 we have (0.094) so … Step 2. Calculate the molarity of the solution using the molarity equation M = # moles = 0.094 mol = 0.047M # liters of sol’n 2 liters

Molarity by dilution Note that it is only possible to go from a more Now what if we had a solution already prepared but Wanted to decrease the molarity to some lower precise Value. How could we use an already prepared higher molarity solution to accomplish this? Note that it is only possible to go from a more conc. (higher molarity) solution to a less conc. (lower molarity) solution. conc dil

Molarity by dilution To make something more dilute we’re going to need to add more solvent to the solution. Recall that Molarity (M) = # moles of solute # liters of solution So we need to dilute the solution by adding solvent But how much more solvent do we need to add?

Molarity by dilution When we add more solvent to a solution do we change the number of moles of solute in the solution? NO! But what does change? We change the # of moles per unit volume of Solvent because the same # of moles are present but now there’s more liquid

Molarity by Dilution Let’s see how this works … So… #moles = M1 x V1 Let’s call the Molarity of the original solution M1 and the New Molarity M2 The number of liters of the original solution and the new Solution will then be designated by V1 and V2 So the molarity of the old solution is… M1 = #moles of solute = #moles #liters of solution V1 So… #moles = M1 x V1

Molarity by Dilution The Molarity of the new solution will then be … M2 = #moles of solute #liters of solution # moles = M2 x V2 Remember that the # of moles does not change when We dilute solutions.

Molarity by dilution So… M1 x V1 = M2 x V2 If we know three values we can calculate the 4th. Let’s see how we use this relationship

Dilution Problem How much of a 3M solution of NiCL2 is needed to Prepare 100 ml of a 1.75M solution of NiCl2? OK how do we solve this? What’s the equation we’re going to use? Right! M1V1 = M2V2 M1 = 3M; M2 = 1.75M; V2 = 0.1L (100ml) So … 3 x V1 = 1.75 x 0.1 V1 = 0.058L or 58ml

Molarity by dilution Add 4.375L H20 to a 0.625L of a 12M sol’n How do we make 5L of A 1.5M sol’n of KCL From 12.0M stock sol’n? M1V1 = M2V2 12 x V1 = 1.5 x 5 12V1= 7.5 V1 = 7.5/12 = 0.625L Add 4.375L H20 to a 0.625L of a 12M sol’n