4.2 - Concentration - DILUTIONS

Stock Solutions A common situation in a chemistry laboratory occurs when a solution of a particular concentration is desired, but only a solution of greater concentration is available: this is called a stock solution.

What is a Dilution? To dilute a solution means to add more solvent without the addition of more solute. You may have heard it referred to as “watering down” This statement is somewhat accurate, as long as your solvent actually is water Of course, the resulting solution is thoroughly mixed so as to ensure that all parts of the solution are homogenous.

We can dilute a solution by adding water to it: Before Dilution: m1 = initial moles of solute After Dilution: m2 = final moles of solute

From previous slide… Notice that we did not add any more solute at all the number of moles will not change. The volume changes, and also the concentration. Final moles of solute Initial moles of solute …is equal to

𝑚𝑜𝑙 𝑏𝑒𝑓𝑜𝑟𝑒 𝑑𝑖𝑙𝑢𝑡𝑖𝑜𝑛 = 𝑚𝑜𝑙 𝑎𝑓𝑡𝑒𝑟𝑑𝑖𝑙𝑢𝑡𝑖𝑜𝑛 Dilution Equation: The fact that the solute amount stays constant allows us to develop calculation techniques. First, we write: 𝑚𝑜𝑙 𝑏𝑒𝑓𝑜𝑟𝑒 𝑑𝑖𝑙𝑢𝑡𝑖𝑜𝑛 = 𝑚𝑜𝑙 𝑎𝑓𝑡𝑒𝑟𝑑𝑖𝑙𝑢𝑡𝑖𝑜𝑛 From the definition of molarity, we know that the moles of solute equals the molarity times the volume. 𝑚𝑜𝑙=𝑀∗𝑉 So we can substitute equation, like this: 𝑀 1 𝑉 1 = 𝑀 2 𝑉 2 M1V1 means the molarity and the volume of the original solution M2V2 means the molarity and the volume of the second solution.

Example #1 53.4 mL of a 1.50 M solution of NaCl is on hand, but you need some 0.800 M solution. How many mL of 0.800 M can you make?

Example #2 100.0 mL of 2.500 M KBr solution is on hand. You need 0.5500 M. What is the final volume of solution which results?

Serial Dilutions The dilution equation is fairly easy to understand – we shouldn’t have any trouble calculating new or desired concentrations or volumes Sometimes however, we will calculate out a value that is too small or unrealistic – this makes it difficult to measure in a practical lab-like setting

Serial Dilutions - Example How could we prepare 200.0 mL of .010 M HCl from 10. M HCl? Calculate the missing value using the dilution equation M1v1 = M2v2 10. ( vi ) = 200.0 (.010) v1 = 0.20 mL It is unrealistic for us to measure .20 mL so we do a series of dilutions so we can maintain our accuracy

Serial Dilutions - Example How do we solve this? Step 1: First pick a practical dilution into an appropriately small volume. This is usually determined by the tools that you have in the lab. For example use 10. ml into 100.0 ml M1V1 = M2V2 10.M ( 10.mL ) = ( M2 ) 100.mL M2 = 1.0 M Step 2: Now you have a 1.0 molar solution so redo the math to find how much of our NEW solution we need to make the concentration we want. (1.0)v1 = 200.0mL ( .010 ) v1 = 2.0 mL 2.0 mL is a much more reasonable amount to measure than 0.20 mL

Serial Dilutions - Example If 2.0 mL is still too small and you don't have the equipment to do it, continue to dilute: Step 1: Do the 10. ml into 100. ml again; this time your initial concentration will be 1.0 M M1v1 = M2v2 (1.0) 10. = (M2) 100. M2 = .10 M Step 2: Now you have a .10 molar solution so redo the math (.10)v1 = 200.0 (.010) v1= 20. mL 20. mL is even easier than 2.0 mL to measure accurately

Serial Dilution Visual Stock Solution M1 = 10.M Volumetric Pipette V1 = 10.0mL X 1st Dilution M2 = ?M Volumetric Flask V2 = 100.0mL M1V1=M2V2 Gives us 1.0M 1st Dilution M1 = 1.0M Graduated Pipette V1=?mL 2nd Dilution M2 = 0.01M Volumetric Flask V2 = 200.0mL X M1V1=M2V2 Gives us 2.0mL