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MEASURING CONCENTRATION OF IONS IN SOLUTION Molarity is ONE way to do this…we will learn others later in the year!!!

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Presentation on theme: "MEASURING CONCENTRATION OF IONS IN SOLUTION Molarity is ONE way to do this…we will learn others later in the year!!!"— Presentation transcript:

1 MEASURING CONCENTRATION OF IONS IN SOLUTION Molarity is ONE way to do this…we will learn others later in the year!!!

2 Molarity Molarity (M) =Mole of solute/liter of solution What do we have to be careful about when solving for molarity? Is amount of solute given in moles? Is volume of solution given in L? How can we manipulate this equation and then solve for: grams of solute? mL of solution?

3 To calculate molarity practice problem: What is the molarity of a solution in which 27 g of NaCl is dissolved in 100 mL of water?

4 To calculate volume practice problem: What is the volume in mL of solution if you want to make a 0.5 M solution by adding 50 g of NaCl?

5 To calculate mass practice problem: If you are making a.33 molar NaCl solution with a 150 mL of NaCL, what mass in g of NaCl must you add?

6 Dilutions Often you start with a solution of a certain molarity and want to “dilute” that solution. Use the following equation: M 1 V 1 = M 2 V 2 (M=molarity; V= volume)

7 How to calculate new molarity if diluting a solution: You have a 0.5 M solution with a volume of 400 mL. You dilute the solution with water to a final volume of 1000mL. What is the resulting molarity?

8 How to determine how much of an original solution to add if you want a certain amount of a solution with a new molarity: If you have a 0.4 M NaCl solution and you want to make 800 mL of a 0.25 M NaCl solution, how much of the original solution should you use?

9 Ion concentration Sometimes it is useful to know the concentration of ions found in a solution. You must multiply the molarity of the solution by the number of the given cation or anion. For a 0.5 M solution of MgCl 2 : Mg +2 : [0.5 M] 2 Cl -1 :2 * [0.5 M] = [1.0 M]

10 Stoichiometry AgNO 3 (aq) + NaCl (aq)  AgCl(s) + NaNO 3 (aq) If 50 mL of a 0.5 M silver nitrate solution is added to a solution of excess NaCl, how much precipitate can be produced? How many mL of a 0.5 M NaCl solution would be needed to run this reaction with no waste?


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