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Preparing Solutions in the Laboratory

Preparing Solutions in the Laboratory. In order to conduct this reaction both the lead (II) nitrate and the potassium iodide must be made into solutions. Pb(NO 3 ) 2 (aq) + 2KI(aq)  PbI 2 (s) + 2KNO 3 (aq). Na +. Na +. Na +. Na +. Na +. Na +. Cl -. Cl -. Cl -. Cl -. Cl -.

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Preparing Solutions in the Laboratory

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  1. Preparing Solutions in the Laboratory

  2. In order to conduct this reaction both the lead (II) nitrate and the potassium iodide must be made into solutions Pb(NO3)2(aq) + 2KI(aq)  PbI2(s) + 2KNO3(aq)

  3. Na+ Na+ Na+ Na+ Na+ Na+ Cl- Cl- Cl- Cl- Cl- When a solute dissolves in a solvent the solute is separated into individual components The intermolecular forces in the solvent are altered The solute and the solvent interact to form the solution Solvent Cl- - - Solution + - + + - + Solute

  4. Preparing a Solution In order to prepare a solution we need to extend the concept of the mole to include the concept of molarity Molarity is defined as moles of solute per volume of solution in liters Solute-a substance that dissolves in a liquid Solution- a homogeneous mixture (moles/liter) Solvent-the dissolving medium in a solution

  5. Understanding Molarity Moles of solute Molarity = Liters of solution What is the molarity of an ascorbic acid solution (C6H8O6) prepared by dissolving 1.80 grams in enough water to make 125 mL of solution. How many milliliters of this solution contain 0.0100 mol ascorbic acid. 1 mol C6H8O6 1.80 g C6H8O6 = 0.0102 mol C6H8O6 176 g C6H8O6 0.0102 mol C6H8O6 Molarity = = 0.0818 M .125 L soln

  6. Understanding Molarity Sample Problem A solution is prepared by mixing 1.00 g ethanol (CH3CH2OH) with 100 ml. of water. Calculate the molarity of this solution.

  7. Preparing a Solution There are two approaches to preparing solutions. The first involves preparing the solution from dried reagent. Let’s assume that the potassium iodide is going to be prepared from dried reagent and that we have decided to prepare a 0.10 Molar solution (0.1 moles/liter)

  8. Preparing a Solution When preparing a solution from dried reagent we simply use dimensional analysis and ask three questions. 1. What volume of solution do you want to use? (let’s assume that you want 50 ml 2. What molarity do you want? ( We have already said that we want 0.10 M KI) 3. What is the molar mass of KI? (ans.: 167 g/mol)

  9. Preparing a Solution 0.100 mol 1 L 167 g KI 50.0 ml = 0.835 g KI 1000 ml 1 L 1 mol KI What volume of solution do you want to use? What molarity do you want? What is the molar mass of KI? In order to prepare the above solution, 0.835 g of KI are dissolved in 50 ml of distilled water

  10. Preparing a Solution There are two approaches to preparing solutions. The second approach involves dilution from a pre-prepared reagent. Let’s assume that a 0.10 M solution of lead (II) nitrate is going to be prepared from a pre-existing 0.75 M solution

  11. Preparing a Solution The second approach involves the use of the equation: M1V1 = M2V2 where M1 = the molarity of the stock solution (0.750 M) V1 = the volume of Pb(NO3)2 that will be removed from the stock solution M2 = the molarity of the solution you want (0.100 M) V2 = the volume of the solution you wish to use (50.0 ml)

  12. Preparing a Solution M1V1 = M2V2 (0.750 M)( V1) = (0.100)(50.0) V1 = 6.67 ml In order to prepare this solution you simple take 6.67 ml of Pb(NO3)2 from the stock solution and bring the volume upto 50 ml with distilled water.

  13. Preparing a Solution Practice Problems

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