Presentation on theme: "Making Molar Solutions"— Presentation transcript:
1 Making Molar Solutions From Liquids(More accurately, from stock solutions)
2 Making molar solutions from liquids Not all compounds are in a solid formAcids are purchased as liquids (“stock solutions”). Yet, we still need a way to make molar solutions of these compounds.The Procedure is similar:Use pipette to measure moles (via volume) Use volumetric flask to measure volumeNow we use the equation M1V1 = M2V21 is starting (concentrated conditions)2 is ending (dilute conditions)
3 Reading a pipette Identify each volume to two decimal places (values tell you how much you have expelled)5.00
4 Practice using a pipette Always keep pipette verticalTo rinse: take up water, remove green filler, rotate pipette, replace filler, expel waterIf filler can not take up or expel enough liquid, remove, place finger over pipette, turn knob, replace filler.Take up water to 0 mark. Measure 3.2 mL into 10 mL cylinder. (one per person)If drop is hanging off, touch to cylinderRepeat with 1.7 mL and 5.1 mL__________________________________If done correctly you should get 10 mL in graduated cylinder
5 The Dilution formulaE.g. if we have 1 L of 3 M HCl, what is M if we dilute acid to 6 L?M1 = 3 mol/L, V1 = 1 L, V2 = 6 LM1V1 = M2V2, M1V1/V2 = M2M2 = (3 mol/L x 1 L) / (6 L) = 0.5 MWhy does the formula work?Because we are equating mol to mol:V1 = 1 LM1 = 3 MV2 = 6 LM2 = 0.5 MM1V1 = 3 molM2V2 = 3 mol
6 Do 1 – 8 on handout. Try 6 two ways Practice problemsQ – What volume of 0.5 M HCl can be prepared from 1 L of 12 M HCl?M1 = 12 mol/L, V1 = 1 L, M2 = 0.5 LM1V1 = M2V2, M1V1/M2 = V2V2 = (12 mol/L x 1 L) / (0.5 L) = 24 LQ – 1 L of a 3 M HCl solution is added to 0.5 L of a 2 M HCl solution. What is the final concentration of HCl? (hint: first calculate total number of moles and total number of L)# mol = (3 mol/L)(1 L) + (2 mol/L)(0.5 L)= 3 mol + 1 mol = 4 mol# L = 1 L L = 1.5 L# mol/L = 4 mol / 1.5 L = 2.67 mol/LDo 1 – 8 on handout. Try 6 two ways
7 1. How many mL of a 14 M stock solution must be used to make 250 mL of a 1.75 M solution? 2. You have 200 mL of 6.0 M HF. What concentration results if this is diluted to a total volume of 1 L?mL of 6.0 M CuSO4 must be diluted to what final volume so that the resulting solution is 1.5 M?4. What concentration results from mixing 400 mL of 2.0 M HCl with 600 mL of 3.0 M HCl?5. What is the concentration of NaCl when 3 L of M NaCl are mixed with 2 L of 0.2 M NaCl?6. What is the concentration of NaCl when 3 L of 0.5 M NaCl are mixed with 2 L of water?7. Water is added to 4 L of 6 M antifreeze until it is 1.5 M. What is the total volume of the new solution?8. There are 3 L of 0.2 M HF L of this is poured out, what is the concentration of the remaining HF?
8 Dilution problems (1-6, 6 two ways) 1.M1 = 14 M, V1 = ?, M2 = 1.75 M, V2 = 250 mLV1 = M2V2 / M1 = (1.75 M)(0.250 L) / (14 M)V1 = L = mL2.M1 = 6 M, V1 = 0.2 L, M2 = ?, V2 = 1 LM2 = M1V1 / V2 = (6 M)(0.2 L) / (1 L)M2 = 1.2 M3.M1 = 6 M, V1 = 100 mL, M2 = 1.5 M, V2 = ?V2 = M1V1 / M2 = (6 M)(0.100 L) / (1.5 M)V2 = 0.4 L or 400 mL
10 Dilution problems (7, 8) 7. M1 = 6 M, V1 = 4 L, M2 = 1.5 M, V2 = ? V2 = M1V1 / M2 = (6 M)(4 L) / (1.5 M)V2 = 16 L8. The concentration remains 0.2 M, both volume and moles are removed when the solution is poured out. Remember M is mol/L. Just like the density of a copper penny does not change if it is cut in half, the concentration of a solution does not change if it is cut in half.
11 Practice making molar solutions Calculate # of mL of 1 M HCl required to make 100 mL of a 0.1 M solution of HClGet volumetric flask, pipette, plastic bottle, mL beaker, 50 mL beaker, eyedropper. Rinse all with tap water. Dry 50 mL beakerPlace about 20 mL of 1 M HCl in 50 mL beakerRinse pipette, with small amount of acidFill flask about 1/4 full with distilled waterAdd correct amount of acid with pipette. Mix.Add water to line (use eyedropper at the end)Place solution in plastic bottleLabel bottle. Place at front of the room.Rinse and return all other equipment.For more lessons, visit