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Making Molar Solutions From Liquids (More accurately, from stock solutions)

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Presentation on theme: "Making Molar Solutions From Liquids (More accurately, from stock 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 form Acids 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 volume Now we use the equation M 1 V 1 = M 2 V 2 1 is starting (concentrated conditions) 2 is ending (dilute conditions)

3 Identify each volume to two decimal places (values tell you how much you have expelled) 4.48 - 4.504.86 - 4.875.00 Reading a pipette

4 Practice using a pipette Always keep pipette vertical To rinse: take up water, remove green filler, rotate pipette, replace filler, expel water If 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 cylinder Repeat with 1.7 mL and 5.1 mL __________________________________ If done correctly you should get 10 mL in graduated cylinder

5 The Dilution formula E.g. if we have 1 L of 3 M HCl, what is M if we dilute acid to 6 L? M 1 = 3 mol/L, V 1 = 1 L, V 2 = 6 L M 1 V 1 = M 2 V 2, M 1 V 1 /V 2 = M 2 M 2 = (3 mol/L x 1 L) / (6 L) = 0.5 M Why does the formula work? Because we are equating mol to mol: M 1 V 1 = 3 mol V 1 = 1 L M 1 = 3 M V 2 = 6 L M 2 = 0.5 M M 2 V 2 = 3 mol

6 Practice problems Q – What volume of 0.5 M HCl can be prepared from 1 L of 12 M HCl? M 1 = 12 mol/L, V 1 = 1 L, M 2 = 0.5 L M 1 V 1 = M 2 V 2, M 1 V 1 /M 2 = V 2 V 2 = (12 mol/L x 1 L) / (0.5 L) = 24 L Q – 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 + 0.5 L = 1.5 L # mol/L = 4 mol / 1.5 L = 2.67 mol/L Do 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? 3.100 mL of 6.0 M CuSO 4 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 0.5 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. 1.7 L of this is poured out, what is the concentration of the remaining HF?

8 Dilution problems (1-6, 6 two ways) 1. M 1 = 14 M, V 1 = ?, M 2 = 1.75 M, V 2 = 250 mL V 1 = M 2 V 2 / M 1 = (1.75 M)(0.250 L) / (14 M) V 1 = 0.03125 L = 31.25 mL 2. M 1 = 6 M, V 1 = 0.2 L, M 2 = ?, V 2 = 1 L M 2 = M 1 V 1 / V 2 = (6 M)(0.2 L) / (1 L) M 2 = 1.2 M 3. M 1 = 6 M, V 1 = 100 mL, M 2 = 1.5 M, V 2 = ? V 2 = M 1 V 1 / M 2 = (6 M)(0.100 L) / (1.5 M) V 2 = 0.4 L or 400 mL

9 Dilution problems (4 - 6) 4. # mol = (2.0 mol/L)(0.4 L) + (3.0 mol/L)(0.6 L) = 0.8 mol + 1.8 mol = 2.6 mol # L = 0.4 L + 0.6 L # mol/L = 2.6 mol / 1 L = 2.6 mol/L 5. # mol = (0.5 mol/L)(3 L) + (0.2 mol/L)(2 L) = 1.5 mol + 0.4 mol = 1.9 mol # mol/L = 1.9 mol / 5 L = 0.38 mol/L 6. # mol = (0.5 mol/L)(3 L) + (0 mol/L)(2 L) = 1.5 mol + 0 mol = 1.5 mol # mol/L = 1.5 mol / 5 L = 0.3 mol/L Or, using M 1 V 1 = M 2 V 2, M 1 = 0.5 M, V 1 = 3 L, M 2 = ?, V 2 = 5 L

10 7. M 1 = 6 M, V 1 = 4 L, M 2 = 1.5 M, V 2 = ? V 2 = M 1 V 1 / M 2 = (6 M)(4 L) / (1.5 M) V 2 = 16 L 8. 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. Dilution problems (7, 8)

11 Practice making molar solutions Calculate # of mL of 1 M HCl required to make 100 mL of a 0.1 M solution of HCl Get volumetric flask, pipette, plastic bottle, 100 mL beaker, 50 mL beaker, eyedropper. Rinse all with tap water. Dry 50 mL beaker Place about 20 mL of 1 M HCl in 50 mL beaker Rinse pipette, with small amount of acid Fill flask about 1/4 full with distilled water Add correct amount of acid with pipette. Mix. Add water to line (use eyedropper at the end) Place solution in plastic bottle Label bottle. Place at front of the room. Rinse and return all other equipment. For more lessons, visit www.chalkbored.com www.chalkbored.com


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