Making Molar Solutions

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

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 M1V1 = M2V2 1 is starting (concentrated conditions) 2 is ending (dilute conditions)

Reading a pipette Identify each volume to two decimal places (values tell you how much you have expelled) 4.48 - 4.50 4.86 - 4.87 5.00

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

The Dilution formula E.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 L M1V1 = M2V2, M1V1/V2 = M2 M2 = (3 mol/L x 1 L) / (6 L) = 0.5 M Why does the formula work? Because we are equating mol to mol: V1 = 1 L M1 = 3 M V2 = 6 L M2 = 0.5 M M1V1 = 3 mol M2V2 = 3 mol

Practice problems Q – 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 L M1V1 = M2V2, M1V1/M2 = V2 V2 = (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

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 mLof 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 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 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 1.5M. Whatisthe totalvolumeofthenew solution? 8. There are 3 L of 0.2 M HF. 1.7 L of this is poured out,whatistheconcentrationoftheremaining HF?

Dilution problems (1-6, 6 two ways) 1. M1 = 14 M, V1 = ?, M2 = 1.75 M, V2 = 250 mL V1 = M2V2 / M1 = (1.75 M)(0.250 L) / (14 M) V1 = 0.03125 L = 31.25 mL 2. M1 = 6 M, V1 = 0.2 L, M2 = ?, V2 = 1 L M2 = M1V1 / V2 = (6 M)(0.2 L) / (1 L) M2 = 1.2 M 3. 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

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 M1V1 = M2V2, M1 = 0.5 M, V1 = 3 L, M2 = ?, V2 = 5 L

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 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.

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

Making Molar Solutions