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 McVc = MdVd c is starting (concentrated conditions) d is ending (dilute conditions)

Dilution Adding more solvent to a “stock” solution to lower it’s molarity McVc = MdVd Really setting moles equal Volume units don’t matter as long as they are the same on both sides Demo with a colored solution what happens when you take some of it out and add water to it. What happens to the concentration of the original solution and what happens to the concentration of the new solution.

Serial Dilution Start with 2 drops of food coloring in 100 mL of water Original Molarity? 2/100 = 0.02M Take 10 mL of original, put in empty beaker, and add 90 mL of water New Molarity? Continue the dilution until you can’t see any more color – what is it’s molarity now? 0.02 M (10 mL) = x M (100mL) X= 0.002 M The second dilution is 0.002 M (10 mL) = x M (100mL) X= 0.0002M The same pattern will continue with each dilution

More practice problems… If you have a 0.82 M stock solution and water, how would you make 10 mL of a 0.5 M solution. How would you prepare 1.5 L of a 0.10 M H2SO4 from a 16 M H2SO4? What would be the new molarity of 125 mL of a 0.2 M HBr solution if 150 mL of water were added? 0.82 M (V1) = 0.5 M (10 mL) V1 = 6.1 mL of the 0.82 M and 3.9 M of water (10-6.1 =3.9) 16 M (V1) = 0.10 M (1.5L) V1 = 0.0094 L which is 9.4 mL of the 16M and about 1.49 L of water 3) 0.2 M (150 mL) = M2 (150+125) M2 = 0.11 M

Practice with dilutions You have a 55 mL of a 4.3 M solution of CaCl2 and you add 100 mL to it. What is the new molarity? You have a 2 M solution of AgNO3, you want to make 20 mL of 1M solution? How will you make it? You make 40 mL of a 0.2 M solution of Na2SO4, you started with 15 mL of a stock solution. What was the molarity of the this stock? Use M1V1 = M2V2 in all of these equations. 55 mL (4.3M) = M2 (55+100) M2 = 1.53M 2M (V1) = 1 M (20mL) V1 = 10 mL of the 2 M and 10 mL of water M1 (15mL) = 0.2 M (40 mL) M1 = 0.53 M

What is the relationship between molarity and volume in dilution? When you make a dilution what always has to happen to the molarity? The total volume? What does this graph look like? By definition, in a dilution the molarity has to go down. In order for that to happen the total volume of the solution has to increase. Volume Molari ty

The Dilution formula 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

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 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 1.5 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?

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 using a pipette Always keep pipette vertical To rinse: rotate pipette, take up water, expel water gently 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

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 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.