Suppose You Work in the Laboratory of a Paint Company Where You Use 100 Ml of a 0.1 M

Dilutions Suppose you work in the laboratory of a paint company where you use 100 mL of a 0.1 M solution of zinc chloride in a quality-control test that you carry out 10 times a day. It would be tedious and time-consuming to continually measure out small amounts of ZnCl2 to make 100 mL of this solution. Of course, you could make many liters of the solution at one time, but that would require several large containers to store the solution. The answer to the problem is to make a much more concentrated solution and then dilute it with water to make the less concentrated solution that you need. The more-concentrated solution is called a stock solution. You could make a 1 M ZnCl2 solution by measuring out 1 mol of zinc chloride, 136.3 g, and dissolving it in enough water to make a liter of solution. This solution is 10 times as concentrated as the solution you need. Every time you need the test solution, you can measure out 10 mL of the 1 M solution and dilute it to 100 mL to yield 100 mL of 0.1 M ZnCl2 solution. To make a solution by dilution, you must determine the volume of a stock solution to use and the amount of solvent needed to dilute to the concentration you need. As you have learned, the molarity of a solution is its concentration in moles of solute per liter of solution. Molarity is found by dividing the moles of solute by the number of liters of solution. moles solute M= liters solution So, for a measured volume of any solution: the amount of solute in mol = molarity X volume of solution If this measured volume of solution is diluted to a new volume by adding solvent, the new, larger volume still contains the same number of moles of solute. Therefore, where 1 and 2 represent the concentrated and diluted solutions: molarity1 X volume1 = moles solute = molarity2 X volume2

Therefore: M1V1 =M2V2 This relationship applies whenever solution 2 is made from solution 1 by dilution.

Example #1 - 53.4 mL of a 1.50 M solution of NaCl is on hand, but you need some 0.800 M solution. How many mL of 0.800 M can you make? Using the dilution equation, we write: (1.50 mol/L) (53.4 mL) = (0.800 mol/L) (x) Solving the equation for x gives an answer of 100. mL. Notice that the volumes need not be converted to liters. Any old volume measurement is fine, just so long as the same one is used on each side.

Example #2 - 100.0 mL of 2.500 M KBr solution is on hand. You need 0.5500 M. What is the final volume of solution which results? Placing the proper values into the dilution equation gives: (2.500 mol/L) (100.0 mL) = (0.5500 mol/L) (x) x = 454.5 mL Sometimes the problem might ask how much more water must be added.

In this last case, the answer is 454.5 - 100.0 = 354.5 mL. Practice 1. What volume of water would be added to 16.5 mL of a 0.0813 M solution of sodium borate in order to get a 0.0200 M solution? ans: 50.6 mLH2O PROBLEMS

1. What is the molarity of a solution of ammonium chloride prepared by diluting 50.00 mL of a 3.79 M NH4Cl solution to 2.00 L?

2. A student takes a sample of KOH solution and dilutes it with 100.00 mL of water. The student determines that the diluted solution is 0.046 M KOH, but has forgotten to record the volume of the original sample. The concentration of the original solution is 2.09 M. What was the volume of the original sample?

3. A chemist wants to prepare a stock solution of H2SO4 so that samples of 20.00 mL will produce a solution with a concentration of 0.50 M when added to 100.0 mL of water. a. What should the molarity of the stock solution be?

b. If the chemist wants to prepare 5.00 L of the stock solution from concentrated H2SO4 , which is 18.0 M, what volume of concentrated acid should be used?

c. The density of 18.0 M H2SO4 is 1.84 g/mL. What mass of concentrated H2SO4 should be used to make the stock solution in (b)?

4. To what volume should 1.19 mL of an 8.00 M acetic acid solution be diluted in order to obtain a final solution that is 1.50 M?

5. What volume of a 5.75 M formic acid solution should be used to prepare 2.00 L of a 1.00 M formic acid solution? 6. A 25.00 mL sample of ammonium nitrate solution produces a 0.186 M solution when diluted with 50.00 mL of water. What is the molarity of the stock solution? Dilutions PROBLEMS

1. What is the molarity of a solution of ammonium chloride prepared by diluting 50.00 mL of a 3.79 M NH4Cl solution to 2.00 L?

0.0948 M NH4Cl 2. A student takes a sample of KOH solution and dilutes it with 100.00 mL of water. The student determines that the diluted solution is 0.046 M KOH, but has forgotten to record the volume of the original sample. The concentration of the original solution is 2.09 M. What was the volume of the original sample?

2.20 mL KOH

3. A chemist wants to prepare a stock solution of H2SO4 so that samples of 20.00 mL will produce a solution with a concentration of 0.50 M when added to 100.0 mL of water.

a. What should the molarity of the stock solution be? 2.5 M H2SO4

b. If the chemist wants to prepare 5.00 L of the stock solution from concentrated H2SO4 , which is 18.0 M,

what volume of concentrated acid should be used? 0.694 L H2SO4

c. The density of 18.0 M H2SO4 is 1.84 g/mL. What mass of concentrated H2SO4 should be used to make the

stock solution in (b)? 1278. g

H2SO4

4. To what volume should 1.19 mL of an 8.00 M acetic acid solution be diluted in order to obtain a final solution that is 1.50 M?

6.34 mL HC2H3O2

5. What volume of a 5.75 M formic acid solution should be used to prepare 2.00 L of a 1.00 M formic acid solution? 0.348 L HCOOH 6. A 25.00 mL sample of ammonium nitrate solution produces a 0.186 M solution when diluted with 50.00 mL of water. What is the

molarity of the stock solution? 0.558 M NH4NO3

Bonus

. Given a solution of known percentage concentration by mass, a laboratory worker can often measure out a calculated mass of the solution in order to obtain a certain mass of solute. Sometimes, though, it is impractical to use the mass of a solution, especially with fuming solutions, such as concentrated HCl and concentrated HNO3 . Measuring these solutions by volume is much more practical. In order to determine the volume that should be measured, a worker would need to know the density of the solution. This information usually appears on the label of the solution bottle.

a. Concentrated hydrochloric acid is 36% HCl by mass and has a density of 1.18 g/mL. What is the volume of 1.0 kg of this HCl solution? What volume contains 1.0 g of HCl? What volume contains 1.0 mol of HCl? b. The density of concentrated nitric acid is 1.42 g/mL, and its concentration is 71% HNO3 by mass. What volume of concentrated HNO3 would be needed to prepare 10.0 L of a 2.00 M solution of HNO3 ? c. What volume of concentrated HCl solution would be needed to prepare 4.50 L of 3.0 M HCl? See (a) for data.