# Dilution of Solutions.

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Dilution of Solutions

Concentration = 1 “fishar”
# of fish volume (L) 1 fish Concentration = 1 (L) Concentration = 1 “fishar” V = 1000 mL V = 1000 mL V = 5000 mL n = 2 fish n = 4 fish n = 20 fish Concentration = 2 “fishar” [ ] = 4 “fishar” [ ] = 4 “fishar”

Concentration = # of moles volume (L) V = 1000 mL V = 1000 mL
n = 8 moles [ ] = 32 molar V = 1000 mL V = 1000 mL V = 5000 mL n = 2 moles n = 4 moles n = 20 moles Concentration = 2 molar [ ] = 4 molar [ ] = 4 molar

Making Molar Solutions
…from liquids (More accurately, from stock solutions)

Concentration…a measure of solute-to-solvent ratio
concentrated vs dilute “lots of solute” “not much solute” “watery” Concentration of a solution describes the quantity of a solute that is contained in a particular quantity of solvent or solution Knowing the concentration of solutes is important in controlling the stoichiometry of reactant for reactions that occur in solution A concentrated solution contains a large amount of solute in a given amount of solution. A 10 mol/L solution would be called concentrated. A dilute solution contains a small amount of solute in a given amount of solution. A 0.01 mol/L solution would be called dilute. Add water to dilute a solution; boil water off to concentrate it.

Making a Dilute Solution
remove sample moles of solute initial solution same number of moles of solute in a larger volume mix Making a Dilute Solution diluted solution Timberlake, Chemistry 7th Edition, page 344

Concentration “The amount of solute in a solution”
A. mass % = mass of solute mass of sol’n B. parts per million (ppm)  also, ppb and ppt – commonly used for minerals or contaminants in water supplies C. molarity (M) = moles of solute L of sol’n – used most often in this class % by mass – medicated creams % by volume – rubbing alcohol mol L M MOLARITY - Most common unit of concentration Most useful for calculations involving the stoichiometry of reactions in solution Molarity of a solution is the number of moles of solute present in exactly 1 L of solution: moles of solute molarity = liters of solution Units of molarity — moles per liter of solution (mol/L), abbreviated as M Relationship among volume, molarity, and moles is expressed as VL M Mol/L = L (mol) = moles (L) There are several different ways to quantitatively describe the concentration of a solution, which is the amount of solute in a given quantity of solution. 1. Molarity – Useful way to describe solution concentrations for reactions that are carried out in solution or for titrations – Molarity is the number of moles of solute divided by the olume of the solution Molarity = moles of solute = mol/L liter of solution – Volume of a solution depends on its density, which is a function of temperature 2. Molality – Concentration of a solution can also be described by its molality (m), the number of moles of solute per kilogram of solvent – Molality = moles of solute kilogram solvent – Depends on the masses of the solute and solvent, which are independent of temperature – Used in determining how colligative properties vary with solute concentrations 3. Mole fraction – Used to describe gas concentrations and to determine the vapor pressures of mixtures of similar liquids – Mole fraction () = moles of component total moles in the solution – Depends on only the masses of the solute and solvent and is temperature independent 4. Mass percentage (%) – The ratio of the mass of the solute to the total mass of the solution – Result can be expressed as mass percentage, parts per million (ppm), or parts per billion (ppb) mass percentage = mass of solute  100% mass of solution parts per million (ppm) = mass of solute  106 parts per billion (ppb) = mass of solute  109 – Parts per million (ppm) and parts per billion (ppb) are used to describe concentrations of highly dilute solutions, and these measurements correspond to milligrams (mg) and micrograms (g) of solute per kilogram of solution, respectively – Mass percentage and parts per million or billion can express the concentrations of substances even if their molecular mass is unknown because these are simply different ways of expressing the ratios of the mass of a solute to the mass of the solution M = mol L D. molality (m) = moles of solute kg of solvent

A part per million may be hard to comprehend or visualize.
Want to see how much it is? It’s a credit card lying in the middle of a football field. A step in a journey of 568 miles. Or one minute in two years. Taking that further, a part per trillion is a million times smaller than that credit card on the football field, for example, or 6 feet out of a journey of six round trips to the sun (Editor's note: 95 million miles one way to the sun). Want to see what taking in a milligram per kilogram of your body weight amounts to? It’s the equivalent of 726 people, each weighing 150 pounds, sharing a chocolate bar (~50 grams).

ppm 1 inch in 16 miles ppb 1 inch in 16,000 miles ppt 1 sec = 32,000 years

Glassware

Glassware – Precision and Cost
beaker vs. volumetric flask When filled to 1000 mL line, how much liquid is present? beaker 5% of 1000 mL = volumetric flask 50 mL 1000 mL mL Range: 950 mL – 1050 mL Range: mL– mL imprecise; cheap precise; expensive

Markings on Glassware Beaker 500 mL + 5% Range = 500 mL + 25 mL
Graduated Cylinder 1000 mL mL Range = mL + 5 mL 475 – 525 mL Volumetric Flask 500 mL mL Range = – mL TC 20oC “to contain at a temperature of 20 oC” 22 TD “to deliver” T s “time in seconds”

Measure to part of meniscus w/zero slope.

How to mix solid chemicals
Lets mix chemicals for the upcoming soap lab. We will need 1000 mL of 3 M NaOH per class. How much sodium hydroxide will I need, for five classes, for this lab? M = mol L 3 M = ? mol 1 L ? = 3 mol NaOH/class x 5 classes 15 mol NaOH How much will this weigh? 1 23g/mol + 16g/mol g/mol MMNaOH = 40g/mol To prepare a solution that contains a specified concentration of a substance, it is necessary to dissolve the desired number of moles of solute in enough solvent to give the desired final volume of solution. Solute occupies space in the solution so the volume of the solvent that is needed is less than the desired volume of solution. To prepare a particular volume of a solution that contains a specified concentration of a solute, calculate the number of moles of solute in the desired volume of solution and then covert the number of moles of solute to the corresponding mass of solute needed. 40.0 g NaOH X g NaOH = mol NaOH = 600 g NaOH 1 mol NaOH FOR EACH CLASS: To mix this, add 120 g NaOH into 1L volumetric flask with ~750 mL cold H2O. Mix, allow to return to room temperature – bring volume to 1 L.

How to mix a Standard Solution
Wash bottle Volume marker (calibration mark) Weighed amount of solute Use a VOLUMETRIC FLASK to make a standard solution of known concentration Step 1> add the weighed amount of solute in the volumetric flask Step 2> add distilled water (about half of final volume) Step 3> cap volumetric flask, and shake to dissolve solute completely Step 4> add distilled water to volume marker (calibration mark) The solution process may be exothermic (release heat). This may cause the liquid to show a larger volume than is real. Allow the solution to return to ambient (room) temperature and check volume again. Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 480

How to mix a Standard Solution
An aqueous solution consists of at least two components, the solvent (water) and the solute (the stuff dissolved in the water). Usually one wants to keep track of the amount of the solute dissolved in the solution. We call this the concentrations. One could do by keeping track of the concentration by determining the mass of each component, but it is usually easier to measure liquids by volume instead of mass. To do this measure called molarity is commonly used. Molarity (M) is defined as the number of moles of solute (n) divided by the volume (V) of the solution in liters. It is important to note that the molarity is defined as moles of solute per liter of solution, not moles of solute per liter of solvent. This is because when you add a substance, perhaps a salt, to some volume of water, the volume of the resulting solution will be different than the original volume in some unpredictable way. To get around this problem chemists commonly make up their solutions in volumetric flasks. These are flasks that have a long neck with an etched line indicating the volume. The solute (perhaps a salt) is added to the flask first and then water is added until the solution reaches the mark. The flasks have very good calibration so volumes are commonly known to at least four significant figures.

Process of Making a Standard Solution from Liquids

How to mix a dilute solution from a concentrated stock solution
A solution of a desired concentration can be prepared by diluting a small volume of a more-concentrated solution, a stock solution, with additional solvent. – Calculate the number of moles of solute desired in the final volume of the more-dilute solution and then calculate the volume of the stock solution that contains the amount of solute. – Diluting a given quantity of stock solution with solvent does not change the number of moles of solute present. – The relationship between the volume and concentration of the stock solution and the volume and concentration of the desired diluted solution is (Vs) (M s) = moles of solute = (Vd) (M d). Copyright © 2007 Pearson Benjamin Cummings. All rights reserved.

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

MConc.VConc. = MDiluteVDilute
Dilution of Solutions Solution Guide Formula Weight Specific Gravity Molarity Reagent Percent To Prepare 1 Liter of one molar Solution Acetic Acid Glacial (CH3COOH) 60.05 1.05 17.45 99.8% 57.3 mL Ammonium Hydroxide (NH4OH) 35.05 0.90 14.53 56.6% 69.0 mL Formic Acid (HCOOH) 46.03 1.20 23.6 90.5% 42.5 mL Hydrochloric Acid (HCl) 36.46 1.19 12.1 37.2% 82.5 mL Hydrofluoric Acid (HF) 20.0 1.18 28.9 49.0% 34.5 mL Nitric Acid (HNO3) 63.01 1.42 15.9 70.0% 63.0 mL Perchloric Acid 60% (HClO4) 100.47 1.54 9.1 60.0% 110 mL Perchloric Acid 70% (HClO4) 1.67 11.7 70.5% 85.5 mL Phosphoric Acid (H3PO4) 97.1 1.70 14.8 85.5% 67.5 mL Potassium Hydroxide (KOH) Sodium Hydroxide (NaOH) 40.0 19.4 45.0% Sulfuric Acid (H2SO4) 98.08 1.84 18.0 50.5% 51.5 mL This chart quickly shows you the amount of concentrated acid needed to make 1 liter of a 1 M solution. If you need a 5 M solution, add 5x the amount of acid in the same volume. MConc.VConc. = MDiluteVDilute

**Safety Tip: When diluting, add acid or base to water.**
Acids (and sometimes bases) are purchased in concentrated form (“concentrate”) and are easily diluted to any desired concentration. Dilutions of Solutions  **Safety Tip: When diluting, add acid or base to water.** C = concentrate D = dilute Dilution Equation: Concentrated H3PO4 is 14.8 M. What volume of concentrate is required to make L of M H3PO4? VC = L = 845 mL

How would you mix the above solution?
1. Measure out L of concentrated H3PO4 . 2. In separate container, obtain ~20 L of cold H2O. 3. In fume hood, slowly pour [H3PO4] into cold H2O. 4. Add enough H2O until L of solution is obtained. Be sure to wear your safety glasses!

You have 75 mL of conc. HF (28.9 M); you need 15.0 L of
0.100 M HF. Do you have enough to do the experiment? MCVC = MDVD 28.9 M (0.075 L) = M (15.0 L) Yes; we’re OK. mol HAVE > 1.50 mol NEED

What do you call a tooth in a glass of water? - A one molar solution.
One mole, in solution. What do you call a tooth in a glass of water?        - A one molar solution.

Dilution Preparation of a desired solution by adding water to a concentrate. Moles of solute remain the same.

Dilution What volume of 15.8M HNO3 is required to make 250 mL of a 6.0M solution? GIVEN: M1 = 15.8M V1 = ? M2 = 6.0M V2 = 250 mL WORK: M1 V1 = M2 V2 (15.8M) V1 = (6.0M)(250mL) V1 = 95 mL of 15.8M HNO3

Preparing Solutions How to prepare 500 mL of 1.54 M NaCl solution
mass 45.0 g of NaCl add water until total volume is 500 mL 500 mL volumetric flask 500 mL mark 45.0 g NaCl solute

Preparing Solutions 500 mL of 1.54M NaCl
molality molarity 1.54m NaCl in kg of water 500 mL of 1.54M NaCl mass 45.0 g of NaCl add kg of water mass 45.0 g of NaCl add water until total volume is 500 mL 500 mL water 500 mL volumetric flask 45.0 g NaCl 500 mL mark

Preparing Solutions measure 95 mL of 15.8M HNO3
250 mL of 6.0M HNO3 by dilution measure 95 mL of 15.8M HNO3 95 mL of 15.8M HNO3 combine with water until total volume is 250 mL 250 mL mark Safety: “Do as you oughtta, add the acid to the watta!” water for safety

Solution Preparation Lab
Turn in one paper per team. Complete the following steps: A) Show the necessary calculations. B) Write out directions for preparing the solution. C) Prepare the solution. For each of the following solutions: 1) mL of 0.50M NaCl 2) 0.25m NaCl in mL of water 3) mL of 3.0M HCl from 12.1M concentrate.

Textbook Problems Textbook problems Textbook problems Keys

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