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![Concentration =
# of fish
volume (L)
Concentration =
V = 1000 mL
n = 2 fish
Concentration = 2 “fishar”
V = 1000 mL
n = 4 fish
[ ] = 4 “fishar”
V = 5000 mL
n = 20 fish
[ ] = 4 “fishar”
1 fish
1 (L)
Concentration = 1 “fishar”](https://image.slidesharecdn.com/dilutionofsolutions-250915032702-deb9bd4e/75/Dilution-of-Solutioasdadasdadasdasdasdans-ppt-1-2048.jpg)
![V = 1000 mL
n = 2 moles
Concentration = 2 molar
V = 1000 mL
n = 4 moles
[ ] = 4 molar
V = 5000 mL
n = 20 moles
[ ] = 4 molar
Concentration =
# of moles
volume (L)
V = 250 mL
n = 8 moles
[ ] = 32 molar](https://image.slidesharecdn.com/dilutionofsolutions-250915032702-deb9bd4e/75/Dilution-of-Solutioasdadasdadasdasdasdans-ppt-2-2048.jpg)
![Be sure to wear your safety glasses
safety glasses!
1. Measure out 0.845 L of concentrated H3
PO4
.
2. In separate container, obtain ~20 L of cold H2
O.
3. In fume hood, slowly pour [H3
PO4
] into cold H2
O.
4. Add enough H2
O until 25.00 L of solution is obtained.
How would you mix the above solution?](https://image.slidesharecdn.com/dilutionofsolutions-250915032702-deb9bd4e/75/Dilution-of-Solutioasdadasdadasdasdasdans-ppt-20-2048.jpg)
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Dilution of Solutioasdadasdadasdasdasdans.ppt
- 1. Concentration = # offish volume (L) Concentration = V = 1000 mL n = 2 fish Concentration = 2 “fishar” V = 1000 mL n = 4 fish [ ] = 4 “fishar” V = 5000 mL n = 20 fish [ ] = 4 “fishar” 1 fish 1 (L) Concentration = 1 “fishar”
- 2. V = 1000mL n = 2 moles Concentration = 2 molar V = 1000 mL n = 4 moles [ ] = 4 molar V = 5000 mL n = 20 moles [ ] = 4 molar Concentration = # of moles volume (L) V = 250 mL n = 8 moles [ ] = 32 molar
- 3. Making Molar Solutions …from liquids (More accurately,from stock solutions)
- 4. Concentration…a measure ofsolute-to-solvent ratio concentrated vs. dilute “lots of solute” “not much solute” “watery” Add water to dilute a solution; boil water off to concentrate it.
- 5. Making a Dilute Solution Timberlake, Chemistry7th Edition, page 344 initial solution remove sample diluted solution same number of moles of solute in a larger volume mix moles of solute
- 6. Concentration “The amount ofsolute in a solution” mol L M 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 D. molality (m) = moles of solute kg of solvent M = mol L % by mass – medicated creams % by volume – rubbing alcohol
- 7. ppm 1 inchin 16 miles ppb 1 inch in 16,000 miles ppt 1 sec = 32,000 years
- 8.
- 9. precise; expensive Range: Glassware –Precision and Cost beaker vs. volumetric flask When filled to 1000 mL line, how much liquid is present? beaker 5% of 1000 mL = 50 mL volumetric flask 1000 mL + 0.30 mL 950 mL – 1050 mL 999.70 mL– 1000.30 mL imprecise; cheap Range:
- 10. Markings on Glassware TC20o C “to contain at a temperature of 20 o C” TD “to deliver” T s “time in seconds” 22 500 mL + 5% Range = 500 mL + 25 mL 475 – 525 mL Beaker Graduated Cylinder Volumetric Flask 500 mL + 0.2 mL Range = 499.8 – 500.2 mL 1000 mL + 5 mL Range = 1000 mL + 5 mL 475 – 525 mL
- 11. ~ ~ ~ ~ ~ ~ ~ ~ waterin grad. cyl. mercury in grad. cyl. Measure to part of meniscus w/zero slope. Copyright © 2007 Pearson Benjamin Cummings. All rights reserved.
- 12. How to mixsolid 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 How much will this weigh? 1 Na @ 23g/mol + 1O @ 16g/mol + 1 H @ 1 g/mol MMNaOH = 40g/mol 40.0 g NaOH 1 mol NaOH X g NaOH = 15.0 mol NaOH = 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. FOR EACH CLASS: x 5 classes 15 mol NaOH 600 g NaOH
- 13. How to mixa Standard Solution Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 480 Wash bottle Volume marker (calibration mark) Weighed amount of solute
- 14. How to mixa Standard Solution http://www.chem.ucla.edu/~gchemlab/soln_conc_web.htm
- 15. Process of Makinga Standard Solution from Liquids Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 483
- 16. How to mixa dilute solution from a concentrated stock solution Copyright © 2007 Pearson Benjamin Cummings. All rights reserved.
- 17. Identify each volumeto two decimal places (values tell you how much you have expelled) 4.48 - 4.50 mL 4.86 - 4.87 mL 5.00 mL Reading a pipette www.chalkbored.com
- 18. Dilution of Solutions SolutionGuide 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) 100.47 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) 60.05 1.05 17.45 99.8% 57.3 mL Sodium Hydroxide (NaOH) 40.0 1.54 19.4 45.0% 85.5 mL Sulfuric Acid (H2SO4) 98.08 1.84 18.0 50.5% 51.5 mL MConc.VConc. = MDiluteVDilute
- 19. D D C C V M V M L) (25.00 M 0.500 ) (V M 14.8 V M V MC D D C C C = concentrate D = dilute Dilutions of Solutions Dilution Equation: Concentrated H3 PO4 is 14.8 M. What volume of concentrate is required to make 25.00 L of 0.500 M H3 PO4 ? VC = 0.845 L = 845 mL Acids (and sometimes bases) are purchased in concentrated form (“concentrate”) and are easily diluted to any desired concentration. **Safety Tip: When diluting, add acid or base to water.**
- 20. Be sure towear your safety glasses safety glasses! 1. Measure out 0.845 L of concentrated H3 PO4 . 2. In separate container, obtain ~20 L of cold H2 O. 3. In fume hood, slowly pour [H3 PO4 ] into cold H2 O. 4. Add enough H2 O until 25.00 L of solution is obtained. How would you mix the above solution?
- 21. Yes; we’re OK. You have75 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? > 2.1675 mol HAVE 1.50 mol NEED MCVC = MDVD 28.9 M (0.075 L) = 0.100 M (15.0 L)
- 22. 2 2 1 1 V M V M Dilution •Preparation of a desired solution by adding water to a concentrate. • Moles of solute remain the same.
- 23. Dilution • What volumeof 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
- 24. Preparing Solutions How toprepare 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
- 25. 500 mL volumetric flask Preparing Solutions 500mL of 1.54M NaCl 500 mL water 45.0 g NaCl – mass 45.0 g of NaCl – add water until total volume is 500 mL – mass 45.0 g of NaCl – add 0.500 kg of water 500 mL mark 1.54m NaCl in 0.500 kg of water molality molarity
Editor's Notes
- #5 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.
- #7 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 mass of solution parts per billion (ppb) = mass of solute 109 mass of solution – 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
- #8 http://www.phthalates.org/graphics/icon_people.jpg http://images.google.com/imgres?imgurl=http://www.phthalates.org/graphics/icon_people.jpg&imgrefurl=http://www.phthalates.org/didyouknow/ppm.asp&h=262&w=316&sz=117&hl=en&start=4&tbnid=BW6jt7Ubr-q9KM:&tbnh=97&tbnw=117&prev=/images%3Fq%3Dpart%2Bper%2Bmillion%26gbv%3D2%26hl%3Den http://www.plymouth.edu/wellness/footsteps.gif http://school.discoveryeducation.com/clipart/images/clock.gif
- #14 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.
- #15 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.
- #16 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. http://www.chem.ucla.edu/~gchemlab/soln_conc_web.htm
- #17 Solutions can be made using liquids or solids (or gases). To make a 5% solution v/v (volume to volume) This means to add 5 mL of solute in 95 mL of solvent. The total is 5 mL / 100 mL or 5%. For the diagram add 25 mL of liquid solute and add water to bring volume to 500 mL (about 475 mL water). SAFETY NOTE: Always add acid concentrate to water…never add water to concentrated acid. If you’ve seen what happens when water or ice crystals hit hot oil…a similar phenomenon occurs when water is added to concentrated acid. The addition of water to concentrated dissipates a large amount of heat. This heat rapidly boils the acid and causes it to spatter. If however, you start with a large volume of water and slowly add acid, the same amount of heat is generated. This time, the large volume of water is capable of absorbing the heat. The solution will not splatter.
- #18 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).
- #20 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.