1 / 15

SOLUTIONS

SOLUTIONS. Concentration Measurement: Percent and PPM. Mr. Shields Regents Chemistry U12 L03. Percent Concentration (V/V). The two “Percent” methods for measuring Concentration are Volume/Volume (V/V) and Mass/Mass (m/m). Let’s look at the V/V method 1 st :.

tamar
Download Presentation

SOLUTIONS

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. SOLUTIONS Concentration Measurement: Percent and PPM Mr. Shields Regents Chemistry U12 L03

  2. Percent Concentration (V/V) The two “Percent” methods for measuring Concentration are Volume/Volume (V/V) and Mass/Mass (m/m). Let’s look at the V/V method 1st: The formula for % Concentration (V/V) is: Percent (V/V) = Volume of solute x 100 Volume of Solution

  3. Percent Concentration (V/V) Using Percent (V/V) to specify Concentration is used when BOTHthe solute and solvent areliquids Example: water + vinegar (H20 + CH3COOH) It also assumes no INTERACTION (i.e. reaction) between The solute and the solvent. And notice that we are dividing the volume of the SOLUTE By the volume of the SOLUTION – not the other way around Percent (V/V) = Volume of Solute x 100 Volume of Solution

  4. Percent Concentration (V/V) V/V solution concentrations tells us what % of solute is present in the solution. Doubling the amount of solute will Not necessarily double the % V/V. As we add solute to the solvent the volume of the solvent also increases. % V/V = 3/9 = 33% % V/V = 6/12 = 50% How much solute and solvent is present in each? AND REMEMBER ! It is essential to express both the Solute and the solution volumes in the SAME UNITS. For example Liters and Liters or Milliliters and Milliliters,

  5. Percent (V/V) Problem 27 ml of ethanol are dissolved in 270 ml of Octane. What is the % Concentration of Ethanol? % Conc. (V/V) = 27 ml x 100 = 9.1% (27 + 270) What is the concentration in % (V/V) if 1.5 ml of Vinegar Is dissolved in 30 ml of water? % conc. of vinegar (V/V) = 1.5 ml x 100 = 4.76 % (1.5 + 30) Note: Vinegar bought at the store is about a 5% sol’n

  6. + mass mass mass = Percent Concentration (m/m) Percent Concentration by mass, % (m/m), is calculated very Much like the % (v/v) method: Percent (m/m) = Mass of solute x 100 Mass of Solution NOTE it’s the mass of the Solution and NOT the just the Solvent. As in v/v calculations the total solution is equal to the sum of Both the Solute and the Solvent

  7. Percent (m/m) Problem 1) What is the % Concentration (m/m) of I2 if 16.7g of I2 Are dissolved in 250g of ethanol? % Conc. (m/m) = 16.7g x 100 = 6.26% (16.7g + 250g) 2) 24g of AgNO3 were dissolved in 125ml of water. What Is the % concentration (m/m) of AgNO3 in solution? Desity of H2O = 1g/ml; the mass of the water = 125g Conc. of AgNO3 (m/m) = 24g x 100 = 16.1% (24 + 125)

  8. PPM The last method of expressing sol’n concentrations Quantitatively is PARTS PER MILLION (i.e. ppm) This method is used extensively in the Environmental Field. Why is this the case? It’s a method that allows one to easily comprehend the Concentration of extremely small amounts of solute in Solvent. So we might find for example that a sample of drinking Water contains 3.2 ppm of Arsenic and compare that to the allowable levels of As in drinking water (0.05 ppm)

  9. PPM If the allowable limits of As in drinking water is 0.05 ppm, What does this mean in terms of concentration? - First, 1 ppm means 1 part in 1 million (106) parts So, if the acceptable level of As is 0.05 ppm then for every 5 Molecules of As molecules allowed in drinking water there must be 100,000,000 (108) molecules of water. Expressed another way we could say… - for every 5g of As there must be 100,000,000g of H20!

  10. PPM Calculations How do we calculate ppm? We’ll use this equation… ppm of solute = Mass of Solute x 106 Mass of Solution Notice that again we are dealing with the TOTAL mass Of the solution. Let’s look at an example…

  11. PPM Calculations A 6.5g sample of well water is found to contain 3.5 x 10-6g of Pb+2. What is the concentration of lead In the well water in ppm? PPM Lead = mass of the solute x 106 mass of the solution PPM Lead = 3.5 x 10-6g x 106 6.5g PPM Lead = 0.538 ppm Is this the total mass? Why can we use it?

  12. PPM Calculations What is the concentration in ppm of NaCl in water if 1.2g is dissolved in 240 ml of water? PPM NaCl = mass of the solute x 106 mass of the solution PPM NaCl = 1.2g x 106 = 4975.1 ppm (1.2 +240)g The salinity of sea water is about 35,000 ppm

  13. PPM Calculations What if we hadn’t added in the 1.2g of NaCl to the Mass of solvent (240g). Would it have made a difference? Let’s check it out … PPM NaCl = mass of the solute x 106 mass of the solution PPM NaCl = 1.2g x 106 = 5000 ppm 240g Mass of solvent ONLY YES! It would have made a difference 5000ppm vs 4975ppm So always add in the mass of the Solute to the solvent if it is a Significant amount

  14. PPM Calculations What if we had only added in 1.2 mg of NaCl to 240 ml Of water. Does it make much difference if we add the Mass of the salt to the solvent? Calculate it both ways and see … PPM NaCl = 0.0012g x 106 = 4.999975ppm (0.0012 + 240)g PPM NaCl = 0.0012g x 106 = 5.000000ppm 240g Are these two numbers significantly different? No, not really! So here we could have ignored the mass Of the salt

  15. Global Warming is blamed on the increase in CO2 Levels in the atmosphere. A) How much did CO2 increase in both ppm and % between 1957 and 2000? 54 ppm increase or B) In 2000 what Was the Percent by Volume of CO2 In the atmosphere? 16.98% increase in CO2 (372-318)/318 x 100 372 X 100 = 106 O.037%

More Related