Solution Concentration.ppt
- 2. Review A solution is a homogeneous mixture. The solvent is the major component of the solution. The solute is the minor component and active ingredient. A saturated solution holds the maximum amount of solute that is theoretically possible for a given temperature.
- 3. How would you describe this picture?
- 4. Solution Concentration Is one glass of tea stronger than the other? – What’s true about the “stronger” glass of tea? – How much tea does it have in it compared to the other glass?
- 5. Solution Concentration Concentration – a ratio comparing the amount of solute to the amount of solution. Many ways of expressing concentration: – Molarity (M) is the one we will be dealing with
- 6. Concentrated vs. Dilute The words “concentrated” and “dilute” are opposites. EX: The dark tea is more concentrated than the light tea. EX: The light tea is more dilute than the dark tea.
- 7. Concentrated vs. Dilute Concentrated solution Dilute solution = solute particles
- 8. Molarity Molarity (M) – UNITS: mol/L or Molar (M) – Example: 0.500 mol/L = 0.500 M solution of Liters solute of moles Molarity
- 9. Molarity What is the Molar concentration of a sol’n if 20.0 grams of KNO3 (MM = 101.11 g/mol) is dissolved in enough water to make 800. mL? – Convert g of KNO3 to mol of KNO3 – Convert mL to L 3 3 3 3 KNO mol 0.198 KNO g 101.11 KNO mol 1 x KNO g 20.0 L 0.800 mL 1 L 0.001 x mL 800.
- 10. Molarity What is the Molar concentration of a sol’n if 0.198 mol KNO3 is dissolved in enough water to make 0.800 L? M 0.248 L 0.800 KNO mol 0.198 Molarity 3
- 11. Molarity What is the Molar concentration of a sol’n if 10.5 grams of glucose (MM = 180.18 g/mol) is dissolved in enough water to make 20.0 mL of sol’n? – Convert g of glucose to mol of glucose. – Convert mL to L. glucose mol 0.0583 glucose g 180.18 glucose mol 1 x glucose g 10.5 L 0.0200 mL 1 L 0.001 x mL 20.0
- 12. Molarity What is the Molar concentration of a sol’n if 0.0583 mol of glucose is dissolved in enough water to make 0.0200 L of sol’n? M 2.92 L 0.0200 glucose mol 0.0583 Molarity
- 13. Calculating Grams How many grams of KI (MM = 166.00 g/mol) are needed to prepare 25.0 mL of a 0.750 M solution? – Convert mL to L. – Solve for moles. – moles of KI = 0.750 M x 0.0250 L = 0.0188 mol KI L 0.0250 mL 1 L 0.001 x mL 25.0 L 0.0250 KI of moles x M 0.750
- 14. Calculating Grams How many grams of KI (MM = 166.00 g/mol) are needed to prepare 25.0 mL of a 0.750 M solution? – Convert 0.0188 mol KI to grams. KI g 3.12 KI mol 1 KI g 166.00 x KI mol 0.0188
- 15. Calculating Grams How many grams of HNO3 (MM = 63.02 g/mol) are present in 50.0 mL of a 1.50 M sol’n? – Convert mL to L. • 50.0 mL = 0.0500 L – Solve for moles: • moles = (1.50 M)(0.0500 L) = 0.0750 mol HNO3 – Convert 0.0750 mol HNO3 to grams: • 0.0750 mol HNO3 = 4.73 g HNO3
- 16. Dilution Dilute (verb) - to add solvent to a solution. – Decreases sol'n concentration. – M1V1 = M2V2 • M1 = initial conc. • V1 = initial volume • M2 = final conc. • V2 = final volume – Assumes no solute is added.
- 17. Dilution Stock Solution Impractically High Concentration Usable Solution Add H2O Question for Consideration: Why do you think chemical supply companies typically sell acids (and other solutions) in extremely high concentrations when it would be safer to ship more dilute solutions?
- 18. Dilution To what volume should 40.0 mL of 18 M H2SO4 be diluted if a concentration of 3.0 M is desired? – What do we want to know? • V2 – What do we already know? • M1 = 18 M • V1 = 40.0 mL • M2 = 3.0 M – (18 M)(40.0 mL) = (3.0 M)V2 – 720 M*mL = (3.0 M)V2 – V2 = 240 mL
- 19. Dilution You are asked to prepare 500. mL of 0.250 M HCl, starting with a 12.0 Molar stock sol'n. How much stock should you use? – What do we want to know? • V1 – What do we already know? • M1 = 12.0 M • M2 = 0.250 M • V2 = 500. mL – (12.0 M) V1 = (0.250 M)(500. mL) – (12.0 M) V1 = 125 M*mL – V1 = 10.4 mL
- 20. To how much water should you add 20.0 mL of 5.00 M HNO3 to dilute it to 1.00 M? – What do we want to know? • How much water to add. (V2 - V1) – What do we already know? • M1 = 5.00 M • V1 = 20.0 mL • M2 = 1.00 M – (5.00 M)(20.0 mL) = (1.00 M) V2 – 100. M*mL = (1.00 M) V2 – V2 = 100. mL – Water added = 100. mL - 20.0 mL = 80. mL Dilution
- 21. Colligative Properties Colligative properties are properties of solutions that are affected by the number of particles but not the identity of the solute