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Solutions

Solutions. Chapter 14. What are Solutions?. Solution : a uniform (homogeneous) mixture that may contain solids, liquids or gases. Parts of a solution: Solute: What gets dissolved in a solution – usually present in smaller amounts

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Solutions

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  1. Solutions Chapter 14

  2. What are Solutions? • Solution : a uniform (homogeneous) mixture that may contain solids, liquids or gases. • Parts of a solution: • Solute: What gets dissolved in a solution – usually present in smaller amounts • Solvent: What does the dissolving in a solution – usually present in larger amounts – WATER is the “UNIVERSAL SOLVENT”

  3. 14.2 - Solution Concentration • Concentration: a measure of how much solute is dissolved in a specific amount of solvent or solution. • We will express concentration in units of Molarity (which will be explored at a later time) • Example: a solution with a concentration of 1.5 M is said to be a 1.5 “Molar” solution

  4. substance being dissolved total combined volume Molarity • Concentration of a solution.

  5. MASS IN GRAMS MOLES NUMBER OF PARTICLES LITERS OF SOLUTION Molarity Calculations 6.02  1023 (particles/mol) molar mass (g/mol) Molarity (mol/L)

  6. Molarity 2M HCl What does this mean?

  7. M = Molarity Calculations • Find the molarity of a 250 mL solution containing 10.0 g of NaF. 10.0 g 1 mol 41.99 g = 0.238 mol NaF 0.238 mol 0.25 L = 0.95M NaF

  8. Solution Concentration (cont’d) • A “Concentrated” solution contains a large amount of solute relative to its solubility • A “dilute” solution contains a small amount of solute relative to its solubility

  9. The Dilution Equation M1V1 = M2V2 M1 = initial molarity (“stock solution”) V1 = initial volume (L) M2 = final (desired) molarity V2 = final volume (L) This equation is used when you have a “stock solution” of higher molarity than you need and you need to dilute it to a lower molarity by adding additional solvent.

  10. Dilution Equation (cont’d) What volume of 3.00M KI stock solution would you use to make 0.300 L of a 1.25M KI solution?

  11. DILUTION PROBLEMS How much concentrated 18 M sulfuric acid is needed to prepare 250 mL of a 6.0 M solution? How much concentrated 12 M hydrochloric acid is needed to prepare 100 mL of a 2.0 M solution? What volume of 3.0 M nitric acid could be made from 25 mL of a 15 M stock solution? If I add 50 mL of water to 130 mL of 12 M hydrochloric acid, what is the new, diluted concentration? If I add 100 mL of water to 100 mL of 6 M nitrous acid, what is the new, diluted concentration?

  12. HOMEWORK • COMPLETE THE LAST 5 DILUTION PROBLEMS AS HOMEWORK • FOUND UNDER THE HOMEWORK TAB • DUE TOMORROW AT THE START OF CLASS: • THE 5 QUESTIONS DONE IN CLASS • THE 5 HOMEWORK QUESTIONS • FOR A TOTAL OF 10 DILUTION PROBLEMS

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