Entry Task: Block 2 Sept 19-20

1. Entry Task: Block 2 Sept 19-20 Question: What volume of a 3.00M KI solution would you use to make 0.300 L of a 1.25 M KI solution? You have 5 minutes

2. Solution Preparation

3. Concentration (Molarity) • Concentration — the amount of solute per unit of solution. • Molarity (M) — expresses concentration:

4. Calculate the molarity of a solution that contains 20.0g copper(I) chloride and has a total volume of 300.0 mL. Must have the correct units.

5. Given: 20.0 grams CuCl; 300.0 ml solution Need: Moles CuCl; L of solution mol CuCl 1 20.0 g CuCl 0.202 mol CuCl = 99.0 g CuCl 1 L 300.0 mL = 0.3000 L 1000 mL

6. 0.673 M CuCl

7. How many grams of NaCl are needed to make 2.5 L of 0.20 molar solution? Given: 0.20 M solution; 2.5 L solution Need: grams of NaCl (L) (L) Must calculate # of moles, and then convert it into grams. mol = ML

8. Mol = ML Mol = 0.20 mol NaCl x 2.5 L = 0.50 mol NaCl L 55.84 g NaCl 0.50 mol NaCl 28 g NaCl = 1 mol NaCl

9. Dilution of Stock Solutions • Chemicals are purchased in concentrated form. They need to be diluted for most lab use. • Formula for dilution: Mi Vi = Mf Vf

10. How much stock (12 M)HCl(aq) is required to make 200.0 mL of 3MHCl (aq)? Mi Vi = Mf Vf Mi = 12 M Vf = 200.0 mL Mf = 3 M Vi = ? mL Must rearrange equation above to solve for initial volume.

11. 50. mL Measure 150 mL of water in a beaker. Slowly add 50.0 mL of 12 M HCl for a final volume of 200.0 mL.

12. Two things to note: • Always add concentrated acid to water, and not the reverse to avoid unwanted splashing due to the heat generated. • 2) When diluting a solution, the amount of solute doesn’t change, only the final volume. WA AW moles before dilution = moles after dilution

13. If diluting a solution other than acids, start with initial volume of concentrated solution, and then dilute with distilled water until you have the desired volume. You try one!

14. We want to prepare 500. mL of 1.00 M acetic acid from a 17.5 M stock solution of acetic acid. What volume of the stock solution is required? Mi Vi = Mf Vf Mi = 17.5 M Vf = 500. mL Mf = 1.00 M Vi = ? mL

15. 28.6 mL Pour 471.4 mL of distilled water into a beaker. Slowly pour the 28.6 ml of acid into the water and swirl. Fill the container with distilled water to 500. mL.

16. There may be times when you must consider the concentration of ions in a solution. (You must consider the subscripts for this) MgCl2 Mg2+ + 2Cl- In a solution of 0.25 M MgCl2 you have: M of Mg2+ = 0.25 M M of Cl- = 2 x 0.25 M = 0.50 M What is the concentration of each ion in the following? 0.15 M Na3P M of Na+ = 0.45 M M of P3- = 0.15 M

17. Titrations • Determining the concentration of an unknown solution. • Use a 2nd solution of known concentration (standard solution) that undergoes a reaction with the unknown solution. • Use the ratios in the balanced equation along with the M = mol/L equation to determine molarity of unknown.

18. The point at which the two solutions are stoichiometrically equal is known as the equivalence point. • The reaction is complete and no excess reactant is present. • How do we know when this occurs during the reaction?

19. In acid base reactions dyes known as indicators are used. • Phenolphthalein is colorless in acid solution, and pink in basic solution. • End point is reached when a drop of the base remains pink. There is no acid for this drop to react with and the solution is now basic.

20. What mass of KI is required to make 500. mL of a 2.80 M KI solution? moles of solute M = molarity = liters of solution volume KI moles KI grams KI 1 L 2.80 mol KI 166 g KI x 500. mL = 232 g KI x x 1000 mL 1 L soln 1 mol KI Solution Stoichiometry The concentration of a solution is the amount of solute present in a given quantity of solvent or solution. M KI M KI 4.5

21. 4.5

22. Acid/Base Titrations • Experimental technique that determines the concentration (in Molarity) of an acid (or base) • This is based upon an acid/base neutralization reaction. • ACID +BASE -> SALT + H2O • Base (or acid) is added until there is the same amount (same # moles) of base and acid.

23. Titrations In a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete. Equivalence point – the point at which the reaction is complete Indicator – substance that changes color at (or near) the equivalence point Slowly add base to unknown acid UNTIL the indicator changes color 4.7

24. Fig. 4.17a,b

25. Acid-Base Titrations Base; (OH)- Acid + Base -> Salt + H2O Acid; H+ Introductory Chemistry 2/e by N Tro, Prentice Hall, 2006, pg 480

26. At the endpoint of an acid/base titration…. • Moles acid = Moles base • (MV)acid = (MV)base • Note • If solid; moles = mass/ MM • If aqueous solution; moles = MV

27. WRITE THE CHEMICAL EQUATION! What volume of a 1.420 M NaOH solution is Required to titrate 25.00 mL of a 4.50 M H2SO4 solution? H2SO4 + 2NaOH 2H2O + Na2SO4 M M rx volume acid moles acid moles base volume base base acid coef. 4.50 mol H2SO4 2 mol NaOH 1000 ml soln 25.00 mL x x x = 159 mL 1000 mL soln 1 mol H2SO4 1.420 mol NaOH 4.7

28. WRITE THE CHEMICAL EQUATION! What volume of a 1.420 MNaOH solution is required to titrate 25.00 mL of a 4.50 M H2SO4 solution? H2SO4 + 2NaOH 2H2O + Na2SO4 Treat as a dilution problem M1V1=M2V2 2(4.50M)(25.00ml) = (1.420M)(Xml) 225= X 225 = 1.420X 1.420 159 mls of NaOH 4.7

29. Titration Problem A 20.0 mL solution of Sr(OH)2 is neutralized after 25.0 mL of standard 0.05 M HCl is added. What is the concentration of Sr(OH)2? 2HCl + Sr(OH)22H2O + SrCl2 0.05Macid25mlacid = 2 X Mbase 20mlbase = 0.0313M base 1.25 40