2. Acids and Bases Notes Part 2

3. Acid Rain Many industrial processes produce gases such as NO, NO 2, CO 2, SO 2, and SO 3. These compounds can dissolve in atmospheric water to produce acidic solutions that fall to the ground in the form of rain or snow. Marble found in many buildings and statues is composed of calcium carbonate, when acid snow or rain falls on these structures a great deal of damage is caused. (page 475)

4. Neutralization Reactions Neutralization – the reaction of an acid with a base to produce water and a salt. Occurs when H 3 O + and OH - ions are supplied in equal numbers by the reactants. Salt- an ionic compound composed of a cation from a base and an anion from an acid.

5. Examples: H 2 SO 4 + 2KOH  K 2 SO 4 + 2H 2 O HCl + NaOH  NaCl + H 2 O H 3 PO 4 + 3NH 4 OH  (NH 4 ) 3 PO 4 + 3H 2 O salts water

6. Titrations Titration – A neutralization reaction of an acid by a base or vice versa; it is usually used to find the concentration, molarity, of an unknown acid or base. The concentration of the other is known. Titrant – solution of known concentration (in buret). The titrant is added to a solution of unknown concentration until the substance being analyzed is just consumed (stoichiometric point or equivalence point).

7. Titrations Indicators-Indicators change color at the end point of a titration. Examples: phenolphthalein and methyl orange Equivalence point-the point at which the two solutions used in a titration are present in chemically equivalent amounts. Strong acid + Strong base at equivalence point pH=7 http://wwwchem.uwimona.edu.jm:1104/software/titr.html http://wwwchem.uwimona.edu.jm:1104/software/titr.html

8. Strong Acid-Strong Base Titrations: Simple reaction H+ + OH-  H 2 0 The pH is easy to calculate because all reactions go to completion. At the equivalence point, the solution is neutral (pH=7.00).

9. Practice Problems: 1. Suppose you know 20. mL of a 0.0050M NaOH solution is required to reach the equivalence point in the titration of 10. mL of an unknown molarity of HCl. What is the molarity of the Start with the balanced equation for the reaction. HCl + NaOH  NaCl + H 2 O

10. Write out all givens and unknowns before beginning problem. ?M of known NaOH  0.0050M ?moles of known NaOH  ?liters of known NaOH  20.mL ?M of unknown HCl  ?moles of unknown HCl  ?liters of unknown HCl  10.mL

11. Determine the moles of known solution used. ?M of known NaOH  0.0050M ?moles of known NaOH  ?liters of known NaOH  20.mL (Remember that Molarity = Moles ) Liters 0.005M NaOH = ?moles.020 L ?moles = (.020 L) (.0050 M) ?moles = 0.00010 moles NaOH

12. So now we know: ?M of known NaOH  0.0050M ?moles of known NaOH  0.00010 moles ?liters of known NaOH  20.mL ?M of unknown HCl  ?moles of unknown HCl  ?liters of unknown HCl  10.mL

13. Determine the moles of unknown solution used. (Using the moles of known solution used and the equation for a mole ratio.) ?moles of unknown HCl  HCl + NaOH  NaCl + H 2 O ? mol HCl  0.00010 mol NaOH X 1 mol HCl = 0.00010mol HCl 1 mol NaOH

14. So now we know: ?M of known NaOH  0.0050M ?moles of known NaOH  0.00010 moles ?liters of known NaOH  20.mL ?M of unknown HCl  ?moles of unknown HCl  0.00010 moles ?liters of unknown HCl  10.mL

15. Determine the molarity of the unknown. ?M of unknown HCl  ?moles of unknown HCl  0.00010 moles ?liters of unknown HCl  10.mL ?M = 0.00010 moles HCl 0.010 L 0.010M HCl

16. So now we know it all!!! ?M of known NaOH  0.0050M ?moles of known NaOH  0.00010 moles ?liters of known NaOH  20.mL ?M of unknown HCl  0.010M ?moles of unknown HCl  0.00010 moles ?liters of unknown HCl  10.mL

17. More Examples Complete examples 8 and 9 from your notes and then complete your homework problems.