1. Ch. 15 pH

2. What is pH? ► pH is a logarithmic measurement of how acidic a solution is. ► pH = “pouvoir hydrogen” = hydrogen power ► If the pH = 1 to 6.9, it’s an acid ► If the pH = 7, it’s neutral ► If the pH = 7.1 to 14, it’s a base

3. The pH Scale:

4. Where pH comes from: ► In water, some water molecules split up into H + and OH - ions like this: ► H 2 O (l) ⇆ H + (aq) + OH - (aq) ► H + could also be H 3 O + or just a proton... ► In pure water, the reverse reaction is favored (a lot!) in equilibrium: [ ] means Molarity or concentration ► [H + ] = 1.0 x 10 -7 moles/L ► [OH - ] = 1.0 x 10 -7 moles/L

5. Acid Neutral Base Acid Neutral Base

6. Fun with LOGS LOGS

7. Logarithm A logarithm is an EXPONENT!!! Definition: - exponent expressing the power to which a fixed number must be raised in order to get a given number - a.k.a log

8. The Relationship The Relationshiplogb(x) = y which is the same as x = by Stated: log-base-b of y equals x

9. Why are we learning about logs? 1) pH is logrithmic

10. The exponent on the 10 is the pH! ► We state the concentration of H + ions in water as its "pH" which is based on the exponent on the 10. ► In pure water, [H + ] = 1.0 x 10 -7 moles/L ► pH = -log [H + ] ► So the pH of pure water is pH = 7.0 ► Remember logs? ie: 10 3 = 1000 ► Then log(1000) = 3

11. How many legs does the elephant have?

12. What’s the pH of 0.005 M HNO 3 ? ► HNO 3  H + + NO 3 - (complete dissociation) ► So, [H + ] = 0.005 M ► pH = -log[H + ] = -log[0.005] = 2.3 (acid range)

13. pH + pOH = 14 ► Or: [H + ]x[OH - ] = 1.0 x 10 -14 ► For a base, pOH = -log [OH - ] ► Ex: What’s the pH of 0.002 M NaOH? ► First, the [OH - ] = 0.002 M ► Then, pOH = -log[0.002] = 2.7 (try it!) ► Then pH = 14 – pOH = 11.3 (basic range)

14. Find the pH of the following: ► 0.025 M HNO 3 ► Acid, [H + ] = 0.025; pH = 1.6 ► 0.00056 M HBr ► Acid, [H + ] = 0.00056; pH = 3.25 ► 0.0301 M KOH ► Base, [OH - ] = 0.0301; pOH = 1.5, pH = 12.5 ► 0.0030 M Sr(OH) 2 ► Note! Sr(OH) 2  Sr 2+ + 2OH ¯ (1 to 2 Ratio) ► Base, [OH - ] = 0.0060; pOH = 2.2, pH = 11.8 ► 0.0044 M H 2 SO 4 ► Acid, [H + ] = 0.0044; pH = 2.35

15. Count Up the Black Dots!

16. Titration: ► A way to figure out what the molarity of an acid is by adding a known molarity base to it until it is neutralized. ► Or Visa Versa: Add a known acid to an unknown molarity base.

17. The pH of some common things.

18. How to do a titration: ► 1. Get a few mL of the unknown solution and add some water and a little bit of an indicator. ► 2. Fill your buret with the “standard” acid or base solution; the one you know the molarity of. ► 3. Slowly dribble from your buret into the unknown solution until the endpoint is reached (shown by the indicator color). Record the mL of how much you added.

19. How a Titration works:

20. Titration in action, continued:

21. Before the HCl is added, the three things in the solution are water molecules, OH - ions and Na + ions.

22. Adding HCl is really adding H + ions which react with the OH - ions to make water. The green spheres are Cl - ions which don’t react.

23. At the endpoint, there are just Cl - ions, Na + ions, and H 2 O molecules in the solution.

24. Example: ► 55.0 mL of 0.5 M NaOH are needed to titrate 20.0 mL of an unknown M HCl solution to the endpoint. What’s the Molarity of the HCl? ► Step 1: Write and balance the reaction ► Step 2: Use DA to find moles of HCl ► Step 3: Divide moles/Liters to get the Molarity

25. Step 1: Write the Reaction ► HCl + NaOH  NaCl + H 2 O

26. Step 2: Find the moles of HCl that were in your unknown M HCl solution. ► The “arrow” always starts with the volume of the standard solution; the one which came from the buret. ► 0.055 L NaOH  ? mols HCl

27. Step 3: Divide the moles of HCl by the Liters of HCl to find the Molarity. ► We found that we had 0.028 mols HCl in our 0.020 L of HCl solution that we started with:

28. A pH graph of this titration.

29. What acid rain did to ol’ George… ► Marble is made of calcium carbonate, a weak base which reacts with the acid rain.

30. Sample Problem: ► 1. You titrate 20 mL of an unknown M sulfuric acid. 15.5 mL of 1.5 M NaOH is required to reach the endpoint. What’s the concentration of the sulfuric acid? ► Follow the last example in the notes. Write the reaction, then find the moles of sulfuric acid with DA, then find the Molarity of it. ► You should have gotten 0.0116 mols of H 2 SO 4, giving [H 2 SO 4 ] = 0.58 M

31. That’s All For Now!