1. Warmup (5 min) 1.Solve: log(3.9 x 10 -4 ) = 2.Name the compound HI and show how it dissociates in solution. 3.How do acids and bases differ? Write down WHATEVER you remember.

2. Acids, Bases, and pH

3. Acid or Base?  Aqueous solution  Have a high pH (above 7-14)  Have a low pH (below 0-7)  Feel slippery  Taste sour  Conduct electricity  Changes the color of pH paper  Corrosive: can damage skin  Neutralize before disposal  Both  Base  Acid  Base  Acid  Both

4. Acids: produce H + ions in water, forming H 3 O + (Arrhenius) or donate H + ions to bases (BL) HNO 3 HCl H 2 SO 4 H3C6H5O7H3C6H5O7 H 3 PO 4 HClO Bases: produce OH - in solution (Arrhenius) OR accept H + from compounds (BL) Mg(OH) 2 KOH CaCO 3 NaHCO 3 NH 3 Any anion

5. Acidic solution: mostly H 2 O, [H 3 O + ] > [OH - ] Neutral solution: mostly H 2 O, [H 3 O + ] = [OH - ] Basic solution: mostly H 2 O, [H 3 O + ] < [OH - ] Water molecules undergo autoionization and split into ions spontaneously: HOH + HOH  H 3 O + + OH - Count each particle and identify each solution as neutral, basic, or acidic H 2 O OH - H 2 O H 2 O H 3 O + H 2 O H 2 O OH - H 2 O H 3 O + H 2 O H 2 O H 3 O + H 2 O OH - H 2 O OH - H 2 O H 2 O H 3 O + H 2 O H 2 O H 3 O + H 2 O H 3 O + H 2 O H 2 O H 3 O + H 2 O H 2 O OH - H 2 O H 2 O H 3 O + H 2 O H 2 O H 3 O + H 2 O H 2 O H 2 O H 2 O H 3 O + H 2 O H 2 O OH - H 2 O H 2 O H 3 O + H 2 O H 2 O H 3 O + H 2 O OH - H 2 O H 2 O OH - H 2 O H 2 O OH - H 2 O H 2 O H 3 O + H 2 O H 2 O OH - H 2 O H 2 O H 2 O H 2 O H 3 O + H 2 O

6. Strong Acids & Bases: (far away from pH 7) NaOH → Na + + OH - HCl → H + + Cl - - completely dissociate in water Weak Acids & Bases: (closer to pH 7) - partially dissociate in water  The pH scale: based on [H + ] or [H 3 O + ]  Water (pH = 7) is neutral

7. Naming and Dissociation Practice: Fill in the blanks! Formula HI HClO 2 HC 2 H 3 O 2 H 2 Se HClO 3 Dissociation  H + + I -  H + + ClO 2 -  H + + C 2 H 3 O 2 –  H + + HSe -  H + + ClO 3 - Name hydroiodic acid chlorous acid acetic acid hydroselenic acid chloric acid

8. Write the complete ion equation and the net ionic equation for the neutralization of nitric acid (strong acid) by sodium hydroxide (strong base). HNO 3 (aq) + NaOH(aq)  H 2 O(l) + NaNO 3 (aq) H + + NO 3 - + Na + + OH -  H 2 O(l) + Na + + NO 3 - Get rid of everything that doesn’t change phase or compound from one side to another! H + (aq) + OH - (aq)  H 2 O(l) If the acid and base are both strong, the net ionic equation will be the same every time. Neutralization acid + base = salt + water

9. Write the molecular and net ionic equation when lithium hydroxide (strong base) is mixed with carbonic acid (weak acid). LiOH(aq) + H 2 CO 3 (aq)  H 2 O(l) + Li 2 CO 3 (aq) Leave weak or insoluble things together. Separate strong or soluble things. Li + + OH - + H 2 CO 3  H 2 O + Li + + CO 3 2- OH - (aq) + H 2 CO 3 (aq)  H 2 O(l) + CO 3 2- (aq) 2 2 2 2

10. Write the molecular and net ionic equations when magnesium hydroxide(weak) is mixed with chlorous acid (weak). Mg(OH) 2 (aq) + HClO 2 (aq)  H 2 O(l) + Mg(ClO 2 ) 2 (aq) *net ionic is the same! Write the molecular and net ionic equations when aluminum hydroxide(weak) is mixed with sulfuric acid (strong). Al(OH) 3 (aq) + H 2 SO 4 (aq)  Al 2 (SO 4 ) 3 (aq) + H 2 O(l) Al(OH) 3 (aq) + H + (aq)  Al 3+ (aq) + H 2 O(l) 2 2 3 6 2 3 3

11. HYDROCATCH The ball represents a H+ “Bronsted-Lowry (BL) Theory”

12. Conjugates Write the conjugate acid of F - H + + F - ↔ HF Conjugate acid: formed after a reaction when a base gains a H + Write the conjugate base of H 2 SO 4 Conjugate base: formed after a reaction when an acid loses a H+ H 2 SO 4 ↔ HSO 4 - + H +

13. HCl + H 2 O  H 3 O + + Cl - gained H + lost H + Acid Conjugate base Conjugate acid Base Let’s identify the acid and base, and their conjugates in the products H 2 O and Cl - are electron pair DONORS in this reaction, so they are both “Lewis bases” The H in HCl and H 3 O + is an electron pair ACCEPTOR in this reaction, so they are both “Lewis acids”

14. NH 3 + HOH  NH 4 + + OH - gained H + lost H + AcidConjugate base Conjugate acid Base

15. H 2 CO 3 + H 2 O  HCO 3 - + H 3 O + gained H + lost H + Acid Conjugate base Conjugate acid Base

16. Calculating pH *let’s practice some logs… Formulas : pH = -log[H + ] pOH = -log[OH - ] [H + ] = 10 -pH [OH - ] = 10 -pOH pOH + pH = 14.00 -log 3.000 = -0.4771 -log(3.9 x 10 -4 ) = 3.4 10 -5.6 = 2.5 x 10 -6

17. 1. A tap water sample is contaminated with acid! If the [H 3 O + ] in the sample is 8.90 x 10 -3 M, calculate the pH of the water. Remember that [H 3 O + ] is the same thing as the [H + ] pH = -log[H 3 O + ] pH = -log(8.90 x 10 -3 ) pH = 2.05 (pH has no units) pH = -log[H + ] pOH = -log[OH - ] [H + ] = 10 -pH [OH - ] = 10 -pOH pOH + pH = 14.00

18. 2. Most tap water samples are slightly basic. If the pH of a sample = 7.9, calculate the [H + ] in the sample. [H + ] = 10 -pH [H + ] = 10 -7.9 [H + ] = 1.3 x 10 -8 M Units for concentration are in M, “molar” pH = -log[H + ] pOH = -log[OH - ] [H + ] = 10 -pH [OH - ] = 10 -pOH pOH + pH = 14.00

19. 3. What is the [OH - ] of a solution that has a pOH of 3.00? [OH - ] = 10 -pOH [OH - ] = 10 -3.00 [OH - ] = 1.00 x 10 -3 M 4. What is the pH of this solution? pOH + pH = 14.00 3.00 + pH = 14.00 pH = 11.00 pH = -log[H + ] pOH = -log[OH - ] [H + ] = 10 -pH [OH - ] = 10 -pOH pOH + pH = 14.00

20. 5. Calculate the [H 3 O + ] AND [OH - ] of human blood (pH =7.40) [H 3 O + ] = 10 -pH [H 3 O + ] = 10 -7.40 [H 3 O + ] = 3.98 x 10 -8 M To find [OH - ], use pOH + pH = 14.00 pOH + 7.40 = 14.00 pOH = 6.60 pH = -log[H + ] pOH = -log[OH - ] [H + ] = 10 -pH [OH - ] = 10 -pOH pOH + pH = 14.00 then [OH - ] = 10 -pOH [OH - ] = 10 -6.60 [OH - ] = 2.51 x 10 -7 M

21. pOH = -log[OH - ] pOH = -log(9.35 x 10 -4 ) pOH = 3.03 pOH + pH = 14.00 pH = 10.97 [H + ] = 10 -pH [H + ] = 10 -10.97 [H + ]= 1.07 x 10 -11 M 6. What is the [H + ] of a solution that has a [OH - ] of 9.35 x 10 -4 M? pH = -log[H + ] pOH = -log[OH - ] [H + ] = 10 -pH [OH - ] = 10 -pOH pOH + pH = 14.00

22. Start on Problem Patent prelab. You may begin brainstorming with your group.