1. Chapter 14 Acids and Bases AP*

2. Section 14.8 Acid-Base Properties of Salts Salts  Ionic compounds.  When dissolved in water, break up into its ions (which can behave as acids or bases).  Review the solubility rules (p. 156, table 4.1)  Salts containing NO 3 -, NH 4 + and group 1A cations are highly soluble.  Example: KCl (s) + H 2 O (l)  K + + Cl - + H 2 O (l) Net equation: KCl (s)  K + + Cl -  Note, this reaction goes to completion, due to the high solubility of KCl. Copyright © Cengage Learning. All rights reserved 2

3. Section 14.8 Acid-Base Properties of Salts Salts  The dissolved ions of salts can affect pH.  The salt of a strong acid and a strong base gives a neutral solution.  For example, KCl & NaNO 3 will NOT change the pH.  Why is this true? KCl (s)  K + + Cl - K + + H 2 O   KOH + H + Cl - + H 2 O   HCl + OH - Since K +, Cl -, Na + & NO 3 - are all conjugates of strong acids/bases, these ions will not change the pH. Copyright © Cengage Learning. All rights reserved 3 1 st identify the ions produced in solution. 2 nd predict how the ions will interact with water. 3 rd any strong acid/base will drive the reaction due to the large K a or K b Only starting ions will be present in significant quantities, so pH will be neutral (due to H2O only)

4. Section 14.8 Acid-Base Properties of Salts Salts  A basic solution is formed if the anion of the salt is the conjugate base of a weak acid.  For example, NaF & KC 2 H 3 O 2 WILL change the pH.  Why is this true? NaF (s)  Na + + F - Na + + H 2 O   NaOH + H + F - + H 2 O   HF + OH - Since Na + is a conjugate of strong base = neutral pH Since F - is a conjugate of weak acid (HF), pH will be >7, because Na +, F -, H 2 O, HF & OH - will all be present in solution. Copyright © Cengage Learning. All rights reserved 4

5. Section 14.8 Acid-Base Properties of Salts Salts  An acidic solution is formed if the cation of the salt is the conjugate acid of a weak base.  NH 4 Cl will produce an acidic solution.  Why is this true? NH 4 Cl (s)  NH 4 + + Cl - NH 4 + + H 2 O   NH 3 + H 3 O + Cl - + H 2 O   HCl + OH - Since NH 4 + is a conjugate of weak base, pH will be <7, because NH 4 +, H 2 O, NH 4 Cl & H 3 O + will all be present in solution. Since Cl - is a conjugate of strong acid = neutral pH Copyright © Cengage Learning. All rights reserved 5

6. Section 14.8 Acid-Base Properties of Salts Copyright © Cengage Learning. All rights reserved 6 *

7. Section 14.8 Acid-Base Properties of Salts Qualitative Prediction of pH of Salt Solutions (from Weak Parents) Copyright © Cengage Learning. All rights reserved 7 If a salt contains BOTH the conjugate of a weak acid & the conjugate of a weak base, then the K a & K b values must be compared to determine the approximate pH. *

8. Section 14.8 Acid-Base Properties of Salts HC 2 H 3 O 2 K a = 1.8 × 10 -5 HCN K a = 6.2 × 10 -10 Calculate the K b values for: C 2 H 3 O 2 − and CN − Don’t forget: K a × K b = K w K b (C 2 H 3 O 2 - ) = 5.6 × 10 -10 K b (CN - ) = 1.6 × 10 -5 Note that as K a increases, a conjugate K b decreases!! Copyright © Cengage Learning. All rights reserved EXERCISE!

9. Section 14.8 Acid-Base Properties of Salts Arrange the following 1.0 M solutions from lowest to highest pH. HBr NaOH NH 4 Cl NaCN NH 3 HCN NaCl HF Justify your answer. HBr, HF, HCN, NH 4 Cl, NaCl, NaCN, NH 3, NaOH Copyright © Cengage Learning. All rights reserved 9 CONCEPT CHECK!

10. Section 14.8 Acid-Base Properties of Salts Consider a 0.30 M solution of NaF. The K a for HF is 7.2 × 10 -4. What are the major species? Na +, F -, H 2 O Copyright © Cengage Learning. All rights reserved 10 CONCEPT CHECK!

11. Section 14.8 Acid-Base Properties of Salts Let’s Think About It…  Q: Why isn’t NaF considered a major species?  Answer: Because NaF is highly soluble. It will dissolve completely into its ions.  What are the possibilities for the dominant reactions?  Answer: examine the next slide... Copyright © Cengage Learning. All rights reserved 11

12. Section 14.8 Acid-Base Properties of Salts Let’s Think About It… The possibilities for the dominant reactions are: 1.F – (aq) + H 2 O(l) HF(aq) + OH – (aq) 2.H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH – (aq) 3.Na + (aq) + H 2 O(l) NaOH + H + (aq) 4.Na + (aq) + F – (aq) NaF Copyright © Cengage Learning. All rights reserved 12

13. Section 14.8 Acid-Base Properties of Salts Let’s Think About It…  How do we decide which reaction controls the pH?  Answer: only reactions 1 & 2 can control pH. Reaction 3 produces a srong base (unreasonable) & reaction 4 will not be in equilibrium, since all the NaF will dissolve.  This leaves: F – (aq) + H 2 O(l) HF(aq) + OH – (aq) H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH – (aq)  Determine the equilibrium constant for each reaction.  The first reaction will dominate, due to the large K a. Copyright © Cengage Learning. All rights reserved 13

14. Section 14.8 Acid-Base Properties of Salts Calculate the pH of a 0.75 M aqueous solution of NaCN. K a for HCN is 6.2 × 10 –10. Copyright © Cengage Learning. All rights reserved 14 EXERCISE!

15. Section 14.8 Acid-Base Properties of Salts Let’s Think About It…  What are the major species in solution? Na +, CN –, H 2 O  Why isn’t NaCN considered a major species? Because it is highly soluble & will dissolve completely. Copyright © Cengage Learning. All rights reserved 15

16. Section 14.8 Acid-Base Properties of Salts Let’s Think About It…  What are all possibilities for the dominant reaction?  The possibilities for the dominant reaction are: 1.CN – (aq) + H 2 O(l) HCN(aq) + OH – (aq) 2.H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH – (aq) 3.Na + (aq) + H 2 O(l) NaOH + H + (aq) 4.Na + (aq) + CN – (aq) NaCN  Which of these reactions really occur? Copyright © Cengage Learning. All rights reserved 16

17. Section 14.8 Acid-Base Properties of Salts Let’s Think About It…  How do we decide which reaction controls the pH? CN – (aq) + H 2 O(l) HCN(aq) + OH – (aq) H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH – (aq) Copyright © Cengage Learning. All rights reserved 17

18. Section 14.8 Acid-Base Properties of Salts Steps Toward Solving for pH K b = 1.6 × 10 –5 pH = 11.54 CN – (aq) +H2OH2OHCN(aq) +OH – (aq) Initial0.75 M0~ 0 Change–x+x Equilibrium0.75–xxx