1. Unit 14 Acids, Bases, Salts and Titrations

2. Acids & Bases Acids :  acids are sour tasting  Arrhenius acid: Any substance that, when dissolved in water will dissociate, increasing the concentration of hydronium ion (H 3 O + )  Bronsted-Lowry acid: A proton (H+)donor  Lewis acid: An electron acceptor Bases :  bases are bitter tasting and slippery  Arrhenius base: Any substance that, when dissolved in water will dissociate, increasing the concentration of hydroxide ion (OH - )  Bronsted-Lowery base: A proton (H+) acceptor  Lewis base: An electron donor

3. 3 Acid/Base Definitions Definition #2: Brønsted – Lowry Acids – proton donor Bases – proton acceptor A “proton” is really just a hydrogen atom that has lost its electron!

4. Lone Hydrogen ions do not exist by themselves in solution. H+ is always bound to a water molecule to form a hydronium ion.

5. Brønsted-Lowry Theory of Acids & Bases Conjugate Acid-Base Pairs General Equation

6. Brønsted-Lowry Theory of Acids & Bases

7. Notice that water is both an acid & a base = amphoteric Reversible reaction

8. 8 Conjugate Pairs **Water is said to be amphoteric because it can act as an acid or a base.

9. 9 Learning Check! Label the acid, base, conjugate acid, and conjugate base in each reaction: HCl + OH -  Cl - + H 2 O H 2 O + H 2 SO 4  HSO 4 - + H 3 O +

10. 10 Learning Check! Label the acid, base, conjugate acid, and conjugate base in each reaction: HCl + OH -  Cl - + H 2 O A B A B CB CA HCl + OH -  Cl - + H 2 O A B A B CB CA H 2 O + H 2 SO 4  HSO 4 - + H 3 O + B A CB CA H 2 O + H 2 SO 4  HSO 4 - + H 3 O + B A CB CA

11. ELECTROLYTES Electrolytes are species which conducts electricity when dissolved in water. Acids, Bases, and Salts are all electrolytes. Salts and strong Acids or Bases form Strong Electrolytes. Salt and strong acids (and bases) are fully dissociated therefore all of the ions present are available to conduct electricity. HCl (s) + H 2 O  H 3 O + + Cl - Weak Acids and Weak Bases for Weak Electrolytes. Weak electrolytes are partially dissociated therefore not all species in solution are ions, most of the molecular form is present. Weak electrolytes have less ions available to conduct electricity. NH 3 + H 2 O  NH 4 + + OH -

13. Acids & Bases STRONG Vs WEAK STRONG Vs WEAK - completely ionized - partially ionized - strong electrolyte - weak electrolyte - ionic/very polar bonds - some covalent bonds : Strong Acids : Strong Bases: HClO 4 LiOH H 2 SO 4 NaOH HIKOH HBrCa(OH) 2 HClSr(OH) 2 HNO 3 Ba(OH) 2

14. Acids & Bases One ionizable proton: monoprotic HCl  H + + Cl - Two ionizable protons: diprotic H 2 SO 4  H + + HSO 4 - HSO 4 -  H + + SO 4 2- Three ionizable protons: triprotic H 3 PO 4  H + + H 2 PO 4 – H 2 PO 4 -  H + + HPO 4 2- HPO 4 2-  H + + PO 4 -3 Combined: H 2 SO 4  2H + + SO 4 2 - Combined: H 3 PO 4  3H + + PO 4 3 -

15. Acids & Bases For the following identify the acid and the base as strong or weak. a. Al(OH) 3 + HCl  b. Ba(OH) 2 + HC 2 H 3 O 2  c. KOH + H 2 SO 4  d. NH 3 + H 2 O  Weak baseStrong acid Weak acid Strong acid Strong base Weak baseWeak acid

16. Acids & Bases For the following predict the product. To check your answer left click on the mouse. a. Al(OH) 3 + HCl  b. Ba(OH) 2 + HC 2 H 3 O 2  c. KOH + H 2 SO 4  d. NH 3 + H 2 O  AlCl 3 + 3 H2OH2O Ba(C 2 H 3 O 2 ) 2 + 2 H2OH2O K 2 SO 4 + 2 H2OH2O NH 4 + + OH - 2 2 3

17. Conjugate Acid-Base Pairs

19. 19 Water Equilibrium K w = [H + ] [OH - ] = 1.0 x 10 -14 Equilibrium constant for water  Water or water solutions in which [H + ] = [OH - ] = 10 -7 M are neutral solutions.  A solution in which [H+] > [OH-] is acidic  A solution in which [H+] < [OH-] is basic

20. 20 pH A measure of the hydronium ion The scale for measuring the hydronium ion concentration [H 3 O + ] in any solution must be able to cover a large range. A logarithmic scale covers factors of 10. The “p” in pH stands for log. A solution with a pH of 1 has [H 3 O + ] of 0.1 mol/L or 10 -1 A solution with a pH of 3 has [H 3 O + ] of 0.001 mol/L or 10 -3 A solution with a pH of 7 has [H 3 O + ] of 0.0000001 mol/L or 10 -7 pH = - log [H 3 O + ]

21. The pH scale The pH scale ranges from 1 to 10 -14 mol/L or from 1 to 14. pH = - log [H 3 O + ] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 acid neutral base

22. Manipulating pH Algebraic manipulation of: pH = - log [H 3 O + ] allows for: [H 3 O + ] = 10 -pH If pH is a measure of the hydronium ion concentration then the same equations could be used to describe the hydroxide (base) concentration. [OH - ] = 10 -pOH pOH = - log [OH - ]thus: pH + pOH = 14 ; the entire pH range!

23. pH + pOH = 14 1 x 10 -5 M Acid 5 pH = -log [H + ]pOH = -log [OH - ] [H + ] = 10 -pH [OH - ] = 10 -pOH K w = [H + ] x [OH - ] 1 x 10 -14 = [H + ] x [OH - ] Equation Summary [H + ] = 10 -pH pH = ___[H + ] = __________ [OH - ] = __________ pOH = ___ [H + ] = 10 -5 pH + pOH = 14 pOH = 14 - pH pOH = 14 - 5 9 [OH - ] = 10 -pOH [OH - ] = 10 -9 1 x 10 -9 M pH = ___[H + ] = __________ [OH - ] = __________ pOH = ___ 6 [OH - ] = 10 -pOH [OH - ] = 10 -6 1 x 10 -6 M pH + pOH = 14 pH = 14 - pOH pH = 14 - 6 8 1 x 10 -8 M [H + ] = 10 -pH [H + ] = 10 -8 Base

24. [H + ] = __________ pH = ___ pH + pOH = 14 1.6 x 10 -3 M Acid 2.8 pH = -log [H + ]pOH = -log [OH - ] [H + ] = 10 -pH [OH - ] = 10 -pOH K w = [H + ] x [OH - ] 1 x 10 -14 = [H + ] x [OH - ] Equation Summary [H + ] = 10 -pH [OH - ] = __________ pOH = ___ [H + ] = 10 -2.8 pH + pOH = 14 pOH = 14 - pH pOH = 14 – 2.8 11.2 [OH - ] = 10 -pOH [OH - ] = 10 -11.2 6.3 x 10 -12 M [H + ] = 0.0016

25. Calculating pH, pOH pH = -log[H 3 O + ] or pH = -log[H + ] pOH = -log[OH - ] Relationship between pH and pOH pH + pOH = 14 Finding [H 3 O + ], [OH - ] from pH, pOH [H 3 O + ] = 10 -pH or [H + ] = 10 -pH [OH - ] = 10 -pOH SUMMARY

26. TITRATION Titration of a strong acid with a strong base ENDPOINT = POINT OF NEUTRALIZATION = EQUIVALENCE POINT At the end point for the titration of a strong acid with a strong base, the moles of acid (H + ) equals the moles of base (OH - ) to produce the neutral species water (H 2 O). If the mole ratio in the balanced chemical equation is 1:1 then the following equation can be used. M A V A = M B V B

27. TITRATION M A V A = M B V B M A V A = M B V B 1. Suppose 75.00 mL of hydrochloric acid was required to neutralize 22.50 mLof 0.52 M NaOH. What is the molarity of the acid? HCl + NaOH  H 2 O + NaCl M A V A = M B V B M A (75.00 mL) = (0.52 M) (22.50 mL) = 0.16 M = 0.16 M Now you try: 2. If 37.12 mL of 0.843 M HNO 3 neutralized 40.50 mL of KOH, what is the molarity of the base? M B = 0.773 mol/L

28. Molarity and Titration

30. TITRATION Titration of a strong acid with a strong base ENDPOINT = POINT OF NEUTRALIZATION = EQUIVALENCE POINT At the endpoint for the titration of a strong acid with a strong base, the moles of acid (H + ) equals the moles of base (OH - ) to produce the neutral species water (H 2 O). If the mole ratio in the balanced chemical equation is NOT 1:1 then you must rely on the mole relationship and handle the problem like any other stoichiometry problem. Use the H + to OH - ratio. (# of H + )(M A )(V A ) = (# of OH - )(M B )(V B )

31. TITRATION 1. If 37.12 mL of 0.543 M LiOH neutralized 40.50 mL of H 2 SO 4, what is the molarity of the acid? 2 LiOH + H 2 SO 4  Li 2 SO 4 + 2 H 2 O It’s a 2:1 mole ratio, this must be incorporated into the formula. (# of H + )(M A )(V A ) = (# of OH - )(M B )(V B ) (2 H + )(M A )(40.50mL) = (1 OH - )(0.543M)(37.12mL) (2 H + )(M A )(40.50mL) = (1 OH - )(0.543M)(37.12mL) = 0.248 M = 0.248 M 2. If 20.42 mL of Ba(OH) 2 solution was used to titrate29.26 mL of 0.430 M HCl, what is the molarity of the barium hydroxide solution? M B = 0.308 mol/L

32. Molarity and Titration A student finds that 23.54 mL of a 0.122 M NaOH solution is required to titrate a 30.00-mL sample of hydrochloric acid solution. What is the molarity of the acid? A student finds that 37.80 mL of a 0.4052 M NaHCO 3 solution is required to titrate a 20.00- mL sample of sulfuric acid solution. What is the molarity of the acid? The reaction equation is: H 2 SO 4 + 2 NaHCO 3  Na 2 SO 4 + 2 H 2 O + 2 CO 2 0.0957 M 0.3829 M

33. PRACTICE PROBLEMS 1. How many milliliters of 1.25 M LiOH must be added to neutralize 34.7 mL of 0.389 M HNO 3 ? 2. How many mL of 0.998 M H 2 SO 4 must be added to neutralize 47.9 mL of 1.233 M KOH? 3. What is the molar concentration of hydronium ion in a solution of pH 8.25? 4. What is the pH of a solution that has a molar concentration of hydronium ion of 9.15 x 10 -5 ? 5. What is the pOH of a solution that has a molar concentration of hydronium ion of 8.55 x 10 -10 ? 10.8 mL 29.6 mL 5.623 x 10 -9 M pH = 4.0 pOH = 4.9