Chapter 16 Acids and Bases

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To become familiar with problems about acid and bases equilibria. TEXTBOOK. Brown, Lemay & Bursten,. Chemistry: The Central Science,. 10th Ed. (Chapter ...
OBJECTIVES 

To become familiar with problems about acid and bases equilibria.

TEXTBOOK Brown, Lemay & Bursten, Chemistry: The Central Science, 10th Ed. (Chapter 16)

Strategy for solving Acid-Base Problems (1) There are two general types of equilibrium problems involving weak acids and their conjugate bases: A. Given equilibrium concentrations, find Ka. B. Given Ka and some concentration information, find the other equilibrium concentrations. First, write the balanced equation and Ka expression; these will tell you what to find. Second, ICE Table: define x as the unknown concentration that changes during the reaction. Frequently, x = [HA]dissoc., the concentration of HA that dissociates which, through the use of certain assumptions, also equals [H3O+] and [A-] at equilibrium.

Strategy for solving Acid-Base Problems (2)

3. Complete the ICE Table incorporating the unknown. 4. Make assumptions that simplify the calculation, usually that x is very small relative to the initial concentration. 5. Substitute the values into the Ka expression and solve for x. 6. Check that the assumptions are justified. We normally apply the 5% rule; if the value of x is greater than 5% of the value it is compared with, you must use the quadratic formula to find x.

Strategy for solving Acid-Base Problems (3) THE NOTATION SYSTEM. Molar concentrations of species are indicated by using square brackets around the species of interest. Brackets with no subscript refer to the molar concentration of the species at equilibrium. THE ASSUMPTIONS. The two key assumptions to simplify the arithmetic are: 1.The [H3O+] from the auto-ionization of water is negligible. In fact, the presence of acid from whatever is put into solution will hinder the autoionization of water, and make it even less important. 2. A weak acid has a small Ka. Therefore, it dissociates to such a small extent that we can neglect the change in its concentration to find its equilibrium concentration.

PROBLEM # 1 A chemist dilutes concentrated hydrochloric acid to make two solutions: (a) 3.0 M and (b) 0.0024 M. Calculate the pH and pOH of the two solutions at 25°C.

PROBLEM # 1. A chemist dilutes concentrated HCl to make two solutions: (a) 3.0 M and (b) 0.0024 M. Calculate the pH and pOH of the solutions at 25°C. Plan: We know that HCl is a strong acid (dissociates completely in water); therefore [H3O+] = [HCl]init.. We will use then [H3O+] to calculate pH and pOH.

Solution: (a) [H3O+] = 3.0 M pH = -log[H3O+] = -0.48 14 K 1 * 10 15 w [ OH  ]    3 . 33 * 10 M  H3O 3.0 pOH = - log(3.33*10-15) = 14.52 +] = 2.62 (b) [H3O+] = 0.0024 M pH = -log[H O 3 14





Kw 1 * 10 12 [ OH ]    4 . 17 * 10 M  H3O 0.0024 





pOH = -log(4.17*10-12) = 11.38

PROBLEM # 2 Calculate the pH of a 1.00 M Solution of Nitrous Acid HNO2 (Ka = 4.0*10-4).

PROBLEM # 2. Calculate the pH of a 1.0 M Solution of Nitrous Acid (Ka = 4.0*10-4). Solution:

HNO2(aq) D H+(aq) + NO2-(aq) Initial: 1.0 M 0 0 Equilibrium: 1.0-x x x   2 [ H ] * [ NO2 ] x 4 Ka   4.0 * 10  [ HNO2 ] 1.0  x Assuming that: 1.0 - x ≈ 1.0 to simplify the problem.

x 2  4.0 * 10 4  x  2 * 10 2  0.02M  [ H  ]  [ NO2 ]

pH = - log[H+] = - log(2.0*10-2) = 1.7

PROBLEM # 3 Calculate the pH of a solution that contains 1.00 M HF (Ka = 7.2*10-4) and 5.00 M HOCl (Ka = 3.5*10-8). Also calculate the concentrations of the Fluoride and Hypochlorite ions at equilibrium.

PROBLEM # 3. Calculate the pH of a solution that contains 1.00 M HF (Ka = 7.2*10-4) and 5.00 M HOCl (Ka = 3.5*10-8). Also calculate the concentrations of the Fluoride and Hypochlorite ions at equilibrium.

Solution: • Three components produce H+: HF(aq) D H+(aq) + F-(aq) Ka = 7.2*10-4 HOCl(aq) D H+(aq) + OCl-(aq) Ka = 3.5*10-8 H2O(aq) D H+(aq) + OH-(aq) Ka = 1.0*10-14 • Even though HF is a weak acid, it has by far the greatest Ka, therefore it will be the   dominate producer of H+. [H ] * [F ] Ka 

[ HF ]

PROBLEM # 3. Calculate the pH of a solution that contains 1.00 M HF (Ka = 7.2*10-4) and 5.00 M HOCl (Ka = 3.5*10-8). Also calculate the concentrations of the Fluoride and Hypochlorite ions at equilibrium.

HF(aq) D H+(aq) + 1.0 0 1.0-x x

F-(aq) 0 x x2   x2 1.0  x

Initial: Equilibrium: [H ] * [F  ] Ka   7.2 * 10  4 [ HF ] x = 2.7*10-2 2 x 2.7 * 10 Using the 5% rule:  * 100  2.7% [ HF ]0

1.00

Therefore: [F- ] = [H+] = x = 2.7*10-2 & pH = 1.57

PROBLEM # 3. Calculate the pH of a solution that contains 1.00 M HF (Ka = 7.2*10-4) and 5.00 M HOCl (Ka = 3.5*10-8). Also calculate the concentrations of the Fluoride and Hypochlorite ions at equilibrium.

Calculating concentration Hypochlorite ions HOCl(aq) D H+(aq) + OCl-(aq) Equilibrium: 5.0-x 2.7*10-2 x [ H  ] * [ OCl  ] Ka  [ HOCl ]

3.5 * 10 8

2.7 * 10 2 * [ OCl  ]   [ OCl  ]  6.48 * 10 6 5.0

PROBLEM # 4 The weak acid hypo-chlorous acid is formed in bleach solutions. If the pH of a 0.12 M solution of HClO is 4.19, what is the value of the Ka of this weak acid?

PROBLEM # 4. The weak acid hypo-chlorous acid is formed in bleach solutions. If the pH of a 0.12 M solution of HClO is 4.19, what is the value of the Ka of this weak acid?

HClO(aq) + H2O(l) D H3O+(aq) + ClO -(aq) Initial: 0.12 -0 0 Equil.: 0.12-x -x x Assumptions: • [H3O+] = [H3O+]HClO • since HClO is a weak acid, we will assume that: 0.12 M - x = 0.12 M Calculating [H3O+] : [H3O+] = 10-pH = 10-4.19 = 6.46*10-5 M

PROBLEM # 4. The weak acid hypochlorous acid is formed in bleach solutions. If the pH of a 0.12 M solution of HClO is 4.19, what is the value of the Ka of this weak acid?

HClO(aq) + H2O(l) D H3O+(aq) + ClO -(aq) Equil.: 0.12-x -x x x = [H3O+] = [ClO-] = 6.46*10-5 M [ H3O  ] * [ ClO  ] ( 6.46 * 10 5 )2 Ka    3.48 * 10 8 [ HClO ] 0.12 Checking if our assumptions are OK: 7 1 * 10 M  For [ H 3O ]  * 100  0.155% 5 6.46 * 10 M 6.46 * 10 5 M For [ HClO ]  * 100  0.054% 0.12 M

PROBLEM # 5 Calculate the pH of a -3 2.0*10 M solution of NaOH.

PROBLEM # 5. Calculate the pH of a 2.0*10-3 M solution of NaOH. Solution: Since NaOH is a strong base, it will dissociate 100% in water.

NaOH(aq) → Na+(aq) + OH-(aq) • Since [NaOH] = 2.0*10-3 M, [OH-] = 2.0*10-3 M, The concentration of [H+] can be calculated from Kw: 14 Kw 1 * 10  12 [H ]    5 * 10  3 [ OH ] 2 * 10

pH = - log [H+] = - log( 5.0*10-12) = 11.3

PROBLEM # 6 Ammonia is commonly used cleaning agent in households and is a weak base, with a Kb of 1.8*10-5. What is the pH of a 1.5 M NH3 solution?

PROBLEM # 6. Ammonia is commonly used cleaning agent in households and is a weak base, with a Kb of 1.8*10-5. What is the pH of a 1.5 M NH3 solution?

Solution: Ammonia reacts with water to form [OH-] and then calculate [H3O+] and the pH. The balanced equation and Kb expression are: NH3(aq) + H2O(l) D NH4+(aq) + OH-(aq) Initial: 1.5 -0 0 Equil.: 1.5-x -x x [ NH4 ] * [ OH  ] Since K is small, we can b Kb  [ NH3 ] assume that: 1.5 - x ≈ 1.5

PROBLEM # 6. Ammonia is commonly used cleaning agent in households and is a weak base, with a Kb of 1.8*10-5. What is the pH of a 1.5 M NH3 solution?

NH3(aq) + H2O(l) D NH4+(aq) + OH-(aq) Equil.:

1.5-x -x x   2 [ NH4 ] * [ OH ] x 5 Kb    1.8 * 10 [ NH3 ] 1.5

x2 = 2.7*10-5 then: x = 5.20*10-3 = [OH-] = [NH4+] Calculating pH: 14 K 1 * 10  12 w [ H3O ]    1.92 * 10  3 [ OH ] 5.2 * 10 pH = -log[H3O+] = - log (1.92*10-12) = 11.72

PROBLEM # 7 Calculate the pH of 1.00 M solution of Acetic Acid -5 (Ka = 1.8*10 ). CH3COOH + H2O D CH3COO- + H3O+

PROBLEM # 7. Calculate the pH of 1 M solution of Acetic Acid (Ka=1.8*10-5). CH3COOH + H2O D CH3COO- + H3O+

Step 1. Define concentrations in ICE Table. Initially Change Equilibrium

[CH3COOH] 1.00 -x 1.00-x

[H3O+] 0 +x x

[CH3COO-] 0 +x x

Step 2. Write Ka expression [ CH3COO  ] * [ H3O  ] x2 Ka    x 2  1.8 * 10 5 [ CH3COOH ] 1 x Step 3. Solve Ka approximate expression

x = [H3O+] = [CH3COO-] = 4.2*10-3 M pH = - log [H3O+] = -log (4.2*10-3) = 2.37

CONCLUSIONS (1) The following steps will be very helpful in solving Acid-Base equilibria problems: 1. List the major species in the solution. 2. Choose the species that can produce H+, and write balanced equations for the reactions producing H+. 3. Comparing the values of the equilibrium constants for the reactions you have written, decide which reaction will dominate in the production of H+. 4. Write the equilibrium expression for the dominant reaction.

CONCLUSIONS (2) 5. Set up ICE Table; first list the initial concentrations of the species participating in the dominate reaction. 6. Set up ICE Table; second, define the change needed to achieve equilibrium; that is x. 7. Set up ICE Table; third, write the equilibrium concentrations in terms of x. 8. Substitute the equilibrium concentrations into the equilibrium expression. 9. Solve for x the “easy” way-that is, by assuming that [HA]0 – x = [HA]0 or [AOH]0 – x = [AOH]0 10.Verify whether the approximation is valid ( the 5% rule is the test in this case). 11.Calculate [H+] or [OH-] and then pH or pOH.