SSS 3 2 nd Class Acid/Base/Salt Equilibrium

Copyright © Cengage Learning. All rights reserved 2 Models of Acids and Bases Arrhenius: Acids produce H + ions in solution, bases produce OH - ions. Brønsted – Lowry: Acids are proton (H + ) donors, bases are proton acceptors. HCl + H 2 O Cl - + H 3 O + acid base conjugate base acid

Copyright © Cengage Learning. All rights reserved 3 Acid in Water HA(aq) + H 2 O(l) H 3 O + (aq) + A - (aq) Conjugate base is everything that remains of the acid molecule after a proton is lost. Conjugate acid is formed when the proton is transferred to the base.

Strong acid: – Ionization equilibrium lies far to the right. – 100% ionization – Yields a weak conjugate base. Weak acid: – Ionization equilibrium lies far to the left. – <100% ionization – Weaker the acid, stronger its conjugate base. Copyright © Cengage Learning. All rights reserved 4

HA(aq) + H 2 O(l) H 3 O + (aq) + A - (aq) acid base conjugate conjugate acid base What is the equilibrium constant expression for an acid acting in water? Copyright © Cengage Learning. All rights reserved 5 Problem 1

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Various Ways to Describe Acid Strength Copyright © Cengage Learning. All rights reserved 7

If the equilibrium lies to the right, the value for K a is __________. large (or >1) If the equilibrium lies to the left, the value for K a is ___________. small (or <1) Copyright © Cengage Learning. All rights reserved 8 Problem 2

Acetic acid (HC 2 H 3 O 2 ) and HCN are both weak acids. Acetic acid is a stronger acid than HCN. Arrange these bases from weakest to strongest and explain your answer: H 2 O Cl – CN – C 2 H 3 O 2 – The bases from weakest to strongest are: Cl –, H 2 O, C 2 H 3 O 2 –, CN – Copyright © Cengage Learning. All rights reserved 9 Problem 3

Water as an Acid and a Base Water is amphoteric: – Behaves either as an acid or as a base. At 25°C: K w = [H + ][OH – ] = 1.0 × 10 –14 No matter what the solution contains, the product of [H + ] and [OH – ] must always equal 1.0 × 10 –14 at 25°C. Copyright © Cengage Learning. All rights reserved 10

pH pOH [OH - ] [H 3 O + ] pH= -log [H 3 O + ] pH+pOH=14 pOH= -log[OH - ] [H 3 O + ] [OH - ]=1x10 14

Three Possible Situations [H + ] = [OH – ]; neutral solution [H + ] > [OH – ]; acidic solution [OH – ] > [H + ]; basic solution Copyright © Cengage Learning. All rights reserved 12

pH = –log[H + ] pH changes by 1 for every power of 10 change in [H + ]. A compact way to represent solution acidity. pH decreases as [H + ] increases. Significant figures: – The number of decimal places in the log is equal to the number of significant figures in the original number. Copyright © Cengage Learning. All rights reserved 13

pH Range pH = 7; neutral pH > 7; basic – Higher the pH, more basic. pH < 7; acidic – Lower the pH, more acidic. Copyright © Cengage Learning. All rights reserved 14

The pH of a solution is What is the [H + ] for this solution? [H + ] = 1.4 × 10 –6 M Copyright © Cengage Learning. All rights reserved 15 Problem 4

Calculate the pH for each of the following solutions. a) 1.0 × 10 –4 M H + pH = 4.00 b) M OH – pH = Copyright © Cengage Learning. All rights reserved 16 Problem 5

Calculate the pOH for each of the following solutions. a) 1.0 × 10 –4 M H + pOH = b) M OH – pOH = 1.40 Copyright © Cengage Learning. All rights reserved 17 Problem 6

The pH of a solution is What is the [OH – ] for this solution? [OH – ] = 7.1 × 10 –9 M Copyright © Cengage Learning. All rights reserved 18 Problem 7

Consider an aqueous solution of 2.0 × 10 –3 M HCl. What is the pH? pH = 2.70 Copyright © Cengage Learning. All rights reserved 19 Problem 8

Weak Acid–Base Problems Solve as any other equilibrium problem 1)Write the balanced equation for the reaction. 2)Write the equilibrium expression using the law of mass action. 3)Solve using ICE List the given information then Solve for the pH if needed. Copyright © Cengage Learning. All rights reserved 20

Writing the Equations Strong acid HX + H 2 O  H X - Strong base NaOH  Na + + OH - Weak acid HX + H 2 O H X - Weak base X - + H 2 O OH - + HX Salt NaX  Na + + X -, then you have a weak base problem (X - + H 2 O OH - + HX) Salt NH 4 Cl  NH Cl - then you have a weak acid problem(NH H 2 O H NH 3 )

Consider a 0.80 M aqueous solution of the weak acid HCN (K a = 6.2 × 10 –10 ). What is the hydronium ion concentration 2.2x10 -5 Copyright © Cengage Learning. All rights reserved 22 Problem 9

Percent Dissociation (Ionization) For a given weak acid, the percent dissociation increases as the acid becomes more dilute. Copyright © Cengage Learning. All rights reserved 23 Amount dissociated is equal to x in ICE

A solution of 8.00 M formic acid (HCHO 2 ) is 0.47% ionized in water. Calculate the K a value for formic acid. K a = 1.8 × 10 –4 Copyright © Cengage Learning. All rights reserved 24 Problem 10

The percent ionization of a 4.00 M formic acid solution should be: higher than lower thanthe same as the percent ionization of an 8.00 M formic acid solution. Explain. Copyright © Cengage Learning. All rights reserved 25 Problem 11

The pH of a 4.00 M formic acid solution should be: higher than lower thanthe same as the pH of an 8.00 M formic acid solution. Explain. Copyright © Cengage Learning. All rights reserved 26 Problem 12

Calculate the percent ionization of a 4.00 M formic acid solution in water. % Ionization = 0.67% Copyright © Cengage Learning. All rights reserved 27 Problem 13

Calculate the pH of a 1.0 × 10 –3 M solution of sodium hydroxide. pH = Copyright © Cengage Learning. All rights reserved 28 Problem 14

Calculate the pH of a 1.0 × 10 –3 M solution of calcium hydroxide. pH = Copyright © Cengage Learning. All rights reserved 29 Problem 15

Calculate the pH of a 1.00 M solution of H 3 PO 4. K a1 = 7.5 × K a2 = 6.2 × K a3 = 4.8 × pH = 1.08 Copyright © Cengage Learning. All rights reserved 30 EXERCISE!

Salts Ionic compounds (cation and anion) When dissolved in water, break up into its ions (which can behave as acids or bases). Copyright © Cengage Learning. All rights reserved 31

Copyright © Cengage Learning. All rights reserved 32 SALTS

HC 2 H 3 O 2 K a = 1.8 × HCN K a = 6.2 × Calculate the K b values for: C 2 H 3 O 2 − and CN − K b (C 2 H 3 O 2 - ) = 5.6 × K b (CN - ) = 1.6 × Copyright © Cengage Learning. All rights reserved EXERCISE!

Important Formula Ka X Kb = 1.0 x

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 35 CONCEPT CHECK!

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 36 Problem 16

Steps Toward Solving for pH K b = 1.6 × 10 –5 pH = CN – (aq) +H2OH2OHCN(aq) +OH – (aq) Initial0.75 M0~ 0 Change–x+x Equilibrium0.75–xxx

Calculate the pH of a 2.0 M solution of ammonia (NH 3 ). (K b = 1.8 × 10 –5 ) pH = Copyright © Cengage Learning. All rights reserved 38 Problem 17

Bond Strengths and Acid Strengths for Hydrogen Halides Copyright © Cengage Learning. All rights reserved 39 What conclusion from this data can you make?

Oxyacids Contains the group H–O–X. For a given series the acid strength increases with an increase in the number of oxygen atoms attached to the central atom. H has less nuclear attractive forces The greater the ability of X to draw electrons toward itself, (Electronegativity)the greater the acidity of the molecule. Copyright © Cengage Learning. All rights reserved 40

Several Series of Oxyacids and Their K a Values Copyright © Cengage Learning. All rights reserved 41

Comparison of Electronegativity of X and K a Value Copyright © Cengage Learning. All rights reserved 42

Multiple Choice 1 Each of the following can act as both a Brönsted acid and a Brönsted base EXCEPT (A) HCO 3 ¯ (B) H 2 PO 4 ¯ (C) NH 4 + (D) H 2 O (E) HS¯

Multiple Choice 2 If the acid dissociation constant, K a, for an acid HA is 8 x 10¯ 4 at 25 °C, what percent of the acid is dissociated in a 0.50-molar solution of HA at 25 °C? (A) 0.08% (B) 0.2% (C) 1% (D) 2% (E) 4%

Multiple Choice 3 As the number of oxygen atoms increases in any series of oxygen acids, such as HXO, HXO 2, HXO 3,...., which of the following is generally true? (A) The acid strength varies unpredictably. (B) The acid strength decreases only if X is a nonmetal. (C) The acid strength decreases only if X is a metal. (D) The acid strength decreases whether X is a nonmetal or a metal. (E) The acid strength increases.

Multiple Choice 4 Which of the following is the correct equilibrium expression for the hydrolysis of CO 3 2 ¯ ? (A) K = [HCO 3 ¯ ] / ( [CO 3 2 ¯ ] [H 3 O + ] ) (B) K = ( [HCO 3 ¯] [OH¯] ) / [CO 3 2 ¯] (C) K = ( [CO 3 2 ¯ ] [OH¯] ) / [HCO 3 ¯] (D) K = [CO 3 2 ¯ ] / ( [CO 2 ] [OH¯] 2 ) (E) K = ( [CO 3 2 ¯ ] [H 3 O + ] ) / [HCO 3

Multiple Choice 5 A solution of calcium hypochlorite, a common additive to swimming-pool water, is (A) basic because of the hydrolysis of the OCl¯ ion (B) basic because Ca(OH) 2 is a weak and insoluble base (C) neutral if the concentration is kept below 0.1 molar (D) acidic because of the hydrolysis of the Ca 2+ ions (E) acidic because the acid HOCl is formed

Multiple Choice 6 What is the H + (aq) concentration in 0.05 M HCN (aq) ? (The K a for HCN is 5.0 x 10¯ 10 ) (A) 2.5 x 10¯ 11 (B) 2.5 x 10¯ 10 (C) 5.0 x 10¯ 10 (D) 5.0 x 10¯ 6 (E) 5.0 x 10¯ 4

Free Response 1996 A HOCl OCl - + H + Hypochlorous acid, HOCl, is a weak acid commonly used as a bleaching agent. The acid-dissociation constant, K a, for the reaction represented above is 3.2x (a) Calculate the [H + ] of a 0.14-molar solution of HOCl. (b) Write the correctly balanced net ionic equation for the reaction that occurs when NaOCl is dissolved in water and calculate the numerical value of the equilibrium constant for the reaction. (c)Household bleach is made by dissolving chlorine gas in water, as represented below. Cl 2 (g) + H 2 O --> H + + Cl - + HOCl(aq) Calculate the pH of such a solution if the concentration of HOCl in the solution is molar.

Answer: (a)K a = = 3.2x10 -8 X = amount of acid that ionizes = [OCl - ] = [H + ] ( X) = [HOCl] that remains unionized 3.2x10 -8 = ; X = 6.7x10 -5 M = [H + ] (b)NaOCl(s) + H 2 O --> Na + (aq) + HOCl(aq) + OH - (aq) K b = = 3.1x10 -7 (C) 1 mol H+ for every 1 mole of HOCl produced [H + ] ~ [HOCl] = M pH = - log (0.065) = 1.2

Review When analyzing an acid-base, or salt equilibrium problems Copyright © Cengage Learning. All rights reserved 51 1)Write the balanced net ionic equation for the reaction. (salts have two) 2)Write the equilibrium expression. 3)Perform ICE 4)Then solve for the concentration, pH, or % ionization