CHEMISTRY - BROWN 13E CH.16 - ACID-BASE EQUILIBRIA - PART 2.

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2 CONCEPT: ph and poh To deal with incredibly small concentration values of [H + ] and [OH - ] we can use the ph scale. Under normal conditions, the ph scale operates within the range of to. By taking the log of [H + ] and [OH - ] we can find ph and poh. ph log[h + ] poh log[oh ] p log By recognizing the relationship between [H + ] and [OH - ] with ph and poh we can create new formula relationships. ph log[h + ] poh log[oh ] A species with a ph greater than 7 is classified as and the [H + ] is than the [OH - ]. The the base then the the ph. A species with a ph less than 7 is classified as and the [H + ] is than the [OH - ]. The the acid then the the ph. A species with a ph equal to 7 is classified as and the [H + ] is than the [OH - ]. By using log with the equilibrium expression for water a relationship between ph and poh can be created. ph + poh 14 Page 2

3 PRACTICE: ph and poh (CALCULATIONS 1) EXAMPLE: What is the hydroxide ion and hydronium ion concentration of an aqueous solution that has a ph equal to 6.12? PRACTICE 1: Which of the following solutions will have the lowest concentration of hydronium ions? a) moles C6H5NH2 b) moles Be(OH)2 c) moles SrH2 d) moles (CH3)2NH PRACTICE 2: Which of the following statements about aqueous solutions is/are true? a) For an basic solution the concentration of H3O + is greater than the concentration of OH. b) The ph of a neutral aqueous solution is 7.00 at all temperatures. c) An acidic solution under normal conditions has a ph value less than d) If the concentration of H3O + decreases then the concentration of OH will also decrease. e) The ph of aqueous solutions is less than 7. Page 3

4 PRACTICE: ph and poh (CALCULATIONS 2) EXAMPLE: A solution is prepared by dissolving mol Sr(OH)2 in water to produce a solution with a volume of 750 ml. a) What is the [OH - ]? b) What is the [H + ]? PRACTICE: What is the Kw of pure water at 20.0 C, if the ph is 7.083? a) b) c) d) Page 4

5 CONCEPT: AUTO IONIZATION OF WATER Water can react with itself in a reaction called self ionization where and are produced. H 2 O (l) + H 2 O (l) This reaction is usually written more simply as: H 2 O (l) The equilibrium equation for water is called the (KW) for water and is given by the following: K W [H + ][OH ] At 25 o C, KW, but remember KW, like all other constants K, is temperature dependent. Increasing the temperature will KW. Constant 0 o C 10 o C 50 o C 100 o C K W 1.14 x x x x EXAMPLE: Determine the concentration of hydronium ions for a neutral solution at 25 o C and at 50 o C. Page 5

6 CONCEPT: CALCULATING ph and poh OF STRONG SPECIES STRONG ACIDS & BASES are considered Electrolytes so they ionize completely in water. HCl (aq) NaOH (aq) H 2 O H 2 O H + (aq) + Cl (aq) Na + (aq) + OH (aq) EXAMPLE 1: Calculate the ph of a M solution of CaH2. EXAMPLE 2: Calculate the ph of a M HBr solution to the correct number of significant figures. a) 3.3 b) 3.26 c) d) e) All are correct PRACTICE: Calculate the ph of ml of 4.3 x 10-7 M H2SO4. Page 6

7 CONCEPT: CALCULATING ph and poh OF WEAK SPECIES WEAK ACIDS & BASES are considered Electrolytes so they don t completely ionize in water. HF + H2O NH3 + H2O F (aq) + H3O + (aq) NH4 + (aq) + OH (aq) EXAMPLE 1: Pryridine, an organic molecule, is a very common weak base. C5H5N (aq) + H2O (l) C5H5NH + (aq) + OH - (g) Assume you have a M aqueous solution of pyridine, C5H5N, determine its ph. The Kb value for the compound is 1.5 x Page 7

8 PRACTICE: CALCULATING ph and poh OF WEAK SPECIES (CALCULATIONS 1) EXAMPLE: An unknown weak base has an initial concentration of M with a ph of Calculate its equilibrium base constant. PRACTICE: Determine the ph of a solution made by dissolving 6.1 g of sodium cyanide, NaCN, in enough water to make a g ml of solution. (MW of NaCN mol ). The Ka value of HCN is 4.9 x Page 8

9 CONCEPT: ACID & BASE CONSTANTS As you might already realize, there are relatively few strong acids. The great majority of acids are weak acids. Consider a weak monoprotic acid, HA, and its ionization in water: HA (aq) + H2O (l) A (aq) + H3O + (aq) The equilibrium expression for this ionization would be: K a Reactan ts Where Ka represents the and it measures the strength of weak acids. When looking at weak bases we don t use Ka, but instead, which represents the. The relationship between Ka and Kb can be expressed with the following equation: K W K a K b In general, the the Ka the stronger the acid and the the concentration of H +. In general, the the pka the stronger the acid and the the concentration of H +. PRACTICE: If the Kb of NH3 is 1.76 x 10-5, determine the acid dissociation constant of its conjugate acid. Page 9

10 PRACTICE: ACID & BASE CONSTANTS (CALCULATIONS 1) EXAMPLE 1: Knowing that HF has a higher Ka value than CH3COOH, determine, if possible, in which direction the following equilibrium lies. HF (aq) + CH3COO (aq) F (aq) + CH3COOH (aq) a) Equilibrium lies to the left. b) Equilibrium lies to the right. c) Equilibrium is equal and balanced. d) Not enough information given. EXAMPLE 2: What is the equilibrium constant for the following reaction and determine if reactants or products are favored. HCN (aq) + ClO2 (aq) CN (aq) + HClO2 (aq) The acid dissociation constant of HCN is 4.9 x and the acid dissociation of HClO2 is 1.1 x HCN (aq) + H2O (aq) CN (aq) + H3O + (aq) HClO2 (aq) + H2O (aq) ClO2 (aq) + H3O + (aq) Page 10

11 PRACTICE: ACID & BASE CONSTANTS (CALCULATIONS 2) EXAMPLE: Which of the following solutions will have the lowest ph? a) 0.25 M HC2F3O2 b) 0.25 M HIO4 c) 0.25 M HC3H5O3 d) 0.25 M H2CO3 e) 0.25 M HSeO4 PRACTICE 1: Which Bronsted-Lowry base has the greatest concentration of hydroxide ions? a) C2H8N2 (Kb 8.3 x 10-5 ) b) C5H5N (Kb 1.7 x 10-9 ) c) (CH3)3N (Kb 1.0 x 10-6 ) d) C3H7NH2 (Kb 3.5 x 10-4 ) e) C6H5NH2 (Kb 3.9 x ) PRACTICE 2: Which Bronsted-Lowry acid has the weakest conjugate base? a) HCNO (Ka 2.0 x 10-4 ) b) HF (Ka 3.5 x 10-4 ) c) HN3 (Ka 2.5 x 10-5 ) d) H2CO3 (Ka 4.3 x 10-7 ) Page 11

12 CONCEPT: DIPROTIC ACIDS Diprotic acids and bases are compounds that can donate or accept H + ion. For diprotic acids their equations can be illustrated by: H2A (aq) + H2O (l) HA (aq) + H3O + (aq) K a1 Reactan ts HA (aq) + H2O (l) A 2 (aq) + H3O + (aq) K a2 Reactan ts For diprotic bases their equations can be illustrated by: A 2 (aq) + H2O (aq) HA (aq) + OH (aq) K b1 Reactan ts HA (aq) + H2O (aq) H2A (aq) + OH (aq) K b2 Reactan ts Based on these equations the relationship between the different forms of diprotic species are: As a result of these equations for diprotic acids and bases the relationship between Ka and Kb will be: K a1 K b2 K w K a2 K b1 K w When dealing with diprotic acids: 1) H2A can be treated as a monoprotic acid and we use can be used to find ph. 2) HA represents the intermediate form and we use can be used to find ph. 3) A 2 represents the basic form and we use can be used to find ph. Page 12

13 PRACTICE: DIPROTIC ACIDS CALCULATIONS 1 EXAMPLE 1: Sulfurous acid, H2SO3, represents a diprotic acid with a Ka1 1.6 x 10-2 and Ka2 4.6 x Calculate the ph and concentrations of H2SO3, HSO3 and SO3 2 when given M H2SO3. EXAMPLE 2: Determine the ph of M Na2S. Hydrosulfuric acid, H2S, contains Ka1 1.0 x 10-7 and Ka2 9.1 x Page 13

14 CONCEPT: POLYPROTIC ACIDS Our understanding of diprotic acids and bases can be used to understand polyprotic acids and bases. For polyprotic acids their equations can be illustrated by: H 3A (aq) + H 2O (l) H 2A (aq) + H 3O + (aq) K a1 Reactan ts H 2A (aq) + H 2O (l) HA 2 (aq) + H 3O + (aq) K a2 Reactan ts HA 2 (aq) + H 2O (l) A 3 (aq) + H 3O + (aq) K a3 Reactan ts For polyprotic bases their equations can be illustrated by: A 3 (aq) + H 2O (l) HA 2 (aq) + OH (aq) K b1 HA 2 (aq) + H 2O (l) H2A (aq) + OH (aq) K b2 H2A (aq) + H 2O (l) H 3A (aq) + OH (aq) K b3 Reactan ts Reactan ts Reactan ts As a result of these equations for polyprotic acids and bases the relationship between Ka and Kb will be: K a1 K b3 K w K a2 K b2 K w K a3 K b1 K w When dealing with polyprotic acids: H3A can be treated as a monoprotic acid and we use can be used to find ph. A 3 represents the basic form and we use can be used to find ph. H 2 A [H + ] K a1 K a2 [ ] 0 + K a1 K w K a1 + [ ] 0 HA 2 [H + ] K a2k a3 [ ] 0 + K a2 K w K a2 + [ ] 0 Page 14

15 PRACTICE: POLYPROTIC ACIDS CALCULATIONS EXAMPLE 1: Determine the ph of M sodium hydrogen phosphate, Na2HPO4. Phosphoric acid, H3PO4, contains Ka1 7.5 x 10-3, Ka2 6.2 x 10-8 and Ka3 4.2 x EXAMPLE 2: Determine the ph of M citric acid, H3C6H5O7 it possesses Ka1 7.4 x 10-4, Ka2 1.7 x 10-5 and Ka3 4.0 x Page 15

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