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chapter_2016 - CHAPTER 16 ACIDS AND BASES 16.3 Table 16.2...

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CHAPTER 16 ACIDS AND BASES 16.3 Table 16.2 of the text contains a list of important Brønsted acids and bases. (a) both (why?), (b) base, (c) acid, (d) base, (e) acid, (f) base, (g) base, (h) base, (i) acid, (j) acid. 16.4 Recall that the conjugate base of a Brønsted acid is the species that remains when one proton has been removed from the acid. (a) nitrite ion: NO 2 (b) hydrogen sulfate ion (also called bisulfate ion): HSO 4 (c) hydrogen sulfide ion (also called bisulfide ion): HS (d) cyanide ion: CN (e) formate ion: HCOO 16.5 In general the components of the conjugate acid base pair are on opposite sides of the reaction arrow. The base always has one fewer proton than the acid. (a) The conjugate acid base pairs are (1) HCN (acid) and CN (base) and (2) CH 3 COO (base) and CH 3 COOH (acid). (b) (1) HCO 3 (acid) and CO 3 2 (base) and (2) HCO 3 (base) and H 2 CO 3 (acid). (c) (1) H 2 PO 4 (acid) and HPO 4 2 (base) and (2) NH 3 (base) and NH 4 + (acid). (d) (1) HClO (acid) and ClO (base) and (2) CH 3 NH 2 (base) and CH 3 NH 3 + (acid). (e) (1) H 2 O (acid) and OH (base) and (2) CO 3 2 (base) and HCO 3 (acid). (f) (1) H 2 O (acid) and OH (base) and (2) CH 3 COO (base) and CH 3 COOH (acid). 16.6 The conjugate acid of any base is just the base with a proton added. (a) H 2 S (b) H 2 CO 3 (c) HCO 3 (d) H 3 PO 4 (e) H 2 PO 4 (f) HPO 4 2 (g) H 2 SO 4 (h) HSO 4 (i) HNO 2 (j) HSO 3 16.7 The conjugate base of any acid is simply the acid minus one proton. (a) CH 2 ClCOO (b) IO 4 (c) H 2 PO 4 (d) HPO 4 2 (e) PO 4 3 (f) HSO 4 (g) SO 4 2 (h) HCOO (i) SO 3 2 (j) NH 3 (k) HS (l) S 2 (m) ClO 16.8 (a) The Lewis structures are C C O O O H and O - C C O O O - O - (b) H + and H 2 C 2 O 4 can act only as acids, HC 2 O 4 can act as both an acid and a base, and C 2 O 4 2 can act only as a base.
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CHAPTER 16: ACIDS AND BASES 351 16.15 Since pH = log[H + ], we write [H + ] = 10 pH (a) [H + ] = 10 2.42 = 3.8 × 10 3 M (c) [H + ] = 10 6.96 = 1.1 × 10 7 M (b) [H + ] = 10 11.21 = 6.2 × 10 12 M (d) [H + ] = 10 15.00 = 1.0 × 10 15 M 16.16 Strategy: Here we are given the pH of a solution and asked to calculate [H + ]. Because pH is defined as pH = log[H + ], we can solve for [H + ] by taking the antilog of the pH; that is, [H + ] = 10 pH . Solution: From Equation (16.5) of the text: (a) pH = log [H + ] = 5.20 log[H + ] = 5.20 To calculate [H + ], we need to take the antilog of 5.20. [H + ] = 10 5.20 = 6.3 × 10 6 M Check : Because the pH is between 5 and 6, we can expect [H + ] to be between 1 × 10 5 M and 1 × 10 6 M . Therefore, the answer is reasonable. (b) pH = log [H + ] = 16.00 log[H + ] = 16.00 [H + ] = 10 16.00 = 1.0 × 10 16 M (c) Strategy: We are given the concentration of OH ions and asked to calculate [H + ]. The relationship between [H + ] and [OH ] in water or an aqueous solution is given by the ion-product of water, K w [Equation (16.4) of the text]. Solution: The ion product of water is applicable to all aqueous solutions. At 25 ° C, K w = 1.0 × 10 14 = [H + ][OH ] Rearranging the equation to solve for [H + ], we write 14 14 9 1.0 10 1.0 10 [OH ] 3.7 10 −− ×× === × 6 [H ] 2.7 10 M +− × Check: Since the [OH ] < 1 × 10 7 M we expect the [H + ] to be greater than 1 × 10 7 M .
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This note was uploaded on 05/22/2008 for the course CHEM 122 taught by Professor Bellew during the Spring '07 term at New Mexico.

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chapter_2016 - CHAPTER 16 ACIDS AND BASES 16.3 Table 16.2...

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