# hw4 soln - CHAPTER 18 ACID-BASE EQUILIBRIA FOLLOW-UP...

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18-1 CHAPTER 18 ACID-BASE EQUILIBRIA FOLLOW-UP PROBLEMS 18.1 a) Chloric acid, HClO 3 , is the stronger acid because acid strength increases as the number of O atoms exceeds the ionizable H atom. b) Hydrochloric acid, HCl , is one of the strong hydrohalic acids whereas acetic acid, CH 3 COOH, is a weak carboxylic acid. c) Sodium hydroxide, NaOH , is a strong base because Na is a Group 1A(1) metal. Methylamine, CH 3 NH 2 , is an organic amine and, therefore, a weak base. 18.2 Plan: The product of [H 3 O + ] and [OH - ] remains constant at 25°C because the value of K w is constant at a given temperature. Use K w = [H 3 O + ] [OH - ] = 1.0 x 10 -14 to solve for [H 3 O + ]. Solution: Calculating [H 3 O + ]: [ H 3 O + ] = K w / [OH - ] = (1.0 x 10 -14 ) / (6.7 x 10 -2 ) = 1.4925 x 10 -13 = 1.5 x 10 -13 M Since [OH - ] > [H 3 O + ], the solution is basic . Check: The solution is neutral when either species concentration is 1 x 10 -7 M . Since the problem specifies an [OH - ] much greater than this, the solution is basic. 18.3 Plan: NaOH is a strong base that dissociates completely in water. Subtract pH from 14.00 to find the pOH, and calculate inverse logs of pH and pOH to find [H 3 O + ] and [OH - ], respectively. Solution: pH + pOH = 14.00 pOH = 14.00 - 9.52 = 4.48 pH = -log [H 3 O + ] [H 3 O + ] = 10 -pH = 10 -9.52 = 3.01995 x 10 -10 = 3.0 x 10 -10 M pOH = -log [OH - ] [ O H - ] = 10 -pOH = 10 -4.48 = 3.3113 x 10 -5 = 3.3 x 10 -5 M Check: At 25°C, [H 3 O + ] [OH - ] should equal 1.0 x 10 -14 . (3.0 x 10 -10 ) (3.3 x 10 -5 ) = 9.9 x 10 -15 1.0 x 10 -14 . The significant figures in the concentrations reflect the number of significant figures after the decimal point in the pH and pOH values. 18.4 Plan: Identify the conjugate pairs by first identifying the species that donates H + (the acid) in either reaction direction. The other reactant accepts the H + and is the base. The acid has one more H and +1 greater charge than its conjugate base. Solution: a) CH 3 COOH has one more H + than CH 3 COO - . H 3 O + has one more H + than H 2 O. Therefore, CH 3 COOH and H 3 O + are the acids, and CH 3 COO - and H 2 O are the bases. The conjugate acid/base pairs are CH 3 COOH/CH 3 COO - and H 3 O + /H 2 O . b) H 2 O donates a H + and acts as the acid. F - accepts the H + and acts as the base. In the reverse direction, HF acts as the acid and OH - acts as the base. The conjugate acid/base pairs are H 2 O/OH - and HF/F - . 18.5 a) The following equation describes the dissolution of ammonia in water: N H 3 ( g ) + H 2 O( l ) ' NH 4 + ( aq ) + OH - ( aq ) weak base weak acid stronger acid strong base Ammonia is a known weak base, so it makes sense that it accepts a H + from H 2 O. The reaction arrow indicates that the equilibrium lies to the left because the question states, “you smell ammonia” (NH 4 + and OH - are odorless). According to Figure 18.9, NH 4 + and OH - are the stronger acid and base, so the reaction proceeds to the formation of the weaker acid and base. b) The addition of excess HCl results in the following equation: N H 3 ( g ) + H 3 O + ( aq ; from HCl) ' NH 4 + ( aq ) + H 2 O( l ) stronger base strong acid weak acid weak base HCl is a strong acid and is much stronger than NH 4 + , according to Figure 18.9. Similarly, NH 3 is a stronger base than H 2 O. The reaction proceeds to produce the weak acid and base, and thus the odor from NH 3 disappears.

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18-2 c) The solution in (b) is mostly NH 4 + and H 2 O.
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hw4 soln - CHAPTER 18 ACID-BASE EQUILIBRIA FOLLOW-UP...

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