chapter_2017

chapter_2017 - CHAPTER 17 ACID-BASE EQUILIBRIA AND...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
CHAPTER 17 ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA 17.5 (a) HCl (hydrochloric acid) is a strong acid. A buffer is a solution containing both a weak acid and a weak base. Therefore, this is not a buffer system. (b) NH 3 (ammonia) is a weak base, and its conjugate acid, NH 4 + is a weak acid. Therefore, this is a buffer system. (c) This solution contains both a weak acid, H 2 PO 4 and its conjugate base, HPO 4 2 . Therefore, this is a buffer system. 17.6 Strategy: What constitutes a buffer system? Which of the preceding solutions contains a weak acid and its salt (containing the weak conjugate base)? Which of the preceding solutions contains a weak base and its salt (containing the weak conjugate acid)? Why is the conjugate base of a strong acid not able to neutralize an added acid? Solution: The criteria for a buffer system are that we must have a weak acid and its salt (containing the weak conjugate base) or a weak base and its salt (containing the weak conjugate acid). (a) HNO 2 (nitrous acid) is a weak acid, and its conjugate base, NO 2 (nitrite ion, the anion of the salt KNO 2 ), is a weak base. Therefore, this is a buffer system. (b) H 2 SO 4 (sulfuric acid) is a strong acid. A buffer is a solution containing both a weak acid and a weak base. Therefore, this is not a buffer system. (c) HCOOH (formic acid) is a weak acid, and its conjugate base, HCOO (formate ion, the anion of the salt HCOOK), is a weak base. Therefore, this is a buffer system. 17.7 H 2 CO 3 ( aq ) U HCO 3 ( aq ) + H + ( aq ) 1 7 a 4.2 10 K 1 a p 6.38 = K 3 a 23 [HCO ] pH p log [H CO ] =+ K 3 [HCO ] 8.00 6.38 log [H CO ] 3 [HCO ] log 1.62 [H CO ] = 3 [HCO ] 41.7 [H CO ] = 3 [H CO ] [HCO ] = 0.024
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
CHAPTER 17: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA 381 17.8 Strategy: The pH of a buffer system can be calculated in a similar manner to a weak acid equilibrium problem. The difference is that a common-ion is present in solution. The K a of CH 3 COOH is 1.8 × 10 5 (see Table 16.3 of the text). Solution: (a) We summarize the concentrations of the species at equilibrium as follows: CH 3 COOH( aq ) U H + ( aq ) + CH 3 COO ( aq ) Initial ( M ): 2.0 0 2.0 Change ( M ): x + x + x Equilibrium ( M ): 2.0 x x 2.0 + x 3 a 3 [H ][CH COO ] [CH COOH] + = K a [H ](2.0 ) [H ](2.0) (2.0 2.0 ++ + =≈ x K x ) K a = [H + ] Taking the log of both sides, p K a = pH Thus, for a buffer system in which the [weak acid] = [weak base], pH = p K a pH = log(1.8 × 10 5 ) = 4.74 (b) Similar to part (a), pH = p K a = 4.74 Buffer (a) will be a more effective buffer because the concentrations of acid and base components are ten times higher than those in (b). Thus, buffer (a) can neutralize 10 times more added acid or base compared to buffer (b). 17.9 NH 4 + ( aq ) U NH 3 ( aq ) + H + ( aq ) K a = 5.6 × 10 10 p K a = 9.25 3 a 4 [NH ] 0.15 p log 9.25 log 0.35 + =+ = + = pH 8.88 M K M 17.10 Step 1: Write the equilibrium that occurs between H 2 PO 4 and HPO 4 2 . Set up a table relating the initial concentrations, the change in concentration to reach equilibrium, and the equilibrium concentrations.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 43

chapter_2017 - CHAPTER 17 ACID-BASE EQUILIBRIA AND...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online