Solutions to Homework Assignment 10 CHM 152 Spring 2002

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1 Solutions to Homework Assignment 10 CHM 152 Spring (a) This is a weak acid problem. Setting up the standard equilibrium table: CH 3 COOH(aq) H + (aq) + CH 3 COO (aq) Initial (M): Equilibrium (M): (0.40 x) +x +x = 2 2 [H ][CH3COO ] x x = [CH3COOH] (0.40 x) 0.40 x = [H + ] = M ph = 2.57 In addition to the acetate ion formed from the ionization of acetic acid, we also have acetate ion formed from the sodium acetate dissolving. CH 3 COONa(aq) CH 3 COO (aq) + Na + (aq) Dissolving 0.20 M sodium acetate initially produces 0.20 M CH 3 COO and 0.20 M Na +. The sodium ions are not involved in any further equilibrium (why?), but the acetate ions must be added to the equilibrium in part (a). CH 3 COOH(aq) H + (aq) + CH 3 COO (aq) Initial (M): Equilibrium (M): (0.40 x) +x ( x) = [H ][CH3COO ] ( x)( x) x(0.20) = [CH3COOH] (0.40 x) 0.40 x = [H + ] = M ph = 4.44 Could you have predicted whether the ph should have increased or decreased after the addition of the sodium acetate to the pure 0.40 M acetic acid in part (a)? An alternate way to work part of this problem is to use the Henderson-Hasselbalch equation. [conjugate base] ph = pk a + log [acid] ph M = log( ) + log 0.40 M = = 4.44

2 16.4 (a) This is a weak base calculation. K b = [NH = 4 ][OH ] ( x)( x) = [NH ] 0.20 x x = M = [OH ] poh = 2.72 ph = NH 4 Cl completely ionizes to give 0.30 M NH 4 + and 0.30 M Cl. NH4 + is the conjugate acid of NH3, and its concentration must be added to the equilibrium from part (a). NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH (aq) Initial (M): Equilibrium (M): (0.20 x) ( x) x 5 ( x)( x) K b = = (. 020 x) x = [OH ] = M poh = 4.92 ph = ( x) 020. Could you have predicted whether the ph should have increased or decreased after the addition of the ammonium chloride to the pure 0.20 M ammonia in part (a)? You could also use the Henderson-Hasselbalch equation to solve part. Table 15.4 gives the value of K a for the ammonium ion. Using this and the given concentrations with the Henderson- Hasselbalch equation gives: [ ] 10 ph = pk a + log ( ) ph = = 9.07 ( ) ( ) conjugate base = log log 0.20 acid 0.30 Is there any difference in the Henderson-Hasselbalch equation in the cases of a weak acid and its conjugate base and a weak base and its conjugate acid? 16.7 The p in front of pk a means "take the negative log" (Think of the meanings of ph and poh). A smaller pk a value will correspond to a larger K a value and hence a stronger acid. Therefore, the acid with pk a of 5.9 is the stronger acid (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. (c) NH 3 (ammonia) is a weak base, and its conjugate acid, NH 4 + is a weak acid. Therefore, this is a buffer system. This solution contains both a weak acid, H 2 PO 4 and its conjugate base, HPO4 2. Therefore, this is a buffer system.

3 (d) 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. (e) (f) 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. 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 (a) This problem is greatly simplified because the concentration of the weak acid (acetic acid) is equal to the concentration of its conjugate base (acetate ion). Let s set up a table of initial concentrations, change in concentrations, and equilibrium concentrations. CH 3 COOH (aq) H + (aq) + CH 3 COO (aq) Initial 2.0 M M Change x +x +x Equilibrium 2.0 M x x 2.0 M + x [H ][CH3COO ] [CH3COOH] + + [H ](2.0 + x) [H ](2.0) (2.0 x) 2.0 [H + ] Taking the log of both sides, p ph Thus, for a buffer system in which the [weak acid] = [weak base], ph = pk a ph = log( ) = 4.74 Similar to part (a), ph = p 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. Thus, buffer (a) can neutralize 10 times more added acid or base compared to buffer H 2 CO 3 (aq) HCO 3 (aq) + H + (aq) K a1 = pk a1 = 6.38 [HCO 3 ] ph = pk a + lo g [H CO ] 2 3

4 8.00 = log [ HCO3 ] [ HCO 2 3] log [ HCO3 ] = 162. [ HCO 2 3] [ HCO3 ] = 41.7 [ HCO 2 3] [ HCO 2 3] = [ HCO3 ] Step 1: Write the equilibrium that occurs between H 2 PO 4 and HPO4 2. Set up a table relating the initial concentrations, the change in concentration to reach equilibrium, and the equilibrium concentrations. H 2 PO 4 (aq) H + (aq) + HPO4 2 (aq) Initial (M): Equilibrium (M): 0.15 x x x Step 2: Write the ionization constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant (K a ), solve for x. + 2 [H ][HPO 4 ] [H2PO 4] You can look up the K a value for dihydrogen phosphate in Table 15.5 of your text = ( x)( x) (0.15 x) ( x)(0.10) (0.15) x = [H + ] = M Step 3: Having solved for the [H + ], calculate the ph of the solution. ph = log[h + ] = log( ) = For the first part we use K a for ammonium ion. (Why?) The Henderson Hasselbalch equation is 10 (0.20 M ) ph = log( ) + log (0.20 M = 9.25 ) For the second part, the acid base reaction is NH 3 (g) + H + + (aq) NH 4 (aq)

5 We find the number of moles of HCl added 1 L 0.10 mol HCl 10.0 ml = mol HCl 1000 ml 1 L + The number of moles of NH 3 and NH 4 originally present are 1 L 0.20 mol 65.0 ml = mol 1000 ml 1 L + Using the acid base reaction, we find the number of moles of NH 3 and NH 4 after addition of the HCl. moles NH 3 = ( ) mol = mol NH moles NH 4 = ( ) mol = mol NH4 We find the new ph: (0.012) ph = log = 9.18 (0.014) As calculated in Problem 16.10, the ph of this buffer system is equal to pk a. ph = p log( ) = 4.74 (a) The added NaOH will react completely with the acid component of the buffer, CH 3 COOH. NaOH ionizes completely; therefore, mol of OH are added to the buffer. Step 1: The neutralization reaction is: CH 3 COOH (aq) + OH (aq) CH3COO (aq) + H 2 O (l) Initial (mol) After reaction (mol) Step 2: Now, the acetic acid equilibrium is reestablished. Since the volume of the solution is 1.00 L, we can convert directly from moles to molar concentration. CH 3 COOH (aq) H + (aq) + CH 3 COO (aq) Initial (M) Change (M) x +x +x Equilibrium (M) 0.92 x x x Write the K a expression, then solve for x. [H ][CH3COO ] [CH3COOH] = ( x )( x ) (1.08) x (0.92 x) 0.92 x = [H + ] = M

6 Step 3: Having solved for the [H + ], calculate the ph of the solution. ph = log[h + ] = log( ) = 4.82 The ph of the buffer increased from 4.74 to 4.82 upon addition of mol of strong base. The added acid will react completely with the base component of the buffer, CH 3 COO. HCl ionizes completely; therefore, 0.12 mol of H + ion are added to the buffer Step 1: The neutralization reaction is: CH 3 COO (aq) + H + (aq) CH3COOH (aq) Initial (mol) After reaction (mol) Step 2: Now, the acetic acid equilibrium is reestablished. Since the volume of the solution is 1.00 L, we can convert directly from moles to molar concentration. CH 3 COOH (aq) H + (aq) + CH 3 COO (aq) Initial (M) Change (M) x +x +x Equilibrium (M) 1.12 x x x Write the K a expression, then solve for x. [H ][CH3COO ] [CH3COOH] = ( x )( x ) (0.88) x (1.12 x) 1.12 x = [H + ] = M Step 3: Having solved for the [H + ], calculate the ph of the solution. ph = log[h + ] = log( ) = 4.64 The ph of the buffer decreased from 4.74 to 4.64 upon addition of 0.12 mol of strong acid Recall that to prepare a solution of a desired ph,, we should choose a weak acid with a pk a value close to the desired ph. Calculating the pk a for each acid: For HA, p log( ) = 2.57 For HB, p log( ) = 5.36 For HC, p log( ) = 8.59 The buffer solution with a pk a closest to the desired ph is HC. Thus, HC is the best choice to prepare a buffer solution with ph = 8.60.

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