CHAPTER 1 ACID-BASE EQUILIBRIA AND SOLUBILITY 1.1 (a) This is a weak acid problem. Setting up the standard equilibrium table: CHCOOH(aq) H (aq) CHCOO (aq) Initial (M): 0.40 0.00 0.00 Change (M): x x x Equilibrium (M): (0.40 x) x x Assuming ideal behavior (that is, activities can be replaced with concentrations): K a [H ][CH COO ] [CH COOH] 5 x x 1.8 10 (0.40 x) 0.40 x [H ].7 10 M ph.57 (b) In addition to the acetate ion formed from the ionization of acetic acid, we also have acetate ion formed from the sodium acetate dissolving. CHCOONa(aq) CHCOO (aq) Na (aq) Dissolving 0.0 M sodium acetate initially produces 0.0 M CHCOO and 0.0 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). CHCOOH(aq) H (aq) CHCOO (aq) Initial (M): 0.40 0.00 0.0 Change (M): x x x Equilibrium (M): (0.40 x) x (0.0 x) K a [H ][CH COO ] [CH COOH] 5 ( x)(0.0 x) x(0.0) 1.8 10 (0.40 x) 0.40 x [H ].6 10 5 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 (b) of this problem is to use the Henderson-Hasselbalch equation. [conjugate base] ph pka log [acid] 5 0.0 ph log(1.8 10 ) log M 4.74 0.0 4.44 0.40 M 1. (a) This is a weak base calculation. NH(aq) HO(l) NH4 (aq) OH (aq) Initial (M): 0.0 0 0 Change (M): x x x Equilibrium (M): 0.0 x x x K b [NH 4 ][OH ] [NH ] 5 ( x)( x) x 1.8 10 0.0 x 0.0 x 1.9 10 M [OH ] poh.7 ph 11.8 (b) The initial concentration of NH4 is 0.0 M from the salt NH4Cl. We set up a table as in part (a). NH(aq) HO(l) NH4 (aq) OH (aq) Initial (M): 0.0 0.0 0 Change (M): x x x Equilibrium (M): 0.0 x 0.0 x x K b [NH 4 ][OH ] [NH ] 5 ( x)(0.0 x) x(0.0) 1.8 10 0.0 x 0.0 x 1. 10 5 M [OH ] poh 4.9 ph 9.08 Alternatively, we could use the Henderson-Hasselbalch equation to solve this problem. Table 1.5 gives the value of Ka for the ammonium ion. Substituting into the Henderson- Hasselbalch equation gives: [conjugate base] 10 (0.0) ph pk a log log(5.6 10 ) log acid (0.0)
ph 9.5 0.18 9.07 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? 1. (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) (c) (d) HSO4 (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. This solution contains both a weak acid, HPO4 and its conjugate base, HPO4. Therefore, this is a buffer system. HNO (nitrous acid) is a weak acid, and its conjugate base, NO (nitrite ion, the anion of the salt KNO), is a weak base. Therefore, this is a buffer system. (e) HSO is a weak acid and SO 4 is a weak base, but they are not conjugates of each other. Hence this is not a buffer system. 1.4 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) (b) (c) (d) HCN is a weak acid, and its conjugate base, CN, is a weak base. Therefore, this is a buffer system. HSO4 is a weak acid, and its conjugate base, SO4 is a weak base (see Table 11.6 of the text). Therefore, this is a buffer system. NH (ammonia) is a weak base, and its conjugate acid, NH4 is a weak acid. Therefore, this is a buffer system. Because HI is a strong acid, its conjugate base, I, is an extremely weak base. This means that the I ion will not combine with a H ion in solution to form HI. Thus, this system cannot act as a buffer system. 1.5 NH4 (aq) NH(aq) H (aq) Ka 5.6 10 10 pka 9.5 [NH ] 0.15 ph M 8.88 M p Ka log [NH 9.5 log 0.5 4 ] 1.6 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 Ka of CHCOOH is 1.8 10 5. Solution:
(a) We summarize the concentrations of the species at equilibrium as follows: CHCOOH(aq) H (aq) CHCOO (aq) Initial (M):.0 0.0 Change (M): x x x Equilibrium (M):.0 x x.0 x K a [H ][CH COO ] [CH COOH] K a [H ](.0 x) [H ](.0) (.0 x).0 Ka [H ] Taking the log of both sides, pka ph Thus, for a buffer system in which the [weak acid] [weak base], ph pka ph log(1.8 10 5 ) 4.74 (b) Similar to part (a), ph pka 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). 1.7 HCO(aq) HCO (aq) H (aq) K a 4. 10 7 1 pk 6.8 a 1 [HCO ] ph pk a log [H CO ] [HCO ] 8.00 6.8 log [H CO ] [HCO ] log 1.6 [H CO ] [HCO ] [H CO ] 41.7
[HCO ] [HCO ] 0.04 1.8 Step 1: Write the equilibrium that occurs between HPO4 and HPO4. Set up a table relating the initial concentrations, the change in concentration to reach equilibrium, and the equilibrium concentrations. HPO4 (aq) H (aq) HPO4 (aq) Initial (M): 0.15 0 0.10 Change (M): x x x Equilibrium (M): 0.15 x x 0.10 x Step : Write the ionization constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant (Ka), solve for x. K a [H ][HPO 4 ] [H PO 4 ] You can look up the Ka value for dihydrogen phosphate in Table 11.6 of your text. 8 ( x)(0.10 x) 6. 10 (0.15 x) 8 ( )(0.10) 6. 10 x (0.15 ) x [H ] 9. 10 8 M Step : Having solved for the [H ], calculate the ph of the solution. ph log[h ] log(9. 10 8 ) 7.0 1.9 Using the HendersonHasselbalch equation: [CHCOO ] ph pk a log [CH COOH] Thus, [CHCOO ] 4.50 4.74 log [CH COOH] [CH COO ] [CH COOH] 0.58 1.10 We can use the Henderson-Hasselbalch equation to calculate the ratio [HCO ]/[HCO]. The Henderson-Hasselbalch equation is: [conjugate base] ph pka log [acid] For the buffer system of interest, HCO is the conjugate base of the acid, HCO. We can write:
The [conjugate base]/[acid] ratio is: 7 [HCO ] ph 7.40 log(4. 10 ) log [H CO ] [HCO ] 7.40 6.8 log [H CO ] [HCO ] log 7.40 6.8 1.0 [H CO ] [HCO ] [H CO ] 101.0 1.0 10 1 = 10. The buffer should be more effective against an added acid because ten times more base is present compared to acid. Note that a ph of 7.40 is only a two significant figure number (Why?); the final result should only have two significant figures. 1.11 For the first part we use Ka for ammonium ion. (Why?) The HendersonHasselbalch equation is 10 (0.0 ) ph log(5.6 10 ) log M 9.5 (0.0 M ) For the second part, the acidbase reaction is NH(g) H (aq) NH4 (aq) We find the number of moles of HCl added 0.10 mol HCl 10.0 ml 0.0010 mol HCl 1000 ml soln The number of moles of NH and NH4 originally present are 0.0 mol 65.0 ml 0.01 mol 1000 ml soln Using the acid-base reaction, we find the number of moles of NH and NH4 after addition of the HCl. NH(aq) H (aq) NH4 (aq) Initial (mol): 0.01 0.0010 0.01 Change (mol): 0.0010 0.0010 0.0010 Final (mol): 0.01 0 0.014 We find the new ph: (0.01) ph 9.5 log 9.18 (0.014) 1.1 As calculated in Problem 1.6, the ph of this buffer system is equal to pka.
ph pka log(1.8 10 5 ) 4.74 (a) The added NaOH will react completely with the acid component of the buffer, CHCOOH. NaOH ionizes completely; therefore, 0.080 mol of OH are added to the buffer. Step 1: The neutralization reaction is: CHCOOH(aq) OH (aq) CHCOO (aq) HO(l) Initial (mol): 1.00 0.080 1.00 Change (mol): 0.080 0.080 0.080 Final (mol): 0.9 0 1.08 Step : 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. CHCOOH(aq) H (aq) CHCOO (aq) Initial (M): 0.9 0 1.08 Change (M): x x x Equilibrium (M): 0.9 x x 1.08 x Write the Ka expression, then solve for x. K a [H ][CH COO ] [CH COOH] 5 ( x)(1.08 x) x(1.08) 1.8 10 (0.9 x) 0.9 x [H ] 1.5 10 5 M Step : Having solved for the [H ], calculate the ph of the solution. ph log[h ] log(1.5 10 5 ) 4.8 The ph of the buffer increased from 4.74 to 4.8 upon addition of 0.080 mol of strong base. (b) The added acid will react completely with the base component of the buffer, CHCOO. HCl ionizes completely; therefore, 0.1 mol of H ion are added to the buffer Step 1: The neutralization reaction is: CHCOO (aq) H (aq) CHCOOH(aq) Initial (mol): 1.00 0.1 1.00 Change (mol): 0.1 0.1 0.1 Final (mol): 0.88 0 1.1 Step : 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. CHCOOH(aq) H (aq) CHCOO (aq) Initial (M): 1.1 0 0.88 Change (M): x x x Equilibrium (M): 1.1 x x 0.88 x
Write the Ka expression, then solve for x. K a [H ][CH COO ] [CH COOH] 5 ( x)(0.88 x) x(0.88) 1.8 10 (1.1 x) 1.1 x [H ]. 10 5 M Step : Having solved for the [H ], calculate the ph of the solution. ph log[h ] log(. 10 5 ) 4.64 The ph of the buffer decreased from 4.74 to 4.64 upon addition of 0.1 mol of strong acid. 1.1 We write K a 1.1 10 1 K a.5 10 6 pka 1.96 pk 5.60 a In order for the buffer solution to behave effectively, the pka of the acid component must be close to the desired ph. Therefore, the proper buffer system is NaA/NaHA. 1.14 Strategy: For a buffer to function effectively, the concentration of the acid component must be roughly equal to the conjugate base component. According to Equation (1.) of the text, when the desired ph is close to the pka of the acid, that is, when ph pka, when [conjugate base] log 0 [acid] [conjugate base] [acid] 1 Solution: To prepare a solution of a desired ph, we should choose a weak acid with a pka value close to the desired ph. Calculating the pka for each acid: For HA, pka log(.7 10 ).57 For HB, pka log(4.4 10 6 ) 5.6 For HC, pka log(.6 10 9 ) 8.59 The buffer solution with a pka closest to the desired ph is HC. Thus, HC is the best choice to prepare a buffer solution with ph 8.60. 1.15 Since the acid is monoprotic, the number of moles of KOH is equal to the number of moles of acid. 0.081 mol Moles acid 16.4 ml 0.001 mol 1000 ml
Molar mass 0.688 g 0.001 mol 0 g mol 1 1.16 We want to calculate the molar mass of the diprotic acid. The mass of the acid is given in the problem, so we need to find moles of acid in order to calculate its molar mass. want to calculate given molar mass of H A g HA mol H A need to find The neutralization reaction is: KOH(aq) HA(aq) KA(aq) HO(l) From the volume and molarity of the base needed to neutralize the acid, we can calculate the number of moles of HA reacted. 1.00 mol KOH 1 mol HA 11.1 ml KOH 5.55 10 mol HA 1000 ml mol KOH We know that 0.500 g of the diprotic acid were reacted (1/10 of the 50 ml was tested). Divide the number of grams by the number of moles to calculate the molar mass. M (H A) 0.500 g H A 5.55 10 mol H A 90.1 g mol 1 1.17 The neutralization reaction is: HSO4(aq) NaOH(aq) NaSO4(aq) HO(l) Since one mole of sulfuric acid combines with two moles of sodium hydroxide, we write: 0.500 mol HSO4 mol NaOH mol NaOH 1.5 ml HSO4 0.015 mol NaOH 1000 ml soln 1 mol H SO 4 concentration of NaOH 0.015 mol NaOH 50.0 10 L soln 0.5 M 1.18 We want to calculate the molarity of the Ba(OH) solution. The volume of the solution is given (19. ml), so we need to find the moles of Ba(OH) to calculate the molarity.
want to calculate need to find M of Ba(OH) mol Ba(OH) L of Ba(OH) soln The neutralization reaction is: given HCOOH Ba(OH) (HCOO)Ba HO From the volume and molarity of HCOOH needed to neutralize Ba(OH), we can determine the moles of Ba(OH) reacted. 0.88 mol HCOOH 1 mol Ba(OH) 0.4 ml HCOOH 9.01 10 mol Ba(OH) 1000 ml mol HCOOH The molarity of the Ba(OH) solution is: 9.01 10 mol Ba(OH) 19. 10 L 0.467 M 1.19 (a) Since the acid is monoprotic, the moles of acid equals the moles of base added. HA(aq) NaOH(aq) NaA(aq) HO(l) 0.06 mol Moles acid 18.4 ml 0.00116 mol 1000 ml soln We know the mass of the unknown acid in grams and the number of moles of the unknown acid. Molar mass 0.176 g 0.00116 mol 1.10 10 g mol 1 (b) The number of moles of NaOH in 10.0 ml of solution is 0.06 mol 4 10.0 ml 6. 10 mol 1000 ml soln The neutralization reaction is: HA(aq) NaOH(aq) NaA(aq) HO(l) Initial (mol): 0.00116 6. 10 4 0 Change (mol): 6. 10 4 6. 10 4 6. 10 4 Final (mol): 5. 10 4 0 6. 10 4 ml. Now, the weak acid equilibrium will be reestablished. The total volume of solution is 5.0
4 5. 10 mol [HA] 0.015 M 0.05 L 4 6. 10 mol [A ] 0.0181 M 0.05 L We can calculate the [H ] from the ph. [H ] 10 ph 10 5.87 1.5 10 6 M HA(aq) H (aq) A (aq) Initial (M): 0.015 0 0.0181 Change (M): 1.5 10 6 1.5 10 6 1.5 10 6 Equilibrium (M): 0.015 1.5 10 6 0.0181 Substitute the equilibrium concentrations into the equilibrium constant expression to solve for Ka. K a [H ][A ] [HA] (1.5 106 )(0.0181) 0.015 1.6 10 6 1.0 The resulting solution is not a buffer system. There is excess NaOH and the neutralization is well past the equivalence point. 0.167 mol Moles NaOH 0.500 L 0.085 mol 1L 0.100 mol Moles CHCOOH 0.500 L 0.0500 mol 1L CHCOOH(aq) NaOH(aq) CHCOONa(aq) HO(l) Initial (mol): 0.0500 0.085 0 Change (mol): 0.0500 0.0500 0.0500 Final (mol): 0 0.05 0.0500 The volume of the resulting solution is 1.00 L (500 ml 500 ml 1000 ml). [OH ] 0.05 mol 1.00 L 0.05 M (0.05 0.0500) mol [Na ] 0.085 M 1.00 L 14 w 1.0 10 [H ] K.0 10 [OH ] 0.05 [CH COO ] 0.0500 mol 1.00 L 0.0500 M 1 M CHCOO (aq) HO(l) CHCOOH(aq) OH (aq) Initial (M): 0.0500 0 0.05
Change (M): x x x Equilibrium (M): 0.0500 x x 0.05 x K b [CH COOH][OH ] [CH COO ] 10 ( x)(0.05 x) ( x)(0.05) 5.6 10 (0.0500 x) (0.0500) x [CHCOOH] 8.4 10 10 M 1.1 HCl(aq) CHNH(aq) CHNH (aq) Cl (aq) Since the concentrations of acid and base are equal, equal volumes of each solution will need to be added to reach the equivalence point. Therefore, the solution volume is doubled at the equivalence point, and the concentration of the conjugate acid from the salt, CHNH, is: 0.0 M 0.10 M The conjugate acid undergoes hydrolysis. CHNH (aq) HO(l) HO (aq) CHNH(aq) Initial (M): 0.10 0 0 Change (M): x x x Equilibrium (M): 0.10 x x x K a [CHNH ] [H O ][CH NH ]. 10 11 x 0.10 x Assuming that, 0.10 x 0.10 x [HO ] 1.5 10 6 M ph 5.8 1. Let's assume we react 1 L of HCOOH with 1 L of NaOH. HCOOH(aq) NaOH(aq) HCOONa(aq) HO(l) Initial (mol): 0.10 0.10 0 Change (mol): 0.10 0.10 0.10 Final (mol): 0 0 0.10 The solution volume has doubled (1 L 1 L L). The concentration of HCOONa is: M 0.10 mol (HCOONa) 0.050 M L
HCOO (aq) is a weak base. The hydrolysis is: HCOO (aq) HO(l) HCOOH(aq) OH (aq) Initial (M): 0.050 0 0 Change (M): x x x Equilibrium (M): 0.050 x x x K b [HCOOH][OH ] [HCOO ] 11 x x 5.9 10 0.050 x 0.050 x 1.7 10 6 M [OH ] poh 5.77 ph 8. 1. The reaction between CHCOOH and KOH is: CHCOOH(aq) KOH(aq) CHCOOK(aq) HO(l) We see that 1 mole CHCOOH 1 mol KOH. Therefore, at every stage of titration, we can calculate the number of moles of acid reacting with base, and the ph of the solution is determined by the excess acid or base left over. At the equivalence point, however, the neutralization is complete, and the ph of the solution will depend on the extent of the hydrolysis of the salt formed, which is CHCOOK. (a) No KOH has been added. This is a weak acid calculation. CHCOOH(aq) HO(l) HO (aq) CHCOO (aq) Initial (M): 0.100 0 0 Change (M): x x x Equilibrium (M): 0.100 x x x K a [H O ][CH COO ] [CH COOH] 5 ( x)( x) x 1.8 10 0.100 x 0.100 x 1.4 10 M [HO ] ph.87 (b) The number of moles of CHCOOH originally present in 5.0 ml of solution is: 0.100 mol CHCOOH 5.0 ml.50 10 mol 1000 ml CH COOH soln The number of moles of KOH in 5.0 ml is:
0.00 mol KOH 5.0 ml 1.00 10 mol 1000 ml KOH soln We work with moles at this point because when two solutions are mixed, the solution volume increases. As the solution volume increases, molarity will change, but the number of moles will remain the same. The changes in number of moles are summarized. CHCOOH(aq) KOH(aq) CHCOOK(aq) HO(l) Initial (mol):.50 10 1.00 10 0 Change (mol): 1.00 10 1.00 10 1.00 10 Final (mol): 1.50 10 0 1.00 10 At this stage, we have a buffer system made up of CHCOOH and CHCOO (from the salt, CHCOOK). We use the Henderson-Hasselbalch equation to calculate the ph. [conjugate base] ph pk a log [acid] 5 1.00 10 ph log(1.8 10 ) log 1.50 10 ph 4.56 (c) This part is solved similarly to part (b). The number of moles of KOH in 10.0 ml is: 0.00 mol KOH 10.0 ml.00 10 mol 1000 ml KOH soln The changes in number of moles are summarized. CHCOOH(aq) KOH(aq) CHCOOK(aq) HO(l) Initial (mol):.50 10.00 10 0 Change (mol):.00 10.00 10.00 10 Final (mol): 0.50 10 0.00 10 At this stage, we have a buffer system made up of CHCOOH and CHCOO (from the salt, CHCOOK). We use the Henderson-Hasselbalch equation to calculate the ph. [conjugate base] ph pk a log [acid] 5.00 10 ph log(1.8 10 ) log 0.50 10 ph 5.4 (d) We have reached the equivalence point of the titration..50 10 mole of CHCOOH reacts with.50 10 mole KOH to produce.50 10 mole of CHCOOK. The only major species present in solution at the equivalence point is the salt, CHCOOK, which contains the conjugate base, CHCOO. Let's calculate the molarity of CHCOO. The volume of the solution is: (5.0 ml 1.5 ml 7.5 ml 0.075 L).
M.50 10 mol (CHCOO ) 0.0667 M 0.075 L We set up the hydrolysis of CHCOO, which is a weak base. CHCOO (aq) HO(l) CHCOOH(aq) OH (aq) Initial (M): 0.0667 0 0 Change (M): x x x Equilibrium (M): 0.0667 x x x K b [CHCOOH][OH ] [CHCOO ] 10 ( x)( x) x 5.6 10 0.0667 x 0.0667 x 6.1 10 6 M [OH ] poh 5.1 ph 8.79 (e) We have passed the equivalence point of the titration. The excess strong base, KOH, will determine the ph at this point. The moles of KOH in 15.0 ml are: 0.00 mol KOH 15.0 ml.00 10 mol 1000 ml KOH soln The changes in number of moles are summarized. CHCOOH(aq) KOH(aq) CHCOOK(aq) HO(l) Initial (mol):.50 10.00 10 0 Change (mol):.50 10.50 10.50 10 Final (mol): 0 0.50 10.50 10 Let's calculate the molarity of the KOH in solution. The volume of the solution is now 40.0 ml 0.0400 L. M 0.50 10 mol (KOH) 0.015 M 0.0400 L KOH is a strong base. The poh is: poh log(0.015) 1.90 ph 1.10 1.4 The reaction between NH and HCl is: NH(aq) HCl(aq) NH4Cl(aq) We see that 1 mole NH 1 mol HCl. Therefore, at every stage of titration, we can calculate the number of moles of base reacting with acid, and the ph of the solution is determined by the excess
base or acid left over. At the equivalence point, however, the neutralization is complete, and the ph of the solution will depend on the extent of the hydrolysis of the salt formed, which is NH4Cl. (a) No HCl has been added. This is a weak base calculation. NH(aq) HO(l) NH4 (aq) OH (aq) Initial (M): 0.00 0 0 Change (M): x x x Equilibrium (M): 0.00 x x x K b [NH 4 ][OH ] [NH ] 5 ( x)( x) x 1.8 10 0.00 x 0.00 x. 10 M [OH ] poh.64 ph 11.6 (b) The number of moles of NH originally present in 10.0 ml of solution is: 0.00 mol NH 10.0 ml.00 10 mol 1000 ml NH soln The number of moles of HCl in 10.0 ml is: 0.100 mol HCl 10.0 ml 1.00 10 mol 1000 ml HClsoln We work with moles at this point because when two solutions are mixed, the solution volume increases. As the solution volume increases, molarity will change, but the number of moles will remain the same. The changes in number of moles are summarized. NH(aq) HCl(aq) NH4Cl(aq) Initial (mol):.00 10 1.00 10 0 Change (mol): 1.00 10 1.00 10 1.00 10 Final (mol):.00 10 0 1.00 10 At this stage, we have a buffer system made up of NH and NH4 (from the salt, NH4Cl). We use the Henderson-Hasselbalch equation to calculate the ph. [conjugate base] ph pk a log [acid] 10.00 10 ph log(5.6 10 ) log 1.00 10 ph 9.55 (c) This part is solved similarly to part (b). The number of moles of HCl in 0.0 ml is:
0.100 mol HCl 0.0 ml.00 10 mol 1000 ml HClsoln The changes in number of moles are summarized. NH(aq) HCl(aq) NH4Cl(aq) Initial (mol):.00 10.00 10 0 Change (mol):.00 10.00 10.00 10 Final (mol): 1.00 10 0.00 10 At this stage, we have a buffer system made up of NH and NH4 (from the salt, NH4Cl). We use the Henderson-Hasselbalch equation to calculate the ph. [conjugate base] ph pk a log [acid] 10 1.00 10 ph log(5.6 10 ) log.00 10 ph 8.95 (d) We have reached the equivalence point of the titration..00 10 mole of NH reacts with.00 10 mole HCl to produce.00 10 mole of NH4Cl. The only major species present in solution at the equivalence point is the salt, NH4Cl, which contains the conjugate acid, NH4. Let's calculate the molarity of NH4. The volume of the solution is: (10.0 ml 0.0 ml 40.0 ml 0.0400 L). M.00 10 mol (NH 4 ) 0.0750 M 0.0400 L We set up the hydrolysis of NH4, which is a weak acid. NH4 (aq) HO(l) HO (aq) NH(aq) Initial (M): 0.0750 0 0 Change (M): x x x Equilibrium (M): 0.0750 x x x K a [H O ][NH ] [NH 4 ] 10 ( x)( x) x 5.6 10 0.0750 x 0.0750 x 6.5 10 6 M [HO ] ph 5.19 (e) We have passed the equivalence point of the titration. The excess strong acid, HCl, will determine the ph at this point. The moles of HCl in 40.0 ml are: 0.100 mol HCl 40.0 ml 4.00 10 mol 1000 ml HClsoln
The changes in number of moles are summarized. NH(aq) HCl(aq) NH4Cl(aq) Initial (mol):.00 10 4.00 10 0 Change (mol):.00 10.00 10.00 10 Final (mol): 0 1.00 10.00 10 Let's calculate the molarity of the HCl in solution. The volume of the solution is now 50.0 ml 0.0500 L. M 1.00 10 mol (HCl) 0.000 M 0.0500 L HCl is a strong acid. The ph is: ph log(0.000) 1.70 1.5 (a) HCOOH is a weak acid and NaOH is a strong base. Suitable indicators are cresol red and phenolphthalein. (b) (c) HCl is a strong acid and KOH is a strong base. Suitable indicators are all those listed with the exceptions of thymol blue, bromophenol blue, and methyl orange. HNO is a strong acid and CHNH is a weak base. Suitable indicators are bromophenol blue, methyl orange, methyl red, and chlorophenol blue. 1.6 CO in the air dissolves in the solution: The carbonic acid neutralizes the NaOH. CO HO HCO 1.7 The weak acid equilibrium is HIn(aq) H (aq) In (aq) We can write a Ka expression for this equilibrium. Rearranging, K a [H ][In ] [HIn] [HIn] [H ] = [In ] K a From the ph, we can calculate the H concentration. [H ] 10 ph 10 4 1.0 10 4 M [HIn] [In ] a 4 [H ] 1.0 10 K 6 1.0 10 100
Since the concentration of HIn is 100 times greater than the concentration of In, the color of the solution will be that of HIn, the nonionized formed. The color of the solution will be red. 1.8 According to Section 1.4 of the text, when [HIn] [In ] the indicator color is a mixture of the colors of HIn and In. In other words, the indicator color changes at this point. When [HIn] [In ] we can write: [In ] Ka [HIn] [H ] 1 [H ] Ka.0 10 6 ph 5.70 1.9 Refer to Table 1. of the text to solve this problem. (a) Ca(PO4) is insoluble. (b) Mn(OH) is insoluble. (c) AgClO is soluble. (d) KS is soluble. 1.0 Strategy: Although it is not necessary to memorize the solubilities of compounds, you should keep in mind the following useful rules: all ionic compounds containing alkali metal cations, the ammonium ion, and the nitrate, bicarbonate, and chlorate ions are soluble. For other compounds, refer to Table 1. of the text. Solution: (a) CaCO is insoluble. Most carbonate compounds are insoluble. (b) ZnSO4 is soluble. Most sulfate compounds are soluble. (c) Hg(NO) is soluble. All nitrate compounds are soluble. (d) HgSO4 is insoluble. Most sulfate compounds are soluble, but those containing Ag, Ca, Ba, Hg, and Pb are insoluble. (e) NH4ClO4 is soluble. All ammonium compounds are soluble. 1.1 (a) Both reactants are soluble ionic compounds. The other possible ion combinations, NaSO4 and Cu(NO), are also soluble. (b) Both reactants are soluble. Of the other two possible ion combinations, KCl is soluble, but BaSO4 is insoluble and will precipitate. Ba (aq) SO4 (aq) BaSO4(s) 1. First, you would accurately measure the electrical conductance of pure water. The conductance of a solution of the slightly soluble ionic compound X should be greater than that of pure water. The increased conductance would indicate that some of the compound X had dissolved. 1. The precipitation reaction is: Ag (aq) Br (aq) AgBr(s) In this problem, the relative amounts of NaBr and CaBr are not known. However, the total amount of Br in the mixture can be determined from the amount of AgBr produced. Let s find the number of moles of Br.
1 mol AgBr 1 mol Br 1.690 g AgBr 9.0149 10 mol Br 187.8 g AgBr 1 mol AgBr The amount of Br comes from both NaBr and CaBr. Let x number of moles NaBr. Then, the 9.0149 10 mol x number of moles of CaBr. The moles of CaBr are divided by, because 1 mol of CaBr produces moles of Br. The sum of the NaBr and CaBr masses must equal the mass of the mixture, 0.9157 g. We can write: mass NaBr mass CaBr 0.9157 g 10.89 g NaBr 9.0149 10 x 199.88 g CaBr mol NaBr mol CaBr x 0.9157 g 1 mol NaBr 1 mol CaBr.95x 0.014751 x 5.000 10 moles NaBr Converting moles to grams: 10.89 g NaBr mass NaBr (5.000 10 mol NaBr) 1 mol NaBr 0.51448 g NaBr The percentage by mass of NaBr in the mixture is: % NaBr 0.51448 g 100% 56.18% NaBr 0.9157 g 1.4 (a) The solubility equilibrium is given by the equation AgI(s) Ag (aq) I (aq) The expression for Ksp is given by Ksp [Ag ][I ] The value of Ksp can be found in Table 1. of the text. If the equilibrium concentration of silver ion is the value given, the concentration of iodide ion must be 17 Ksp 8. 10 [I ] 9.1 10 9 [Ag ] 9.1 10 9 M (b) The value of Ksp for aluminum hydroxide can be found in Table 1. of the text. The equilibrium expressions are: Al(OH)(s) Al (aq) OH (aq) Ksp [Al ][OH ] Using the given value of the hydroxide ion concentration, the equilibrium concentration of aluminum ion is:
Ksp 1.8 10 8 9 [Al ] 7.4 10 [OH ] (.9 10 ) What is the ph of this solution? Will the aluminum concentration change if the ph is altered? 1.5 Strategy: In each part, we can calculate the number of moles of compound dissolved in one liter of solution (the molar solubility). Then, from the molar solubility, s, we can determine Ksp. Solution: M (a) 7. 10 g SrF 1 L soln 1 mol SrF 15.6 g SrF 5.8 10 4 mol L 1 = s Consider the dissociation of SrF in water. Let s be the molar solubility of SrF. SrF(s) Sr (aq) F (aq) Initial (M): 0 0 Change (M): s s s Equilibrium (M): s s Ksp [Sr ][F ] (s)(s) 4s The molar solubility (s) was calculated above. Substitute into the equilibrium constant expression to solve for Ksp. Ksp [Sr ][F ] 4s 4(5.8 10 4 ) 7.8 10 10 (b) 6.7 10 g Ag PO 4 1 L soln 1 mol Ag PO 4 418.7 g Ag PO 4 1.6 10 5 mol L 1 = s (b) is solved in a similar manner to (a) The equilibrium equation is: AgPO4(s) Ag (aq) PO4 (aq) Initial (M): 0 0 Change (M): s s s Equilibrium (M): s s 1.6 For MnCO dissolving, we write Ksp [Ag ] [PO4 ] (s) (s) 7s 4 7(1.6 10 5 ) 4 1.8 10 18 MnCO(s) Mn (aq) CO (aq) For every mole of MnCO that dissolves, one mole of Mn will be produced and one mole of CO will be produced. If the molar solubility of MnCO is s mol L 1, then the concentrations of Mn and CO are: [Mn ] [CO ] s 4. 10 6 M Ksp [Mn ][CO ] s (4. 10 6 ) 1.8 10 11
For MX dissolving, we write MX(s) M x (aq) X x (aq) For every mole of MX that dissolves, one mole of M x will be produced and one mole of X x will be produced. (Note: we do not need to know the charge x). To find K sp we must convert the solubility of MX from g L 1 to mol L 1 : molar solubility = s = 4.6 10 g L 1 The concentrations of M x and X x are: [M x ] [X x ] s 1.4 10 5 M 46 g mol 1 1.4 10 5 mol L 1 Ksp [M x ][ X x ] s (1.4 10 5 ) 1.80 10 10 1.7 The charges of the M and X ions are and, respectively (are other values possible?). We first calculate the number of moles of MX that dissolve in 1.0 L of water. We carry an additional significant figure throughout this calculation to minimize rounding errors. 17 1 mol 19 Moles MX (.6 10 g) 1.5 10 mol 88 g The molar solubility, s, of the compound is therefore 1. 10 19 M. At equilibrium the concentration of M must be s and that of X must be s. (See Table 1.4 of the text.) Ksp [M ] [X ] [s] [s] 108s 5 Since these are equilibrium concentrations, the value of Ksp can be found by simple substitution Ksp 108s 5 108(1.5 10 19 ) 5. 10 9 1.8 Strategy: We can look up the Ksp value of CaF in Table 1. of the text. Then, setting up the dissociation equilibrium of CaF in water, we can solve for the molar solubility, s. Solution: Consider the dissociation of CaF in water. CaF(s) Ca (aq) F (aq) Initial (M): 0 0 Change (M): s s s Equilibrium (M): s s Recall, that the concentration of a pure solid does not enter into an equilibrium constant expression. Therefore, the concentration of CaF is not important. Substitute the value of Ksp and the concentrations of Ca and F in terms of s into the solubility product expression to solve for s, the molar solubility. Ksp [Ca ][F ]
4.0 10 11 (s)(s) 4.0 10 11 4s s molar solubility. 10 4 mol L 1 The molar solubility indicates that. 10 4 mol of CaF will dissolve in 1 L of an aqueous solution. 1.9 Let s be the molar solubility of Zn(OH). The equilibrium concentrations of the ions are then [Zn ] s and [OH ] s Ksp [Zn ][OH ] (s)(s) 4s 1.8 10 14 s 1 14 5 1.8 10 1.7 10 4 [OH ] s.4 10 5 M and poh 4.47 ph 14.00 4.47 9.5 If the Ksp of Zn(OH) were smaller by many more powers of ten, would s still be the hydroxide ion concentration in the solution? 1.40 First we can calculate the OH concentration from the ph. The equilibrium equation is: poh 14.00 ph poh 14.00 9.68 4. [OH ] 10 poh 10 4. 4.8 10 5 M MOH(s) M (aq) OH (aq) From the balanced equation we know that [M ] [OH ] Ksp [M ][OH ] (4.8 10 5 ). 10 9 1.41 According to the solubility rules, the only precipitate that might form is BaCO. Ba (aq) CO (aq) BaCO(s) The number of moles of Ba present in the original 0.0 ml of Ba(NO) solution is 0.10 mol Ba 0.0 ml.0 10 mol Ba 1000 ml soln The total volume after combining the two solutions is 70.0 ml. The concentration of Ba in 70 ml is
.0 10 mol Ba [Ba ].9 10 70.0 10 L M The number of moles of CO present in the original 50.0 ml NaCO solution is 0.10 mol CO 50.0 ml 5.0 10 mol CO 1000 ml soln The concentration of CO in the 70.0 ml of combined solution is 5.0 10 mol CO [CO ] 7.1 10 M 70.0 10 L Now we must compare Q and Ksp. From Table 1. of the text, the Ksp for BaCO is 8.1 10 9. As for Q, Q [Ba ]0[CO ]0 (.9 10 )(7.1 10 ).1 10 Since (.1 10 ) > (8.1 10 9 ), Q > Ksp. Therefore, BaCO will precipitate. 1.4 The net ionic equation is: Sr (aq) F (aq) SrF(s) Let s find the limiting reagent in the precipitation reaction. 0.060 mol Moles F 75 ml 0.0045 mol 1000 ml soln 0.15 mol Moles Sr 5 ml 0.008 mol 1000 ml soln From the stoichiometry of the balanced equation, twice as many moles of F are required to react with Sr. This would require 0.0076 mol of F, but we only have 0.0045 mol. Thus, F is the limiting reagent. Let s assume that the above reaction goes to completion. Then, we will consider the equilibrium that is established when SrF partially dissociates into ions. Sr (aq) F (aq) SrF(s) Initial (mol): 0.008 0.0045 0 Change (mol): 0.005 0.0045 0.005 Final (mol): 0.00155 0 0.005 Now, let s establish the equilibrium reaction. The total volume of the solution is 100 ml 0.100 L. Divide the above moles by 0.100 L to convert to molar concentration. SrF(s) Sr (aq) F (aq) Initial (M): 0.05 0.0155 0 Change (M): s s s Equilibrium (M): 0.05 s 0.0155 s s
Write the solubility product expression, then solve for s. Ksp [Sr ][F ].0 10 10 (0.0155 s)(s) (0.0155)(s) s 5.7 10 5 M [F ] s 1.1 10 4 M [Sr ] 0.0155 s 0.016 M Both sodium ions and nitrate ions are spectator ions and therefore do not enter into the precipitation reaction. [NO ] (0.008) mol 0.10 L 0.076 M [Na ] 0.0045 mol 0.10 L 0.045 M 1.4 (a) The solubility product expressions for both substances have exactly the same mathematical form and are therefore directly comparable. The substance having the smaller Ksp (AgI) will precipitate first. (Why?) (b) When CuI just begins to precipitate the solubility product expression will just equal Ksp (saturated solution). The concentration of Cu at this point is 0.010 M (given in the problem), so the concentration of iodide ion must be: Ksp [Cu ][I ] (0.010)[I ] 5.1 10 1 1 5.1 10 10 [I ] 5.1 10 M 0.010 Using this value of [I ], we find the silver ion concentration 17 Ksp 8. 10 [Ag ] 1.6 10 10 [I ] 5.1 10 7 M (c) The percent of silver ion remaining in solution is: 7 1.6 10 M % Ag ( aq) 100% 0.0016% or 1.6 10 % 0.010 M Is this an effective way to separate silver from copper? 1.44 For Fe(OH), Ksp 1.1 10 6. When [Fe ] 0.010 M, the [OH ] value is: Ksp [Fe ][OH ] [OH ] Ksp [Fe ] 1
1 6 1 1.1 10 [OH ] 4.8 10 0.010 This [OH ] corresponds to a ph of.68. In other words, Fe(OH) will begin to precipitate from this solution at ph of.68. For Zn(OH), Ksp 1.8 10 14. When [Zn ] 0.010 M, the [OH ] value is: M [OH ] Ksp [Zn ] 1 1 14 6 1.8 10 [OH ] 1. 10 0.010 This corresponds to a ph of 8.11. In other words Zn(OH) will begin to precipitate from the solution at ph 8.11. These results show that Fe(OH) will precipitate when the ph just exceeds.68 and that Zn(OH) will precipitate when the ph just exceeds 8.11. Therefore, to selectively remove iron as Fe(OH), the ph must be greater than.68 but less than 8.11. 1.45 When there is a common ion already present in the solution, less of the solid will dissolve. In the reaction that corresponds to the Ksp of the solid, the common ion is a product. The common ion s presence pushes the equilibrium toward the reactants (the solid) consistent with Le Châtelier s principle. For example, CaCO will be less soluble in a solution of Na CO, than in pure water. 1.46 (a) This is not true. We know that Ksp [Ag ][Cl ] = (solubility) (b) This is a reasonable assumption. (c) This is not true. Silver is a common ion and will affect the solubility of AgCl. (d) The Ksp of AgCl is very small (1.6 10 10 ) making this a reasonable assumption. (e) This is not a reasonable assumption. The K sp of AgCl is very small. Because of the common ion effect the Ag + concentration in a solution of AgCl(aq) in AgNO will be significantly different than that in pure water. 1.47 First let s be the molar solubility of CaCO in this solution. CaCO(s) Ca (aq) CO (aq) Initial (M): 0.050 0 Change (M): s s s Equilibrium (M): (0.050 s) s Ksp [Ca ][CO ] (0.050 s)s 8.7 10 9 We can assume 0.050 s 0.050, then 9 8.7 10 7 s 1.7 10 M 0.050 The mass of CaCO can then be found. M 7 1.7 10 mol 100.1 g CaCO (.0 10 ml) 1000 ml soln 1 mol 6 5.1 10 g CaCO
1.48 Strategy: In parts (b) and (c), this is a common-ion problem. In part (b), the common ion is Br, which is supplied by both PbBr and KBr. Remember that the presence of a common ion will affect only the solubility of PbBr, but not the Ksp value because it is an equilibrium constant. In part (c), the common ion is Pb, which is supplied by both PbBr and Pb(NO). Solution: (a) Set up a table to find the equilibrium concentrations in pure water. PbBr(s) Pb (aq) Br (aq) Initial (M) 0 0 Change (M) s s s Equilibrium (M) s s Ksp [Pb ][Br ] 8.9 10 6 (s)(s) s molar solubility 0.01 M (b) Set up a table to find the equilibrium concentrations in 0.0 M KBr. KBr is a soluble salt that ionizes completely giving an initial concentration of Br 0.0 M. PbBr(s) Pb (aq) Br (aq) Initial (M) 0 0.0 Change (M) s s s Equilibrium (M) s 0.0 s Ksp [Pb ][Br ] 8.9 10 6 (s)(0.0 s) 8.9 10 6 (s)(0.0) s molar solubility. 10 4 M Thus, the molar solubility of PbBr is reduced from 0.01 M to. 10 4 M as a result of the common ion (Br ) effect. (c) Set up a table to find the equilibrium concentrations in 0.0 M Pb(NO). Pb(NO) is a soluble salt that dissociates completely giving an initial concentration of [Pb ] 0.0 M. PbBr(s) Pb (aq) Br (aq) Initial (M): 0.0 0 Change (M): s s s Equilibrium (M): 0.0 s s Ksp [Pb ][Br ] 8.9 10 6 (0.0 s)(s) 8.9 10 6 (0.0)(s) s molar solubility. 10 M
Thus, the molar solubility of PbBr is reduced from 0.01 M to. 10 M as a result of the common ion (Pb ) effect. Check: You should also be able to predict the decrease in solubility due to a common-ion using Le Châtelier's principle. Adding Br or Pb ions shifts the system to the left, thus decreasing the solubility of PbBr. 1.49 We first calculate the concentration of chloride ion in the solution. 10.0 g CaCl 1 mol CaCl mol Cl [Cl ] 0.180 M 1 L soln 111.0 g CaCl 1 mol CaCl AgCl(s) Ag (aq) Cl (aq) Initial (M): 0.000 0.180 Change (M): s s s Equilibrium (M): s (0.180 s) If we assume that (0.180 s) 0.180, then Ksp [Ag ][Cl ] 1.6 10 10 sp 10 1.6 10 10 K [Ag ] 8.9 10 M [Cl ] 0.180 s The molar solubility of AgCl is 8.9 10 10 M. 1.50 (a) The equilibrium reaction is: BaSO4(s) Ba (aq) SO4 (aq) Initial (M): 0 0 Change (M): s s s Equilibrium (M): s s Ksp [Ba ][SO4 ] 1.1 10 10 s s 1.0 10 5 M The molar solubility of BaSO4 in pure water is 1.0 10 5 mol L 1. (b) The initial concentration of SO4 is 1.0 M. BaSO4(s) Ba (aq) SO4 (aq) Initial (M): 0 1.0 Change (M): s s s Equilibrium (M): s 1.0 s Ksp [Ba ][SO4 ] 1.1 10 10 (s)(1.0 s) (s)(1.0)
s 1.1 10 10 M Due to the common ion effect, the molar solubility of BaSO4 decreases to 1.1 10 10 mol L 1 in 1.0 M SO4 (aq) compared to 1.0 10 5 mol L 1 in pure water. 1.51 When the anion of a salt is a base, the salt will be more soluble in acidic solution because the hydrogen ion decreases the concentration of the anion (Le Châtelier s principle): B (aq) H (aq) HB(aq) (a) (b) (c) (d) BaSO4 will be slightly more soluble because SO4 is a base (although a weak one). The solubility of PbCl in acid is unchanged over the solubility in pure water because HCl is a strong acid, and therefore Cl is a negligibly weak base. Fe(OH) will be more soluble in acid because OH is a base. CaCO will be more soluble in acidic solution because the CO ions react with H ions to form CO and HO. The CO escapes from the solution, shifting the equilibrium. Although it is not important in this case, the carbonate ion is also a base. 1.5 (b) SO4 (aq) is a weak base (c) (d) (e) OH (aq) is a strong base CO4 (aq) is a weak base PO4 (aq) is a weak base. The solubilities of the above will increase in acidic solution. Only (a), which contains an extremely weak base (I is the conjugate base of the strong acid HI) is unaffected by the acid solution. 1.5 In water: Mg(OH) Mg OH s s Ksp 4s 1. 10 11 s 1.4 10 4 M In a buffer at ph 9.0 [H ] 1.0 10 9 [OH ] 1.0 10 5 1. 10 11 (s)(1.0 10 5 ) s 0.1 M 1.54 From Table 1., the value of Ksp for iron(ii) is 1.6 10 14. (a) At ph 8.00, poh 14.00 8.00 6.00, and [OH ] 1.0 10 6 M
sp 14 1.6 10 6 K [Fe ] 0.016 M [OH ] (1.0 10 ) The molar solubility of iron(ii) hydroxide at ph 8.00 is 0.016 M (b) At ph 10.00, poh 14.00 10.00 4.00, and [OH ] 1.0 10 4 M 14 Ksp 1.6 10 6 4 [Fe ] 1.6 10 [OH ] (1.0 10 ) M The molar solubility of iron(ii) hydroxide at ph 10.00 is 1.6 10 6 M. 1.55 The solubility product expression for magnesium hydroxide is Ksp [Mg ][OH ] 1. 10 11 We find the hydroxide ion concentration when [Mg ] is 1.0 10 10 M 1 11 1. 10 [OH ] 0.5 M 10 1.0 10 Therefore the concentration of OH must be slightly greater than 0.5 M. 1.56 We first determine the effect of the added ammonia. Let's calculate the concentration of NH. This is a dilution problem. MiVi MfVf (0.60 M)(.00 ml) Mf(100 ml) Mf 0.001 M NH Ammonia is a weak base (Kb 1.8 10 5 ). NH HO NH4 OH Initial (M): 0.001 0 0 Change (M): x x x Equil. (M): 0.001 x x x K b [NH 4 ][OH ] [NH ] 5 1.8 10 x (0.001 x) Solving the resulting quadratic equation gives x 0.00014, or [OH ] 0.00014 M This is a solution of iron(ii) sulfate, which contains Fe ions. These Fe ions could combine with OH to precipitate Fe(OH). Therefore, we must use Ksp for iron(ii) hydroxide. We compute the value of Qc for this solution. Fe(OH)(s) Fe (aq) OH (aq)
Q [Fe ]0[OH ]0 (1.0 10 )(0.00014).0 10 11 Note that when adding.00 ml of NH to 1.0 L of FeSO4, the concentration of FeSO4 will decrease slightly. However, rounding off to significant figures, the concentration of 1.0 10 M does not change. Q is larger than Ksp [Fe(OH)] 1.6 10 14. The concentrations of the ions in solution are greater than the equilibrium concentrations; the solution is saturated. The system will shift left to reestablish equilibrium; therefore, a precipitate of Fe(OH) will form. 1.57 First find the molarity of the copper(ii) ion 1 mol Moles CuSO4.50 g 0.0157 mol 159.6 g 0.0157 mol [Cu ] 0.0174 M 0.90 L As in Example 1.16 of the text, the position of equilibrium will be far to the right. We assume essentially all the copper ion is complexed with NH. The NH consumed is 4 0.0174 M 0.0696 M. The uncombined NH remaining is (0.0 0.0696) M, or 0. M. The equilibrium concentrations of Cu(NH)4 and NH are therefore 0.0174 M and 0. M, respectively. We find [Cu ] from the formation constant expression. 1.58 The reaction K [Cu(NH ) 4 ] 1 0.0174 f 5.0 10 4 4 [Cu ][NH ] [Cu ](0.) [Cu ] 1. 10 1 M Al(OH)(s) OH (aq) Al(OH)4 (aq) is the sum of the two known reactions Al(OH)(s) Al (aq) OH (aq) Ksp 1.8 10 Al (aq) 4OH (aq) Al(OH)4 (aq) Kf.0 10 The equilibrium constant is [Al(OH) 4 ] K KspK f (1.8 10 )(.0 10 ).6 [OH ] When ph 14.00, [OH ] 1.0 M, therefore [Al(OH)4 ] K[OH ] (.6)(1 M).6 M This represents the maximum possible concentration of the complex ion at ph 14.00. Since this is much larger than the initial 0.010 M, the complex ion will be the predominant species. 1.59 Silver iodide is only slightly soluble. It dissociates to form a small amount of Ag and I ions. The Ag ions then complex with NH in solution to form the complex ion Ag(NH). The balanced equations are:
AgI(s) Ag (aq) I (aq) Ag (aq) NH(aq) Ag(NH) (aq) Ksp [Ag ][I ] 8. 10 17 K [Ag(NH ) ] f [Ag ][NH ] 1.5 10 Overall: AgI(s) NH(aq) Ag(NH) (aq) I (aq) K Ksp Kf 1. 10 9 If s is the molar solubility of AgI then, AgI(s) NH(aq) Ag(NH) (aq) I (aq) Initial (M): 1.0 0.0 0.0 Change (M): s s s s Equilibrium (M): (1.0 s) s s Because Kf is large, we can assume all of the silver ions exist as Ag(NH). Thus, [Ag(NH) ] [I ] s We can write the equilibrium constant expression for the above reaction, then solve for s. 7 K 9 ( s)( s) ( s)( s) 1. 10 (1.0 s) (1.0) s.5 10 5 M At equilibrium,.5 10 5 moles of AgI dissolves in 1 L of 1.0 M NH solution. 1.60 The balanced equations are: Ag (aq) NH(aq) Ag(NH) (aq) Zn (aq) 4NH(aq) Zn(NH)4 (aq) Zinc hydroxide forms a complex ion with excess OH and silver hydroxide does not; therefore, zinc hydroxide is soluble in 6 M NaOH. 1.61 (a) The equations are as follows: CuI(s) Cu (aq) I (aq) Cu (aq) 4NH(aq) [Cu(NH)4] (aq) The ammonia combines with the Cu ions formed in the first step to form the complex ion [Cu(NH)4], effectively removing the Cu ions, causing the first equilibrium to shift to the right (resulting in more CuI dissolving). (b) Similar to part (a): AgBr(s) Ag (aq) Br (aq) Ag (aq) CN (aq) [Ag(CN)] (aq) (c) Similar to parts (a) and (b).
HgCl(s) Hg (aq) Cl (aq) 1.6 Silver chloride will dissolve in aqueous ammonia because of the formation of a complex ion. Lead chloride will not dissolve; it does not form an ammonia complex. 1.6 Since some PbCl precipitates, the solution is saturated. From Table 1., the value of Ksp for lead(ii) chloride is.4 10 4. The equilibrium is: PbCl(aq) Pb (aq) Cl (aq) We can write the solubility product expression for the equilibrium. Ksp [Pb ][Cl ] Ksp and [Cl ] are known. Solving for the Pb concentration, sp 4.4 10 K [Pb ] 0.011 M [Cl ] (0.15) 1.64 Ammonium chloride is the salt of a weak base (ammonia). It will react with strong aqueous hydroxide to form ammonia (Le Châtelier s principle). NH4Cl(s) OH (aq) NH(g) HO(l) Cl (aq) The human nose is an excellent ammonia detector. Nothing happens between KCl and strong aqueous NaOH. 1.65 Chloride ion will precipitate Ag but not Cu. So, dissolve some solid in HO and add HCl. If a precipitate forms, the salt was AgNO. A flame test will also work. Cu gives a green flame test. 1.66 According to the Henderson-Hasselbalch equation: [conjugate base] ph pk a log [acid] If: [conjugate base] 10, then: [acid] ph pka 1 If: [conjugate base] 0.1, then: [acid] ph pka 1 Therefore, the range of the ratio is: [conjugate base] 0.1 10 [acid] 1.67 We can use the Henderson-Hasselbalch equation to solve for the ph when the indicator is 90% acid and 10% conjugate base and when the indicator is 10% acid and 90% conjugate base.
[conjugate base] ph pk a log [acid] Solving for the ph with 90% of the indicator in the HIn form: [10] ph.46 log.46 0.95.51 [90] Next, solving for the ph with 90% of the indicator in the In form: [90] ph.46 log.46 0.95 4.41 [10] Thus the ph range varies from.51 to 4.41 as the [HIn] varies from 90% to 10%. 1.68 Referring to Figure 1.6, at the half-equivalence point, [weak acid] [conjugate base]. Using the Henderson-Hasselbalch equation: [conjugate base] ph pk a log [acid] so, ph pka 1.69 First, calculate the ph of the.00 M weak acid (HNO) solution before any NaOH is added. HNO(aq) H (aq) NO (aq) Initial (M):.00 0 0 Change (M): x x x Equilibrium (M):.00 x x x K a [H ][NO ] [HNO ] 4 x x 4.5 10.00 x.00 x [H ] 0.00 M ph log(0.00) 1.5 Since the ph after the addition is 1.5 ph units greater, the new ph 1.5 1.50.0. From this new ph, we can calculate the [H ] in solution. [H ] 10 ph 10.0 9.55 10 4 M When the NaOH is added, we dilute our original.00 M HNO solution to: MiVi MfVf (.00 M)(400 ml) Mf(600 ml) Mf 1. M Since we have not reached the equivalence point, we have a buffer solution. The reaction between HNO and NaOH is:
HNO(aq) NaOH(aq) NaNO(aq) HO(l) Since the mole ratio between HNO and NaOH is 1:1, the decrease in [HNO] is the same as the decrease in [NaOH]. We can calculate the decrease in [HNO] by setting up the weak acid equilibrium. From the ph of the solution, we know that the [H ] at equilibrium is 9.55 10 4 M. HNO(aq) H (aq) NO (aq) Initial (M): 1. 0 0 Change (M): x x Equilibrium (M): 1. x 9.55 10 4 x We can calculate x from the equilibrium constant expression. K a [H ][NO ] [HNO ] 4 4 (9.55 10 )( x) 4.5 10 1. x x 0.46 M Thus, x is the decrease in [HNO] which equals the concentration of added OH. However, this is the concentration of NaOH after it has been diluted to 600 ml. We need to correct for the dilution from 00 ml to 600 ml to calculate the concentration of the original NaOH solution. MiVi MfVf Mi(00 ml) (0.46 M)(600 ml) [NaOH] Mi 1.8 M 1.70 The Ka of butyric acid is obtained by taking the antilog of 4.7 (10 4.7 ) which is 10 5. The value of Kb is: a 14 w 1.0 10 K K b 5 10 K 5 10 1.71 The resulting solution is not a buffer system. There is excess NaOH and the neutralization is well past the equivalence point. 10 0.167 mol Moles NaOH 0.500 L 0.085 mol 1L 0.100 mol Moles HCOOH 0.500 L 0.0500 mol 1L HCOOH(aq) NaOH(aq) HCOONa(aq) HO(l) Initial (mol): 0.0500 0.085 0 Change (mol): 0.0500 0.0500 0.0500 Final (mol): 0 0.05 0.0500 The volume of the resulting solution is 1.00 L (500 ml 500 ml 1000 ml).
[OH ] 0.05 mol 1.00 L 0.05 M (0.05 0.0500) mol [Na ] 0.085 M 1.00 L 14 w 1.0 10 [H ] K.0 10 [OH ] 0.05 0.0500 mol [HCOO ] 0.0500 M 1.00 L 1 M HCOO (aq) HO(l) HCOOH(aq) OH (aq) Initial (M): 0.0500 0 0.05 Change (M): x x x Equilibrium (M): 0.0500 x x 0.05 x K b [HCOOH][OH ] [HCOO ] 11 ( x)(0.05 x) ( x)(0.05) 5.9 10 (0.0500 x) (0.0500) x [HCOOH] 8.8 10 11 M 1.7 Most likely the increase in solubility is due to complex ion formation: Cd(OH)(s) OH Cd(OH)4 (aq) This is a Lewis acid-base reaction. 1.7 The number of moles of Ba(OH) present in the original 50.0 ml of solution is: 1.00 mol Ba(OH) 50.0 ml 0.0500 mol Ba(OH) 1000 ml soln The number of moles of HSO4 present in the original 86.4 ml of solution, assuming complete dissociation, is: 0.494 mol HSO4 86.4 ml 0.047 mol HSO4 1000 ml soln The reaction is: Ba(OH)(aq) HSO4(aq) BaSO4(s) HO(l) Initial (mol): 0.0500 0.047 0 Change (mol): 0.047 0.047 0.047 Final (mol): 0.007 0 0.047 Thus the mass of BaSO4 formed is:.4 g BaSO4 0.047 mol BaSO4 9.97 g BaSO 4 1 mol BaSO 4
The ph can be calculated from the excess OH in solution. First, calculate the molar concentration of OH. The total volume of solution is 16.4 ml 0.164 L. mol OH 0.007 mol Ba(OH) 1 mol Ba(OH) [OH ] 0.11 M 0.164 L poh log(0.11) 0.96 ph 14.00 poh 14.00 0.96 1.04 1.74 A solubility equilibrium is an equilibrium between a solid (reactant) and its components (products: ions, neutral molecules, etc.) in solution. Only (d) represents a solubility equilibrium. Consider part (b). Can you write the equilibrium constant for this reaction in terms of Ksp for calcium phosphate? 1.75 First, we calculate the molar solubility of CaCO. CaCO(s) Ca (aq) CO (aq) Initial (M): 0 0 Change (M): s s s Equil. (M): s s The moles of CaCO in the kettle are: Ksp [Ca ][CO ] s 8.7 10 9 s 9. 10 5 M 9. 10 5 mol L 1 1 mol CaCO 116 g 1.16 mol CaCO 100.1 g CaCO The volume of distilled water needed to dissolve 1.16 moles of CaCO is: 1L 4 1.16 mol CaCO 1. 10 L 5 9. 10 mol CaCO The number of times the kettle would have to be filled is: 4 1 filling (1. 10 L).0 L 6.0 10 fillings Note that the very important assumption is made that each time the kettle is filled, the calcium carbonate is allowed to reach equilibrium before the kettle is emptied. 1.76 Since equal volumes of the two solutions were used, the initial molar concentrations will be halved. 0.1 M [Ag ] 0.060 M (0.14 M ) [Cl ] 0.14 M
Let s assume that the Ag ions and Cl ions react completely to form AgCl(s). Then, we will reestablish the equilibrium between AgCl, Ag, and Cl. Ag (aq) Cl (aq) AgCl(s) Initial (M): 0.060 0.14 0 Change (M): 0.060 0.060 0.060 Final (M): 0 0.080 0.060 Now, setting up the equilibrium, AgCl(s) Ag (aq) Cl (aq) Initial (M): 0.060 0 0.080 Change (M): s s s Equilibrium (M): 0.060 s s 0.080 s Set up the Ksp expression to solve for s. Ksp [Ag ][Cl ] 1.6 10 10 (s)(0.080 s) s.0 10 9 M [Ag ] s.0 10 9 M [Cl ] 0.080 M s 0.080 M [Zn ] 0.14 M 0.070 M 1.77 First we find the molar solubility and then convert moles to grams. The solubility equilibrium for silver carbonate is: AgCO(s) Ag (aq) CO (aq) Initial (M): 0 0 Change (M): s s s Equilibrium (M): s s Ksp [Ag ] [CO ] (s) (s) 4s 8.1 10 1 s 1 1 4 8.1 10 1. 10 4 M Converting from mol L 1 to g L 1 : 1. 10 4 mol 1 L soln 75.8 g 1 mol 0.06 g L 1 0.1 M [NO ] 0.060 M 1.78 For Mg(OH), Ksp 1. 10 11. When [Mg ] 0.010 M, the [OH ] value is