15.116 First we must calculate the molarity of the trifluoromethane sulfonic acid. (Molar mass 150.1 g/mol) 1 mol 150.1 g 0.0164 M 0.250 L 0.616 g Molarity Since trifluoromethane sulfonic acid is a strong acid and is 100% ionized, the [H ] is 0.0165 M. pH log(0.0164) 1.79 15.126 0.100 M Na2CO3 0.200 M Na 0.100 M CO32 First stage: Initial (M): Change (M): Equilibrium (M): CO32 (aq) H2O(l ) 0.100 x 0.100 x K1 K1 HCO3 (aq) OH(aq) 0 0 x x x x Kw 1.0 1014 2.1 104 K2 4.8 1011 [HCO3 ][OH ] [CO32 ] 2.1 104 x2 x2 0.100 x 0.100 x 4.6 103 M [HCO3 ] [OH] Second stage: HCO3 (aq) H2O(l) Initial (M): Change (M): Equilibrium (M): H2CO3(aq) OH(aq) 4.6 103 y 0 y 4.6 103 y (4.6 103) y y (4.6 103) y K2 [H2CO3 ][OH ] [HCO3 ] 2.4 108 y[(4.6 103 ) y] (4.6 103 ) y ( y)(4.6 103 ) (4.6 103 ) y 2.4 108 M At equilibrium: [Na] 0.200 M [CO23 ] = 0.100 M 4.6 103 M = 0.095 M [HCO 3 ] (4.6 103) M (2.4 108) M 4.6 103 M [H2CO3] 2.4 108 M [OH] (4.6 103) M (2.4 108] M 4.6 103 M CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA [H ] 1.0 1014 4.6 10 3 543 2.2 1012 M 15.129 When the pH is 10.00, the pOH is 4.00 and the concentration of hydroxide ion is 1.0 104 M. The concentration of HCN must be the same. (Why?) If the concentration of NaCN is x, the table looks like: Initial (M): Change (M): Equilibrium (M): Kb CN(aq) H2O(l) x 1.0 104 HCN(aq) 0 1.0 104 (x 1.0 104) (1.0 104) OH(aq) 0 1.0 104 (1.0 104) [HCN][OH ] [CN ] 2.0 105 (1.0 104 )2 ( x 1.0 104 ) x 6.0 104 M [CN]0 Amount of NaCN 250 mL 16.14 6.0 104 mol NaCN 49.01 g NaCN 7.4 103 g NaCN 1000 mL 1 mol NaCN Step 1: Write the equilibrium that occurs between H2 PO4 and HPO24 . Set up a table relating the initial concentrations, the change in concentration to reach equilibrium, and the equilibrium concentrations. Initial (M): Change (M): Equilibrium (M): H2 PO4 (aq) 0.15 x 0.15 x H(aq) HPO24 (aq) 0 0.10 x x x 0.10 x Step 2: Write the ionization constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant (Ka), solve for x. Ka [H ][HPO24 ] [H2 PO4 ] You can look up the Ka value for dihydrogen phosphate in Table 15.5 of your text. 6.2 108 ( x)(0.10 x) (0.15 x) 6.2 108 ( x)(0.10) (0.15) x [H] 9.3 108 M Step 3: Having solved for the [H], calculate the pH of the solution. 544 CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA pH log[H] log(9.3 108) 7.03 16.15 Using the HendersonHasselbalch equation: pH pKa log [CH3COO ] [CH3COOH] 4.50 4.74 log [CH3COO ] [CH3COOH] Thus, [CH3COO ] 0.58 [CH3COOH] 16.18 As calculated in Problem 16.12, the pH of this buffer system is equal to pKa. pH pKa log(1.8 105) 4.74 (a) The added NaOH will react completely with the acid component of the buffer, CH 3COOH. NaOH ionizes completely; therefore, 0.080 mol of OH are added to the buffer. Step 1: The neutralization reaction is: Initial (mol): Change (mol): Final (mol): CH3COOH(aq) OH(aq) CH3COO(aq) H2O(l) 1.00 0.080 1.00 0.080 0.080 0.080 0.92 0 1.08 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. CH3COOH(aq) Initial (M): 0.92 Change (M): x Equilibrium (M): 0.92 x H(aq) CH3COO(aq) 0 1.08 x x x 1.08 x Write the Ka expression, then solve for x. Ka [H ][CH3COO ] [CH3COOH] 1.8 105 ( x)(1.08 x) x(1.08) (0.92 x) 0.92 x [H] 1.5 105 M Step 3: Having solved for the [H], calculate the pH of the solution. pH log[H] log(1.5 105) 4.82 The pH of the buffer increased from 4.74 to 4.82 upon addition of 0.080 mol of strong base. (b) The added acid will react completely with the base component of the buffer, CH 3COO. HCl ionizes completely; therefore, 0.12 mol of H ion are added to the buffer CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA 545 Step 1: The neutralization reaction is: Initial (mol): Change (mol): Final (mol): CH3COO(aq) H(aq) CH3COOH(aq) 1.00 0.12 1.00 0.12 0.12 0.12 0.88 0 1.12 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. H(aq) CH3COO(aq) 0 0.88 x x x 0.88 x CH3COOH(aq) Initial (M): 1.12 Change (M): x Equilibrium (M): 1.12 x Write the Ka expression, then solve for x. Ka [H ][CH3COO ] [CH3COOH] 1.8 105 ( x)(0.88 x) x(0.88) (1.12 x) 1.12 x [H] 2.3 105 M 546 CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA Step 3: Having solved for the [H], calculate the pH of the solution. pH log[H] log(2.3 105) 4.64 The pH of the buffer decreased from 4.74 to 4.64 upon addition of 0.12 mol of strong acid. 16.29 The neutralization reaction is: H2SO4(aq) 2NaOH(aq) Na2SO4(aq) 2H2O(l) Since one mole of sulfuric acid combines with two moles of sodium hydroxide, we write: mol NaOH 12.5 mL H 2SO4 concentration of NaOH 16.53 (a) 0.500 mol H2SO4 2 mol NaOH 0.0125 mol NaOH 1000 mL soln 1 mol H 2SO4 0.0125 mol NaOH 50.0 103 L soln 0.25 M 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 16.2 of the text. If the equilibrium concentration of silver ion is the value given, the concentration of iodide ion must be [I ] Ksp [Ag ] (b) 8.3 1017 9.1 10 9 9.1 109 M The value of Ksp for aluminum hydroxide can be found in Table 16.2 of the text. The equilibrium expressions are: Al(OH)3(s) Al3(aq) 3OH(aq) Ksp [Al3][OH]3 Using the given value of the hydroxide ion concentration, the equilibrium concentration of aluminum ion is: Ksp 1.8 1033 [Al 3 ] 7.4 108 M 3 9 3 [OH ] (2.9 10 ) What is the pH of this solution? Will the aluminum concentration change if the pH is altered? CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA 16.54 547 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: (a) 7.3 102 g SrF2 1 mol SrF2 5.8 104 mol/L = s 1 L soln 125.6 g SrF2 Consider the dissociation of SrF2 in water. Let s be the molar solubility of SrF2. SrF2(s) Initial (M): Change (M): Equilibrium (M): s Sr2(aq) 2F(aq) 0 0 s 2s s 2s Ksp [Sr2][F]2 (s)(2s)2 4s3 The molar solubility (s) was calculated above. Substitute into the equilibrium constant expression to solve for Ksp. Ksp [Sr2][F]2 4s3 4(5.8 104)3 7.8 1010 (b) 6.7 103 g Ag3PO4 1 mol Ag3PO4 1.6 105 mol/L = s 1 L soln 418.7 g Ag3PO4 (b) is solved in a similar manner to (a) The equilibrium equation is: Ag3PO4(s) Initial (M): Change (M): Equilibrium (M): s 3Ag(aq) PO34 (aq) 0 0 3s s 3s s Ksp [Ag]3 [PO34 ] (3s)3(s) 27s4 27(1.6 105)4 1.8 1018