CHEM 1412 Zumdahl & Zumdahl Practice Exam II (Ch. 14, 15 & 16) Multiple Choices: Please select one best answer. Answer shown in bold.

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1 CHEM 1412 Zumdahl & Zumdahl Practice Exam II (Ch. 14, 15 & 16) Multiple Choices: Please select one best answer. Answer shown in bold. 1. Consider the equilibrium: PO -3 4 (aq) + H 2 O (l) HPO 2-4 (aq) + OH - (aq). Which of the following pair is a conjugate acid-base pair for the above equilibrium? (A) HPO 2-4 (aq), PO 3-4 (aq) (B) HPO 2-4 (aq), OH - (aq) (C) OH - (aq), HPO 2-4 (aq) (D) HPO - 4 (aq), H 2 O(l) (E) PO -3 4 (aq), H 2 O (l) Hint: For 9 th ed., see Section 14.1: p Also see End-of-Chapter Exercises 37 & 38. What is the conjugate acid for HN 3-? (A) N 3- (B) HN 2- (C) H 2 N - (D) H 3 N (E) H 2 N 2- Hint: For 9 th ed., see Section 14.1: p Also see End-of-Chapter Exercises 37 & Which of the following lies predominately to the right upon equilibrium? (A) SO 4 2- (aq) + HCO 3 - (aq) HSO 4 - (aq) + CO 3 2- (aq) (B) NO 2 - (aq) + HCO 3 - (aq) HNO 2 (aq) + CO 3 2- (aq) (C) CH 3 COOH(aq) + HCOO - (aq) CH 3 COO - (aq) + HCOOH(aq) (D) HCOOH (aq) + CN - (aq) HCOO - (aq) + HCN (aq) (E) HCN (aq) + F - (aq) CN - (aq) + HF (aq) Hint: For 9 th ed., see Section 14.2: p.p The reaction direction for acid and base is always from stronger acid + stronger base weaker conjugate base + weaker conjugate acid. You may or may not be given K a from Tables 14.2, 14.4 and A5.1 in p. A22 as reference. For (A), HSO 4 - (K a = 1.2x10-2 ); HCO 3 - (K a = 5.6x10-11 ); For (B), HNO 2 (K a = 4.0x10-4 ); HCO 3 - (K a = 5.6x10-11 ); For (C), CH 3 COOH (K a = 1.8x10-5 ); HCOOH (K a = 1.8x10-4 ); For (D), HCOOH (K a = 1.8x10-4 ); HCN (K a = 6.2x10-10 ); For (E), HCN (K a = 6.2x10-10 ); HF (K a = 7.2x10-4 ); 3. Which of the following is the weakest acid? (A) CCl 3 COOH (B) CBr 3 COOH (C) CI 3 COOH (D) CClBrICOOH (E) CClF 2 COOH Hint: For 9 th ed., see Section 14.2: p. 657: carboxylic acid and its derivatives. Inductive effect for organic acids derived from acetic acid, CH 3 COOH: (i) More halogens, same kind, attached to C of CH 3, the stronger the acid. (ii) More electronegative halogens attached to C of CH 3, the stronger the acid. Electronegativity: F (4.0), O (3.5), N (3.0), Cl (3.0), Br (2.8), C (2.5), H(2.1) Which of the following is the strongest acid? (A) HNO 3 (B) HNO 2 (C) H 2 O (D) HIO (E) HClO 2 1

2 Hint: For 9 th ed., see Sections 14.2 & Comparing the oxoacids (or oxyacids) and remembering the periodic trend. 4. What is the ph of M of Ca(OH) 2? What is the ph of 1.0x10-10 M HCl? (A) 1.66; 10 (B) 1.96; 4 (C) 7; 7 (D) 12.04; 7 (E) 12.34; 7 Hint: For 9 th ed., see Section 14.3: Interactive Examples 14.5 and There is a very important concept in Interactive Example 14.7(b). See Section 14.6: Bases: 2 nd paragraph of p Also see End-of-Chapter Exercise: Solution of Bases: 90 (a). Since Ca(OH) 2 is a strong base and thus the [OH - ] = 2 x M = M poh = - log = 1.66 ph = 14 poh = What is the ph of an aqueous solution of M NaOH? (A) 9.96 (B) 7.00 (C) 4.04 (D) 2.66 (E) None of the above Hint: The same concept as Interactive Example 14.7(b). 5. The K b for NH 3 at 25 o C is What s the ph for 0.4 M NH 4 NO 3 (the conjugate acid of ammonia) aqueous solution? (A) 2.57 (B) 4.83 (C) 5.44 (D) (E) Hint: For 9 th ed., see Section 14.8: Interactive Example (calculate the ph of conjugate acid) or (calculate the ph of conjugate base). For (1) any weak base and its conjugate acid or (2) any weak acid and its conjugate base, the relationship of K a x K b = K w is always valid. Here, K a, NH4NO3 = K w / K b = / = Using equilibrium table (textbook method) to calculation the concentration of proton: + NH 4 H + + NH 3 [initial] [change] -x +x +x 0.4-x +x +x K a = [H + ][NH 3 ]/[ NH 4 + ] = (+x)(+x)/0.4-x = Assuming that x << 0, then K a = [H + ][NH 3 ]/[ NH 4 + ] = (+x)(+x)/0.4 = Thus, x = square root of (0.4 x ) = = [H + ] ph = -log = For mono-protic acid: Use equilibrium table once to calculate [H+]. Short-cut: [H+] = {K a x [HA]} 1/2 = square root of {K a x [HA]}; this is only valid if [HA]initial K a x Which of the following substances will be more soluble in basic solution than in acidic solution? (A) KAl(SO 4 ) 2 (B) NaCl (C) K 2 CO 3 (D) CH 3 COONH 4 (E) KNO 3 Hint: For 9 th ed., see Section Interactive example Also see lab manual. Consider the acidity or basicity of solutions. To be more soluble in basic solution, the acidic substance will be more soluble than neutral substance, which in turns is more 2

3 soluble than the basic substance. That is, the basic substance is the least one to be soluble in basic solution. For (A): IA & IIA metal ions are neutral; anions from strong acids are neutral. K + : neutral; Al 3+ : acidic; SO 4 2- : neutral. Thus, overall, KAl(SO 4 ) 2, is acidic. For (B): IA & IIA metal ions are neutral; anions from strong acids are neutral. Na + : neutral; Cl - : neutral. Thus, overall, NaCl, is neutral. For (C): Anions from weak acids are weak bases; K + : neutral; CO 3 2- : weak base. Thus, overall, K 2 CO 3, is basic. For (D): Anion from a weak acid is basic; conjugate acid of a weak base is a weak acid. So we need to see the K b of CH 3 COO - (K b = K w / K a, CH3COOH = / = ) and the K a of NH 4 + (K a = K w / K b, NH3 = / = ). Since the K b of CH 3 COO - is same as the K a of NH 4 +, and thus CH 3 COONH 4 is neutral. For (E): K + : neutral; NO 3 - : neutral. Thus, KNO 3 is neutral. 7. Which of the following is a not a Lewis base? - - (A) BF 3 (B) AlCl 4 (C) BF 4 (D) O 2- (E) PF 3 Hint: For 9 th ed., see Section 14.11: Definition. Lewis acid is an electron-pair acceptor; while the Lewis base is an electron-pair donor. This question is asking for Lewis acid. Q8-10. A buffer contains 0.30 mole HC 2 H 3 O 2 and 0.20 mole NaC 2 H 3 O 2 in 1 L. K a of HC 2 H 3 O 2 is What is the ph of this buffer? (A) 4.57 (B) 4.64 (C) 4.74 (D) 4.88 (E) 4.49 Hint: For 9 th ed., see Section 15.2: Interactive Example Use equilibrium table or Short-cut: Henderson-Hasselbalch equation: [HC 2 H 3 O 2 ] = 0.30 mol/1 L = 0.30 M; [NaC 2 H 3 O 2 ] = 0.20 mol/1 L = 0.20 M ph = pka + log [weak base] /[weak acid] = pka + log [salt] /[weak acid] = -log log [0.2] /[0.3] = ( ) = What is the ph of this buffer after the addition of 0.02 mole of KOH? (A) 3.66 (B) 4.64 (C) 4.74 (D) 4.88 (E) 4.49 Hint: For 9 th ed., see Section 15.2: Interactive Example First, use stoichiometric table, and then use the equilibrium table (TEXTBOOK METHOD) or Henderson- Hasselbalch equation to solve it. Or use short-cut (if you want to use it, you need to memorize it): the modified Henderson-Hasselbalch equation. Use Modified Henderson-Hassbach Equations: a. Add strong acid into a buffer ph = pka + log { [weak base] [strong acid] }/{ [weak acid] + [strong acid] } b. Add strong base into a buffer ph = pka + log { [weak base] + [strong base] }/{ [weak acid] - [strong base] 3

4 Since this is adding the strong base (limiting reagent: used up) to the buffer, ph = pka + log { [weak base] + [strong base] }/{ [weak acid] - [strong base] = -log log {( )/( )} = ( ) = What is the ph of this buffer after the addition of 0.02 mole of HCl? (A) 3.66 (B) 3.77 (C) 4.74 (D) 4.88 (E) 4.49 Hint: For 9 th ed., see Interactive Example 15.6, and Example First, use stoichiometric table, and then use the equilibrium table (TEXTBOOK METHOD) or Henderson-Hasselbalch equation to solve it. Or use short-cut (if you want to use it, you need to memorize it): the modified Henderson-Hasselbalch equation. Since this is adding the strong acid (limiting reagent: used up) to the buffer, ph = pka + log { [weak base] - [strong acid] }/{ [weak acid] + [strong acid] = -log log {( )/( )} = ( ) = Which of the following can make a buffer solution? (Multiple Answers) (A) C 4 H 9 COOH and CH 3 COH (B) H 2 SO 4 and NaOH (C) HS and NaF (D) H 2 SO 4 and CH 3 COOH (E) NaOH and NH 3 (F) CH 3 COH and CH 3 OH (G) NH 4 Cl (in excess) and NaOH (H) NH 3 (in excess) and HCl Hint: For 9 th ed., see Section 15.2, p.p There are FOUR ways to make a buffer solution (memorize them): (1) weak acid and its conjugated base; (2) weak base and its conjugate acid; (3) strong acid and excessive weak base; and (4) strong base and excessive weak acid. CH 3 COH is aldehyde; C 4 H 9 COOH is carboxylic acid; CH 3 OH is alcohol. More practice on ph of buffer calculation: What is the ph for a buffer that contains 0.46 M HNO 2 and 0.40 M KNO 2? The K a for HNO 2 is (A) 3.14 (B) 3.29 (C) 3.95 (D) 4.04 (E) 4.44 Hint: Similar to Q 8. For 9 th ed., see Section 15.2: Interactive Examples 15.2, 15.3, 15.4, and Two methods to solve it: (1) use the equilibrium table or (2) Henderson- Hasselbalch equation (memorize it). ph = pka + log [weak base] /[weak acid] = pka + log [salt] /[weak acid] = -log log [0.4] /[0.46] = ( ) = What is the ph for a buffer that is prepared containing 0.25 M HCOOH (formic acid) and 0.10 M HCOONa (sodium formate)? The K a for HCOOH is (A) 3.62 (B) 3.57 (C) 3.35 (D) 3.29 (E) 3.10 Hint: Similar to Q 8. For 9 th ed., see Section 15.2: Interactive Examples 15.2, 15.3, 15.4, and Two methods to solve it: (1) use the equilibrium table or (2) Henderson- Hasselbalch equation (memorize it). ph = pka + log [weak base] /[weak acid] = pka + log [salt] /[weak acid] 4

5 = -log log [0.1] /[0.25] = ( ) = Q Strong Acid - Strong Base Titration: Basic type of titration without buffer and hydrolysis regions to consider. For 9 th ed., see Section 15.4: p.p Calculate the ph when the following quantities of 0.1 M HCl have been added to 25.0 ml of 0.1 M NaOH solution: (a) 0.0 ml, (b) 24.9 ml, (c) 25.0 ml, (d) 25.1 ml Answer: (a) Before the titration: NaOH is in the beaker and HCl is in the buret; as no HCl has been added into the beaker, and the ph meter is placed in the beaker, the ph of the solution is determined by the substance in the beaker. As NaOH is a strong base, so poh = - log[oh - ] = -log[naoh] = -log 0.1 = 1. Thus, ph = 14 poh = (b) Before the equivalence point: At the equivalence point, the mole of hydrogen ions must be equal to the mole of hydroxide ions. Therefore, to reach the equivalence point, the volume of HCl should be 25.0 ml. Thus, this is the region before the equivalence point, which the NaOH is in excess. The ph of the solution is determined by the substance in excess in the beaker. [OH - ] new = [OH - ] left over = (0.1x x24.9)mmol/( )mL ={0.1x( )/( )} = M, so poh = -log = 3.70; ph = 14 poh = (c) At the equivalence point: For a strong acid-base (vise versa) titration, the ph at the equivalence point is 7 as it is the hydrolysis of a neutral salt. No calculation is needed. 14. (d) After the equivalence point: This is a region after the equivalence point. Thus, HCl is in excess. [H + ] new = [H + ] left over = (0.1x x25.0)mmol/( )mL ={0.1x( )/( )} = M, so ph = -log = Q A 50.0 ml of M HC 7 H 5 O 2 (K a = ) is titrated with M KOH solution. 15. What is the ph after adding 10 ml of base is added? (A) 3.32 (B) 4.01 (C) 6.43 (D) 8.20 (E) Hint: For 9 th ed., see Section 15.4: p.p : Titration of Weak Acids with Strong Bases. This is before the equivalence point. So the solution is a buffer. First, use 5

6 stoichiometric table, and then use the equilibrium table or Henderson-Hasselbalch equation to solve it. Moles of HC 7 H 5 O 2 = x (50.0/1000) = 1.25x10-3 = mole of KOH = x (10/1000) = Before the equivalence point: Note that this is a buffer region as shown below: To reach an equivqlence point, the volume of KaOH should be 1.25/0.05 = 25.0 ml. Since now KOH only has 10 ml < 25.0 ml, and thus, this is the region before the equivalence point, which acetic acid is in excess. HC 7 H 5 O 2 + KOH KC 7 H 5 O 2 + H 2 O Before reaction mol mol 0 mol Change mol mol mol After reaction mol 0 mol mol Weak acid Conjugate base Total Volume = 50 ml + 10 ml = 60 ml So, apply ph = pka + log {[weak base]/[weak acid]}; since the volume is the same, we can use mole to represent the concentration; it is because the ratio is the same. ph = pka + log [weak base] /[weak acid] = pka + log [salt] /[weak acid] = -log log [0.0005] /[ ] = ( ) = What is the ph after adding 25 ml of base is added? (A) 3.32 (B) 4.01 (C) 6.43 (D) 8.20 (E) Hint: This is at the equivalence point. So the ph of the solution is determined by the hydrolysis of the salt (product). Mole KOH = 0.05 x (25/1000) = HC 7 H 5 O 2 + KOH KC 7 H 5 O 2 + H 2 O Before reaction mol mol 0 mol Change mol mol mol After reaction mol mol mol Conjugate base Total Volume = 50 ml + 25 ml = 75 ml = L [NaC 7 H 5 O 2 ] = /0.075 = M The salt will hydrolyze: - C 7 H 5 O 2 + H 2 O HC 7 H 5 O 2 + OH - [initial] [change] -x +x +x x +x +x K b = K w /K a = 1x10-14 / = x = (x)(x)/ x Assuming x << ; then x = (x)(x)/0.0167; 6

7 Thus x = square root of (1.538 x )(0.0167) = x 10-6 poh = -log x = - log ( x 10-6 ) = So ph = 14 poh = = Or use the short-cut, that is the formula (memorize): poh = square root of (K b x C) = square root of {(K w /K a ) x C} = square root of (1.538 x )(0.0167) = x 10-6 poh = -log x = - log ( x 10-6 ) = So ph = 14 poh = = What is the ph after adding 50 ml of base is added? (A) 3.32 (B) 4.74 (C) 6.43 (D) 8.81 (E) Hint: This is after the equivalence point. So the ph is determined by the excess amount of the strong base. First, use stoichiometric table, and then use the poh = log [OH - ] and the ph = 14 poh. Mole of KOH = x (50/1000) = HC 7 H 5 O 2 + KOH KC 7 H 5 O 2 + H 2 O Before reaction mol mol 0 mol Change mol mol mol After reaction mol mol mol Strong base Weak base Total Volume = 50 ml + 50 ml = 100 ml = 0.1 L The ph is determined by the excess amount of the KOH. [KOH] = /0.1 = M poh = -log = ph = 14 poh = Q Calculate the ph when the following quantities of 0.1 M HCl have been added to 25.0 ml of 0.1 M NH 3 solution: (a) 0.0 ml, (b) 24.9 ml, (c) 25.0 ml, (d) 25.1 ml Hint: For 9 th ed., see Section 15.4, p.p : Titration of Weak Bases with Strong Acids. Answer: 18. (a) Before the titration: NH 3 is in the beaker and HCl is in the buret; as no HCl has been added into the beaker, and the ph meter is placed in the beaker, the ph of the solution is determined by the substance in the beaker, that is, ammonia. As ammonia is a weak base, the equilibrium problem at this point is the familiar one of calculating the poh of a solution of a weak base. Thus, the [OH - ] can be obtained from either of the two methods: (1) equilibrium table or (2) short-cut formula [OH - ] = square root of (K b x C) = square root of {(1.8x10-5 ) x (0.1)} = 1.34x10-3. poh = -log 1.34x10-3 = Thus, ph = 14 poh =

8 19. (b) Before the equivalence point: Note that this is a buffer region as shown below: To reach an equivqlence point, the volume of HCl should be 25.0 ml. Thus, this is the region before the equivqlence point, which acetic acid is in excess. + NH 3 + HCl NH 4 + Cl - Before reaction mol mol 0 mol Change mol mol mol After reaction mol 0 mol mol Weak base Conjugate acid [NH 3 ] = mol/ L = M [NH + 4 ] = / = M So, apply ph = pka + log {[weak base]/[weak acid]} = (14 pkb) + log {[weak base]/[weak acid]} = {14 (-log 1.8x10-5 )}+ log / = ( ) + ( ) = (c) At the equivalence point: This is the region of hydrolysis of salt. NH 3 + HCl + NH 4 Before reaction mol mol 0 mol Change mol mol mol After reaction mol mol mol + Cl - 8

9 [NH + 4 ] = / = M + NH 4 + H 2 O NH 3 + H 3 O + [Initial] [Change] -x +x +x [Equilibrium] x +x +x So, Ka = [H 3 O + ][NH 3 ]/[NH + 4 ] = K w /K b = 1.0x10-14 /1.8x10-5 = 5.56 x Thus, (x)(x)/(0.050 x) = 5.56 x Assume 0 < x << and thus the above quadric equation can be simplified as (x)(x)/(0.050) = 5.56 x Therefore, x = square root of x (5.56 x )= 5.27 x 10-6 Thus, ph = -log 5.27 x 10-6 = (d) After the equivalence point: Thus, HCl is in excess. [H+] left over = (0.1x x25.0)/( ) ={0.1x( )/( )} = M, so ph = -log = Additional Practice: A 50.0 ml of 0.1 M NH 3 (K b = 1.8x10-5 ) solution is titrated with 0.25 M HCl solution. (See 9 th ed., p.p : same weak base, same strong acid) *What is the ph after adding 0.0 ml of acid is added? (A) (B) 9.26 (C) 5.28 (D) 4.74 (E) 2.87 Hint: Before adding any strong acid, the ph of the solution is solely determined by the WEAK base, NH 3. Use equilibrium table to calculate the hydroxide ion concentration and then calculate the poh and thus ph = 14 poh. *What is the ph after adding 15 ml of acid is added? (A) 8.78 (B) 4.74 (C) 7.0 (D) 9.26 (E) Hint: This is before the equivalence point and thus the solution is a buffer solution. To check whether the titration reaches the equivalence point, you need to calculate the moles or milli-moles of both acid and base. First, use stoichiometric table to calculate how much ammonia is left (1.25 milli-moles) and how much ammonium ion (3.75 milli-moles) is produced. Then use the equilibrium table or modified Henderson- Hasselbalch equation to solve it. See 9 th ed. lecture notes. If you use equilibrium table, you need to remember that Kw = K a x K b = 1.00 x to calculate K a for ammonium ion who will undergo hydrolysis. Hydrogen ion concentration calculated is 1.67 x Thus ph = *What is the ph after adding 20 ml of acid is added? (A) 1.51 (B) 2.87 (C) 5.19 (D) 7.0 (E) Hint: This is at the equivalence point. The ph is determined by the hydrolysis of the salt, ammonium chloride. Chloride ion is neutral and thus it does not hydrolyze but ammonium does as it s a weak acid. Total volume of solution now is 70 ml. The 9

10 hydrogen ion concentration calculated is 6.3 x Since this number, 6.3 x 10-6, is much smaller than x 10-2 M of hydrogen ion concentration calculated by stoichiometry. Thus we can drop the x. However, when calculating the total hydrogen ion concentration = 6.3 x x 10-7 = 6.4 x 10-6 M. Thus, ph = log 6.4 x 10-6 = Therefore, poh = 14 ph = See 9 th ed. lecture notes. *What is the ph after 30 ml of acid is added? (A) 1.51 (B) 2.87 (C) 8.81 (D) 7.0 (E) 9.26 Hint: This is after the equivalence point. Thus, the ph of the solution is solely determined by the excess strong acid, HCl. Use stoichiometric table to calculate how much of the HCl left. Then calculate the new concentration of HCl. The total volume of the solution now is = 80 ml. The new HCl concentration = x 10-2 M = new hydrogen ion concentration. Thus, the ph = The K sp for Ag 2 CrO 4 in water at 25 o C is What is the molar concentration for CrO 2-4 and Ag +, respectively? (A) and (B) and (C) and (D) and (E) and Hint: For 9 th ed., see Section 16.1: Interactive Exercise Ag 2 CrO 4 (s) 2Ag + (aq) + CrO 4 2- (aq) [Equilibrium] --- 2x x K sp = [Ag + ] 2 [CrO 4 2- ] = (2x) 2 (x) = 4x 3 = x 3 = ( )/4 = x = cubic root of = ( ) 1/3 = M = [CrO 4 2- ] 2x = [Ag + ] = 2x( )= M 23. What is the molar solubility for Cu(OH) 2 at ph 10.5? K sp for Cu(OH) 2 is (A) (B) (C) (D) (E) Hint: For 9 th ed., see Section 16.1: p.p This is the common-ion effect concerning ph and solubility. The question indirectly gives you the concentration of hydroxide ion via ph. Given ph 10.5 poh = = 3.5 [OH - ] = = M Cu(OH) 2(s) Cu OH - [Equilibrium] --- x 2x K sp = [Cu +2 ][OH - ] 2 = (x)(2x ) 2 = Assuming that 2x << and thus 2x = K sp = [Cu +2 ][OH - ] 2 = (x)( ) 2 = x = ( )/( ) 2 = ( )/( ) = M 10

11 Additional Practice: *The molar solubility for CaF 2 in water at 25 o C is M. What is its K sp at this temperature? (A) (B) (C) (D) (E) Hint: For 9 th ed., see Section 16.1: Interactive Examples 16.1 and *What is the molar solubility for cerium (III) hydroxide, Ce(OH) 3, in 0.2 M Ba(OH) 2 solution? K sp for Ce(OH) 3 is (A) (B) (C) (D) (E) Hint: For 9 th ed., see Section 16.1: p.p see Q. 23; This is the common-ion effect. The question indirectly gives you the concentration of hydroxide ion via concentration of the strong base, di-basic base. Thus, [OH - ] = 0.2 x 2 = 0.4 M. Ce(OH) 3 (s) Ce +3 (aq) + 3 OH - (aq) [Equilibrium] --- x 3x+0.4 Thus, K sp = [Ce +3 ][OH - ] 3 = (x)(3x+0.4) 3 = Assuming 3x << 0.4, then K sp = [Ce +3 ][OH - ] 3 = (x)(0.4) 3 = x = /(0.4) 3 = M *Will Mn(OH) 2 to precipitate when the ph of a M solution of MnCl 2 is adjusted to 10.00? K sp for Mn(OH) 2 is (A) Yes (B) No (C) Not sure Hint: For 9 th ed., see section The question is indirectly given the [OH - ]. Calculate Q and compare it with K sp. If Q is greater than K sp, then there will be a solid formed. 24. Which of the following action may/will NOT decrease the solubility of a precipitate? (A) Add the common ion into the solution. (B) Add a complexing agent into the solution. (C) Decrease the temperature of the solution. (D) Add a precipitating agent into the solution. (E) Increase the concentration of solution. Hint: For 9 th ed., see Section The solubility for SrF 2 in water at 45 o C is found 2 g/100 ml. What is its K sp? The equilibrium is SrF 2 (s) Sr 2+ (aq) + 2F - (aq) (A) (B) (C) (D) (E) Hint: For 9 th ed., see Section 16.1: The definition of K sp is made up by the molar concentration product of each ion involved and raised the coefficient as its exponent. 11

12 The molar concentration is molarity (mole/liter). So the solubility in g/ml must be converted to mole/l. Thus, K sp = [Sr 2+ ][F - ] 2 = (0.159)(2x0.159) 2 = NOTE: K sp and other equilibrium constants do not have units. Additional information: [H + ] = [H + ] solvent + [H + ] solute [OH - ] = [OH - ] solvent + [OH - ] solute ph = -log[h + ] poh = -log[oh - ] ax 2 + bx + c = 0 where x = {-b (b 2-4ac) 1/2 }/2a K p = K c (RT) n(g) 12

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