HW 16-10: Review from textbook (p.725 #84, 87, 88(mod), 89, 95, 98, 101, 102, 110, 113, 115, 118, 120, SG#23,A)

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1 HW 6: Review from textbook (p.75 #84, 87, 88(mod), 89, 95, 98,,,, 3, 5, 8,, SG#3,A) 6.84 The pk a of the indicator methyl orange is Over what ph range does this indicator change from 9 percent HIn to 9 percent In? Acid Ionization Rxn for HIn: HIn (aq) + H O (l) In (aq) + H 3O + (aq) Orange yellow HendersonHasselbalch Eq: (no numbers yet.) ph pk a + log [In ] [HIn] Solving for the ph with 9% of the indicator in the HIn form and % in the I form: ph log [] (mostly HIn, orange solution) [9] Solving for the ph with % of the indicator in the HIn form and 9% of the indicator in the In form: ph log [9] (mostly In, yellow solution) [] Thus the ph range varies from.5 to 4.4 as the [HIn] varies from 9% to % The pk a of butyric acid (HBut) is 4.7. Calculate K b for the butyrate ion (But ) K The K a of butyric acid is obtained by setting a K a. The value of K b is: K b K w. 4 K a A solution is made by mixing exactly 5 ml of.67 M NaOH with exactly 5 ml. M HCOOH. Calculate the ph of the solution. Moles NaOH.5 L. mol Moles HCOOH.5 L.5 mol HCOOH HCOOH(aq) + OH (aq) HCOO (aq) + H O(l) Initial (mol): Change (mol): Final (mol): There is excess OH, so the ph is determined by this OH concentration. (The weak base, HCOO does not significantly affect the ph.) Thus, find the [OH ] and then the ph: The volume of the resulting solution is. L (5 ml + 5 ml ml). [OH ].335 mol. L 6.89 Cd(OH) is an insoluble compound. However, it dissolves in excess NaOH in solution. Write a balanced ionic equation for the reaction that must occur in excess NaOH. Most likely the increase in solubility is due to complex ion formation: Cd(OH) (s) + OH Cd(OH) 4 (aq) 6.98 When a KI solution was added to a HgCl solution, a precipitate, HgI, formed. A student plotted the mass of the precipitate versus the volume of the KI solution added and obtained the following graph. Explain the appearance of the graph. As KI is added, more and more of the precipitate, HgI is formed: Hg + (aq) + I (aq) HgI (s) However, with further addition of I, a soluble complex ion is formed and the precipitate dissolves. HgI (s) + I (aq).67 mol [HgI 4] (aq) p mol NaOH.335 M; poh.475 ; ph.55

2 6.95 Find the approximate ph range suitable for separating Mg + from Zn + given a solution that is initially. M in Mg + and. M Zn +. To separate the ions, an NaOH solution is added. K sp of Mg(OH),. K sp of Zn(OH).8 4 For Mg(OH), K sp [Mg + ][OH ] [OH K sp ] [Mg + ] For Zn(OH), K sp [Zn + ][OH ] [OH K sp ] [Zn + ] When [Mg + ]. M, the [OH ] value is When [Zn + ]. M, the [OH ] value is Thus, Zn(OH) will precipitate first at ph 8.. Mg(OH) will precipitate when the ph just exceeds Therefore, to selectively precipitate Zn(OH), the ph must be greater than 8. but less than (It is best to be as close to ph 9.54 as possible without exceeding it. This would cause as much Zn(OH) as possible to precipitate without having any Mg(OH) precipitate.) 6. Cacodylic acid (CacH) is (CH 3) AsO H. Its acid ionization constant is (a) Calculate the ph of 5. ml of a. M solution of CacH. CacH(aq) Cac (aq) + H + (aq) Initial (M):. Change (M): x +x +x Equilibrium (M):. x x x x.5 4 M [H + ], so ph log(.5 4 ) 3.6 (b) Calculate the ph of 5. ml of.5 M NaCac, (CH 3) AsO Na. Cac (aq) + H O(l) CacH(aq) + OH (aq) Initial (M):.5 Change (M): x +x +x Equilibrium (M):.5 x x x The ionization constant, K b, for Cac is: K a [H+ ][Cac ] CacH M.3 6 M x. x x K b K w. 4 K a ; K b [CacH][OH ] x [Cac ].5 x x x [OH ] M; poh log(4.9 5 ) 4.3, so ph (c) Mix the solutions in part (a) and part (b). Calculate the ph of the resulting solution. Number of moles of CacH from (a) is:. mol CacH 5. ml CacH 5. 3 mol CacH or could get [CacH] 5. 3 mol ml.75l Number of moles of Cac from (b) is:.5 mol CacNa 5. ml CacNa mol Cac or could get [Cac ] mol ml.75l At this point we have a buffer solution, so use HendersonHasselbalch to determine ph: ph pk a + log [Cac ] [CacH] log(6.4 7 ) + log mol 5. 3 mol 6.7 [Cac ].5 M or ph pka + log log 6.7 [CacH].67 M Thus, poh 4.47 and ph Mg(OH) will begin to precipitate from this solution at ph of Thus, poh 5.89 and ph 8.. Zn(OH) will begin to precipitate from the solution at ph M.5M

3 3 6. Solid NaBr is slowly added to a solution that is. M in Cu + and. M in Ag +. K sp of CuBr 4. 8 K sp of AgBr (a) Which compound will begin to precipitate first? The solubility product expressions for both substances have exactly the same mathematical form and are therefore directly comparable. The substance having the smaller K sp (AgBr) will precipitate first because it is the least soluble. (b) Calculate [Ag + ] right before CuBr begins to precipitate. When CuBr just begins to precipitate the solubility product expression will just equal K sp (saturated solution). The concentration of Cu + at this point is. M (given in the problem), so the concentration of bromide ion must be: For CuBr: K sp [Cu + ][Br ] (.)[Br ] 4. 8 [Br 4. 8 ] 4. 6 M is [Br ] right before CuBr begins to precipitate.. For AgBr: K sp [Ag + ][Br ] K sp so, [Ag ] [Br ] M is [Ag + ] right before CuBr precipitates. (c) What percent of Ag + remains in solution at this point? % Ag + (aq).8 7 M %.8%. M 6. What reagent (give anion of reagent) would you employ to separate the following pairs of ions in solution? a) Na + and Ba + b) Ca + and Cu + c) Zn + and Pb + SO 4 or many others SO 4 or F Cl, Br, I 6.3 CaSO 4 (K sp.4 5 ) has a larger K sp value than that of Ag SO 4 (K sp.4 5 ). Does it follow that CaSO 4 also has a greater solubility (g/l)? (In other words, find the solubility of each substance.) CaSO 4 Ca + (aq) + SO 4 (aq) I (M) solid C (M) s +s +s E (M) s s K sp [Ca + ][SO 4 ] s.4 5 s ; s M; Solubility mol 36. g mol.67 g / L Ag SO 4 Ag + (aq) + SO 4 (aq) I (M) solid C (M) s +s +s E (M) s s K sp [Ag + ] [SO 4 ] (s) (s).4 5 4s 3 ; s.5 M;.5 mol Solubility 3. g 4.7 g / L mol Ag SO 4 has a larger solubility, due both to the form of K sp (higher molar solubility) and to its larger molar mass. 6.8 Which of the following solutions has the highest [H + ]? Justify. (a). M HF (b). M HF in. M NaF (Not answer b/c added F shifts equilibrium to the left. Less H + (c). M HF in. M SbF 5. (Hint: SbF 5 reacts with F to form the complex ion [SbF 6].) HF (aq) H + (aq) + F (aq). M HF in. M SbF 5. has the highest [H + ] because the formation of complex ion removes F. Thus, ionization equilibrium shifts right to replenish lost F causing further ionization of HF

4 4 6.5 How many milliliters of. M NaOH must be added to a ml of. M NaH PO 4 to make a buffer solution with a ph of 7.5? Some K a values are given at right. Acid K a H 3PO To help you solve this question, do the following steps: H PO HPO a) In the desired buffer, what is the weak acid? H PO 4 What is the weak base? HPO 4 b) Write the acid ionization reaction for the acid and use HH to solve for the needed [base]/[acid] ratio. H PO 4 H + + HPO 4 where K a 6. 8 so, pk a log [HPO 4 ] ; so [H PO 4 ] c) Write the net ionic equation for the NaOH reacting with the weak acid. Do an ICF chart and use the variable, x, to represent the moles of NaOH needed. Do the ICF chart in moles since you don t know the total volume of the buffer. Initially there was. L. M. mol NaH PO 4 present. Let x the amount of NaOH added: H PO 4 + OH HPO 4 + H O Initial (mol):. x Change (mol): x x +x Final (mol):. x x d) Use your [base]/[acid] ratio to solve for x [HPO 4 ] x. [H PO 4 ]. x [HPO 4 ] [H PO 4 ]. so, x.3 mol of NaOH e) Determine the milliliters of. M NaOH needed to make the buffer at the desired ph. V NaOH mol NaOH M NaOH.3 mol. mol/l.3 L 3 ml 6. Water containing Ca + and Mg + ions is called hard water and is unsuitable for some household and industrial uses because these ions react with soap to form insoluble salts. One way to remove the Ca + ions from hard water is to add washing soda (Na CO 3 H O). (a) The molar solubility of CaCO 3 in water is M. What is its molar solubility in a.5 M Na CO 3 (aq)? The K sp of CaCO The dissociation of Na CO 3 is: Na CO 3(s) Na + (aq) + CO 3 (aq) (.5 M).5 M H O CaCO 3(s) Ca + (aq) + CO 3 (aq) Initial (M):..5 Change (M): +s +s Equil. (M): +s.5 + s K sp [Ca + ][CO 3 ] s(.5 + s).5s Molar solubility s.7 7 M (b) Why are Mg + ions not removed from the solution by this procedure? K sp of MgCO Mg + is not removed by this procedure, because MgCO 3 is fairly soluble.

5 5 Study Guide p. 36 #3. mol of AgNO3 solid and. mol of NaCl solid are both dissolved into water to make a. L solution. AgCl solid precipitates. Then, an aqueous NH3 solution is slowly added until all of the AgCl(s) is just able to all dissolve. What is the concentration of NH3 in this final solution? Ksp of AgCl.6 Kf of [Ag(NH3)]+.5 7 Complete these steps to answer the question: a) This question couples the slightly soluble salt, AgCl, with the formation of the complex ion, [Ag(NH3)]+. Determine the net overall reaction and its corresponding equilibrium constant, K. Solubility Equil: AgCl (s) Ag+ (aq) + Cl (aq) Ksp.6 Complex Ion Ag+ (aq) + NH3 (aq) [Ag(NH3)]+ Kf.5 7 Formation: K Ksp Kf Net Overall Rxn: AgCl (s) NH3 (aq) [Ag(NH3)]+ + Cl (aq).4 3 b) To get all of the AgCl solid to dissolve, essentially all of the Ag+ ions (in AgCl) must react and become part of the complex ion, [Ag(NH3)]+, in the final solution. Determine the [Ag(NH3)]+ and [Cl ] in the final solution. + [Ag(NH 3 ) ] mol Ag + volume solution. mol. L All of the Cl comes from NaCl: [Cl ]. M. mol NaCl mol Cl. M. L mol NaCl c) Write the equilibrium constant expression for the overall reaction. Calculate the [NH3] in the final solution. [Ag(NH 3 )+ ][Cl ] (.)(.) (.)(.) K ;.4 3 ; Thus, [NH 3 ].9 M 3 [NH 3 ] [NH 3 ].4 A. The titration curve shown to the right is obtained when 5. ml of a solution of a diprotic acid, HA, is titrated with.m NaOH. nd equiv. pt. a) What is the ph at the first equivalence pt? ~5. b) What is the predominant species in the solution at the first equivalence point? HA First equiv. pt. c) What is the ph at the nd equivalence pt? ~. d) What is the predominant species in the solution at the nd equivalence point? A e) What is the pka of HA? ~.5 Justify. Acid Ionization Eq: HA H+ + HA At first halfequivalence point, [HA] [HA ], so using HH equation, ph pka of HA f) What is pka of HA? ~7.5 Justify. Acid Ionization Eq: HA H+ + A At nd halfequivalence point, [HA ] [A ], so using HH equation, ph pka of HA g) What is the concentration of the initial HA solution? Justify. It takes. ml of.m NaOH to titrate HA to its endpoint (The nd equivalence pt.). mol H A. mol NaOH mol H A. ml NaOH. mol H A;. M H A ml mol NaOH.5 L H A Or, determine initial moles of HA using an ICF chart HA(aq) + NaOH (aq) NaA (aq) + HO(l) I. mol. mol. mol H A. M H A +. mol C. mol. mol.5 L H A F. mol