Practice Worksheet - Answer Key. Solubility #1 (KEY)

Practice Worksheet - Answer Key Solubility #1 (KEY) 1 Indicate whether the following compounds are ionic or covalent a) NaCl ionic f) Sr(OH) 2 ionic b) CaBr 2 ionic g) MgCO 3 ionic c) SO 2 covalent h) AgOH ionic d) NH 3 covalent i) ZnCl 2 ionic e) H 2 O covalent j) CH 3 OH covalent 2 An ionic compound can be separated into positive and negative ions, and therefore they can conduct electricity A covalent compound cannot 3 Dissociation equations a) KCl(s) K + (aq) + Cl (aq) b) MgF 2 (s) Mg 2+ (aq) + 2 F (aq) c) FeS(s) Fe 2+ (aq) + S 2 (aq) d) Hg 2 SO 4 (s) Hg 2+ 2 (aq) + SO 2 4 (aq) *** Hg 2+ 2 is a diatomic ion e) Ca 3 (PO 4 ) 2 (s) 3 Ca 2+ (aq) + 2 PO 3 4 (aq) f) CCl 4 (l) does not dissociate CCl 4 is a covalent compound g) Ag 2 S(s) 2 Ag + (aq) + S 2 (aq) h) NH 4 Cl(s) NH + 4 (aq) + Cl (aq) i) (NH 4 ) 2 Cr 2 O 7 (s) 2 NH + 4 (aq) + Cr 2 O 2 7 (aq) j) K 2 CrO 4 (s) 2 K + (aq) + CrO 2 4 (aq) 4 More soluble in water: a) Na 2 CO 3 e) KOH b) ZnSO 4 f) Na 2 S c) NH 4 Cl g) K 2 CO 3 d) AlCl 3 h) CuSO 4 5 For each of the following pairs of chemicals: (i) = Molecular equation (aka Formula equation) = Complete ionic equation = Net ionic equation a) (i) Na 2 SO 4 (aq) + Pb(NO 3 ) 2 (aq) 2 NaNO 3 (aq) + PbSO 4 (s) 2 Na + (aq) + SO 4 2 (aq) + Pb 2+ (aq) + 2 NO 3 (aq) 2 Na + + 2 NO 3 (aq) + PbSO 4 (s) Pb 2+ (aq) + SO 4 2 (aq) PbSO 4 (s) Key page 1

b) (i) c) (i) d) (i) e) (i) f) (i) g) (i) h) (i) i) (i) j) (i) AgNO 3 (aq) + KI(aq) KNO 3 (aq) + AgI(s) Ag + (aq) + NO 3 (aq) + K + (aq) + I (aq) K + (aq) + NO 3 (aq) + AgI(s) Ag + (aq) + I (aq) AgI(s) BaCl 2 (aq) + K 2 SO 4 (aq) 2 KCl(aq) + BaSO 4 (s) Ba 2+ (aq) + 2 Cl (aq) + 2 K + (aq) + SO 2 4 (aq) 2 K + (aq) + 2 Cl (aq) + BaSO 4 (s) Ba 2+ (aq) + SO 2 4 (aq) BaSO 4 (s) CuSO 4 (aq) + CaCl 2 (aq) CuCl 2 (aq) + CaSO 4 (s) Cu 2+ (aq) + SO 2 4 (aq) + Ca 2+ (aq) + 2 Cl (aq) Cu 2+ (aq) + 2 Cl (aq) + CaSO 4 (s) Ca 2+ (aq) + SO 2 4 (aq) CaSO 4 (s) Pb(NO 3 ) 2 (aq) + K 2 S(aq) 2 KNO 3 (aq) + PbS(s) Pb 2+ (aq) + 2 NO 3 (aq) + 2 K + (aq) + S 2 (aq) 2 K + (aq) + 2 NO 3 (aq) + PbS(s) Pb 2+ (aq) + S 2 (aq) PbS(s) 2 AgNO 3 (aq) + BaCl 2 (aq) Ba(NO 3 ) 2 (aq) + 2 AgCl(s) 2 Ag + (aq) + 2 NO 3 (aq) + Ba 2+ (aq) + 2 Cl (aq) Ba 2+ (aq) + 2 NO 3 (aq) + 2 AgCl(s) Ag + (aq) + Cl (aq) AgCl(s) Pb(NO 3 ) 2 (aq) + 2 NaCl(aq) 2 KNO 3 (aq) + PbCl 2 (s) Pb 2+ (aq) + 2 NO 3 (aq) + 2 Na + (aq) + 2 Cl (aq) 2 K + (aq) + 2 NO 3 (aq) + PbCl 2 (s) Pb 2+ (aq) + 2 Cl (aq) PbCl 2 (s) Na 2 SO 3 (aq) + Ba(NO 3 ) 2 (aq) 2 NaNO 3 (aq) + BaSO 3 (s) 2 Na + (aq) + SO 2 3 (aq) + Ba 2+ (aq) + 2 NO 3 (aq) 2 Na + + 2 NO 3 (aq) + BaSO 3 (s) Ba 2+ (aq) + SO 2 3 (aq) BaSO 3 (s) K 2 S(aq) + ZnCl 2 (aq) 2 KCl(aq) + ZnS(s) 2 K + (aq) + S 2 (aq) + Zn 2+ (aq) + 2 Cl (aq) 2 K + (aq) + 2 Cl (aq) + ZnS(s) Zn 2+ (aq) + S 2 (aq) ZnS(s) 2 NH 4 Br(aq) + Pb(C 2 H 3 O 2 ) 2 (aq) 2 NH 4 C 2 H 3 O 2 (aq) + PbBr 2 (s) 2 NH + 4 (aq) + 2 Br (aq) + Pb 2+ (aq) + 2 C 2 H 3 O 2 (aq) 2 NH + 4 (aq) + 2 C 2 H 3 O 2 (aq) + PbBr 2 (s) Pb 2+ (aq) + 2 Br (aq) PbBr 2 (s) Key page 2

k) (i) l) (i) m) (i) n) (i) o) (i) 2 NH 4 OH(aq) + Cu(NO 3 ) 2 (aq) 2 NH 4 NO 3 (aq) + Cu(OH) 2 (s) 2 NH + 4 (aq) + 2 OH (aq) + Cu 2+ (aq) + 2 NO 3 (aq) 2 NH + 4 (aq) + 2 NO 3 (aq) + Cu(OH) 2 (s) Cu 2+ (aq) + 2 OH (aq) Cu(OH) 2 (s) (NH 4 ) 2 S(aq) + 2 NaCl(aq) 2 NH 4 Cl(aq) + Na 2 S(aq) ***not a true reaction*** 2 NH + 4 (aq) + S 2 (aq) + 2 Na + + 2 Cl (aq) 2 NH + 4 (aq) + 2 Cl (aq) + 2 Na + + S 2 (aq) None, as there is no precipitate formed Cr 2 (SO 4 ) 3 (aq) + 3 K 2 CO 3 (aq) 3 K 2 SO 4 (aq) + Cr 2 (CO 3 ) 3 (s) 2 Cr 3+ (aq) + 3 SO 2 4 (aq) + 6 K + (aq) + 3 CO 2 3 (aq) 6 K + (aq) + 3 SO 2 4 (aq) + Cr 2 (CO 3 ) 3 (s) 2 Cr 3+ (aq) + 3 CO 2 3 (aq) Cr 2 (CO 3 ) 3 (s) Sr(OH) 2 (aq) + MgCl 2 (aq) SrCl 2 (aq) + Mg(OH) 2 (s) Sr 2+ (aq) + 2 OH (aq) + Mg 2+ (aq) + 2 Cl (aq) Sr 2+ (aq) + 2 Cl (aq) + Mg(OH) 2 (s) Mg 2+ (aq) + 2 OH (aq) Mg(OH) 2 (s) Ba(NO 3 ) 2 (aq) + Na 2 SO 3 (aq) 2 NaNO 3 (aq) + BaSO 3 (s) Ba 2+ (aq) + 2 NO 3 (aq) + 2 Na + (aq) + SO 2 3 (aq) 2 Na + (aq) + 2 NO 3 (aq) + BaSO 3 (s) Ba 2+ (aq) + 2 SO 2 3 (aq) BaSO 3 (s) Key page 3

Solubility #2 (KEY) 1 Table 1 FeCl 3 CoCl 2 Co(NO 3 ) 2 NaOH KOH NaNO 3 FeCl 3 Fe(OH) 3 Fe(OH) 3 CoCl 2 Co(OH) 2 Co(OH) 2 Co(NO 3 ) 2 Co(OH) 2 Co(OH) 2 NaOH Fe(OH) 3 Co(OH) 2 Co(OH) 2 KOH Fe(OH) 3 Co(OH) 2 Co(OH) 2 NaNO 3 Table 2 NiCl 2 MgCl 2 Na 2 SO 4 NaOH Sr(OH) 2 MgSO 4 NiCl 2 Ni(OH) 2 Ni(OH) 2 MgCl 2 Mg(OH) 2 Mg(OH) 2 Na 2 SO 4 SrSO 4 NaOH Ni(OH) 2 Mg(OH) 2 Mg(OH) 2 Sr(OH) 2 Ni(OH) 2 Mg(OH) 2 SrSO 4 Mg(OH) 2 & SrSO 4 MgSO 4 Mg(OH) 2 Mg(OH) 2 & SrSO 4 2 Washing soda dissociates in water as follows: Na 2 CO 3 10H 2 O(aq) 2 Na + (aq) + CO 2 3 (aq) Hard water contains Ca 2+ ions These calcium ions can be removed by adding washing soda, through a precipitation reaction with carbonate ions (CO 3 2 ): Ca 2+ (aq) + CO 3 2 (aq) CaCO 3 (s) Once CaCO 3 (s) is filtered off, the water is free of Ca 2+ ions Key page 4

3 ***note: There are more than one procedure to separate the following groups of ions: a) Add X(NO 3 ) 2 (aq) to precipitate PO 4 3 (aq) ions, where X can be Fe 2+, Ca 2+, Mg 2+ ions, etc, (except H +, alkali, NH 4 +, Ag +, Cu +, Pb 2+ ions) Filter off the precipitate Add CuNO 3 (or AgNO 3 or Pb(NO 3 ) 2 ) to precipitate Cl (aq) ions, and then filter off the 2 nd precipitate b) Add X(NO 3 ) 2 (aq) to precipitate SO 4 2 (aq) ions, where X can be Ca 2+, Sr 2+, or Ba 2+ ions Filter off the precipitate Add AgNO 3 (or Pb(NO 3 ) 2 ) to precipitate S 2 (aq) ions, and then filter off the 2 nd precipitate c) Add X(NO 3 ) 2 (aq) to precipitate OH (aq) ions, where X can be Fe 2+, Ca 2+, Mg 2+ ions Filter off the precipitate Add CuNO 3 (or AgNO 3 or Pb(NO 3 ) 2 ) to precipitate Br (aq) ions, and then filter off the 2 nd precipitate d) Add NaCl(aq), NaBr(aq), or NaI(aq) to precipitate Pb 2+ (aq) ions Filter off the precipitate Add Na 2 SO 4 to precipitate Sr 2+ (aq) ions, and then filter off the 2 nd precipitate Add Na 2 S, or NaOH, or Na 2 CO 3 to precipitate Cr 2+ (aq) ions Filter off the precipitate e) Add Na 2 SO 4 (aq) to precipitate Sr 2+ (aq) ions Filter off the precipitate Add Na 2 S to precipitate Cu 2+ (aq) ions, and then filter off the 2 nd precipitate Add NaOH, or Na 2 CO 3 to precipitate Mg 2+ (aq) ions Filter off the precipitate f) Add NaCl(aq), NaBr(aq), or NaI(aq) to precipitate Ag + (aq) ions Filter off the precipitate Add Na 2 SO 4 to precipitate Ca 2+ (aq) ions, and then filter off the 2 nd precipitate Add Na 2 S, or NaOH, or Na 2 CO 3 to precipitate Fe 2+ (aq) ions Filter the precipitate Na + ions will remain in the resulting solution 4 Soluble lead compounds dissociate completely to give Pb 2+ (aq) ions in the bloodstream Lead ions (Pb 2+ ) is poisonous Sodium sulfate and/or magnesium sulfate contain sulfate ions (SO 4 2 ), which can combine with Pb 2+ ions to yield lead(ii) or lead(iv) sulfates as precipitates: Pb 2+ (aq) + SO 4 2 (aq) PbSO 4 (s) Pb 4+ (aq) + 2 SO 4 2 (aq) Pb(SO 4 ) 2 (s) Once these precipitates are removed from the bloodstream, lead poisoning will no longer be an issue Key page 5

Solubility #3 (KEY) 1 a) KCl(aq) K + (aq) + Cl (aq) b) Ca(C 2 H 3 O 2 ) 2 (aq) Ca 2+ (aq) + 2 C 2 H 3 O 2 (aq) 2 *** Dissociate, and then calculate molarities based on mole ratios in the balanced equation a) K 2 SO 4 (aq) 2 K + (aq) + SO 4 2 (aq) 020 M 040 M 020 M [K + ] = 040 M [SO 4 2 ] = 020 M b) Na 3 PO 4 (aq) 3 Na + (aq) + PO 4 3 (aq) 0300 M 0900 M 030 M [Na + ] = 0900 M [PO 4 3 ] = 0300 M c) [MnCl 2 ] = = 0020 M MnCl 2 MnCl 2 (aq) Mn 2+ (aq) + 2 Cl (aq) 0020 M 0020 M 0040 M d) Al 2 (SO 4 ) 3 (aq) 2 Al 3+ (aq) + 3 SO 4 2 (aq) 0235 M 0470 M 0705 M [Mn 2+ ] = 0020 M [Cl ] = 0040 M [Al 3+ ] = 0470 M [SO 4 2 ] = 0705 M 3 a) FeCl 3 (aq) Fe 3+ (aq) + 3 Cl (aq) b) [FeCl 3 ] = = 0074 M FeCl 3 Then by 1:3 ratio [Fe 3+ ] = 0074 M and [Cl ] = 022M 4 [see #3 for sample calculations] [NH 4 + ] = 00400 M and [SO 4 2 ] = 00200 M 5 *** Always write the dissociation equilibrium and the K sp expression first! a) Molar solubility of CaSO 4 is 50 10 3 mol/l CaSO 4 (s) Ca 2+ (aq) SO 2 4 (aq) I 50 10 3 M 0 0 C (1:1) 50 10 3 M + 50 10 3 M + 50 10 3 M E 0 50 10 3 M 50 10 3 M K sp = [Ca 2+ ][SO 4 2 ] = (50 10 3 ) 2 = 25 10 5 Key page 6

b) Molar solubility of MgF 2 is 27 10 3 M MgF 2 (s) Mg 2+ (aq) 2 F (aq) I 27 10 3 M 0 0 C (1:2) 27 10 3 M + 27 10 3 M + 54 10 3 M E 0 27 10 3 M 54 10 3 M K sp = [Mg 2+ ][F ] 2 = (27 10 3 )(54 10 3 ) 2 = 79 10 8 c) solubility of AgC 2 H 3 O 2 is 102 g per 1000 ml Molar solubility = = 00611 M AgC 2 H 3 O 2 (s) Ag + (aq) C 2 H 3 O 2 (aq) I 00611 M 0 0 C (1:1) 00611M + 00611 M + 00611 M E 0 00611 M 00611 M K sp = [Ag + ][C 2 H 3 O 2 ] = (00611) 2 = 373 10 3 d) solubility of SrF 2 is 00122 g per 0100 L Molar solubility = = 971 10 4 M SrF 2 (s) Sr 2+ (aq) 2 F (aq) I 971 10 4 M 0 0 C (1:2) 971 10 4 M + 971 10 4 M + 194 10 3 M E 0 971 10 4 M 194 10 3 M K sp = [Sr 2+ ][F ] 2 = (971 10 4 )(194 10 3 ) 2 = 365 10 9 6 *** Let x be the molar solubility of the given compound a) AgCN(s) Ag + (aq) CN (aq) C (1:1) x M + x M + x M E 0 x M x M K sp = [Ag + ][CN ] = x 2 = 2 10 12 x = 1 10 6 M Molar solubility of AgCN = 1 10 6 mol/l; [Ag + ] = 1 10 6 M ; [CN ] = 1 10 6 M Key page 7

b) BaSO 4 (s) Ba 2+ (aq) SO 2 4 (aq) C (1:1) x M + x M + x M E 0 x M x M K sp = [Ba 2+ ][SO 4 2 ] = x 2 = 11 10 10 from the data booklet x = 10 10 5 M Molar solubility of BaSO 4 = 10 10 5 mol/l; [Ba 2+ ] = 10 10 5 M; [SO 4 2 ] = 10 10 5 M c) FeS(s) Fe 2+ (aq) S 2 (aq) C (1:1) x M + x M + x M E 0 x M x M K sp = [Fe 2+ ][S 2 ] = x 2 = 60 10 19 from the data booklet x = 77 10 10 M Molar solubility of FeS = 77 10 10 mol/l; [Fe 2+ ] = 77 10 10 M; [S 2 ] = 77 10 10 M d) Mg(OH) 2 (s) Mg 2+ (aq) 2 OH (aq) C (1:2) x M + x M + 2x M E 0 x M 2x M K sp = [Mg 2+ ][OH ] 2 = (x)(2x) 2 = 56 10 12 from the data booklet 4x 3 = 56 10 12 x = 11 10 4 M (divide by 4, then take cubic root) Molar solubility of Mg(OH) 2 = 11 10 4 mol/l; [Mg 2+ ] = 11 10 4 M; [OH ] = 22 10 4 M e) Ag 2 S(s) 2 Ag + (aq) S 2 (aq) C (1:2) x M + x M + 2x M E 0 x M 2x M K sp = [Ag + ] 2 [S 2 ] = (2x) 2 (x)= 16 10 49 4x 3 = 16 10 49 x = 34 10 17 M (divide by 4, then take cubic root) Molar solubility of Ag 2 S = 34 10 17 mol/l; [Ag + ] = 68 10 17 M; [S 2 ] = 34 10 17 M Key page 8

7 *** Let x be the molar solubility of AgI a) AgI(s) Ag + (aq) I (aq) C (1:1) x M + x M + x M E 0 x M x M K sp = [Ag + ][I ] = x 2 = 85 10 17 x = 92 10 9 M Molar solubility of AgI = 92 10 9 mol/l; Multiply by molar mass of AgI (2348 g/mol): 22 10 6 g/l *** Let x be the molar solubility of CuI b) CuI(s) Ag + (aq) I (aq) C (1:1) x M + x M + x M E 0 x M x M K sp = [Cu + ][I ] = x 2 = 13 10 12 x = 11 10 6 M Molar solubility of CuI = 11 10 6 mol/l; Multiply by molar mass of CuI (1904 g/mol): 22 10 4 g/l 8 [see #7 for sample calculations] molar solubility = 756 10 3 mol/l; solubility = 133 g/l 9 K sp for CaF 2 = 834 10 14 10 K sp for PbSO 4 = 18 10 8 11 K sp for Ag 2 CrO 4 = 14 10 11 12 K sp = [Ag + ][Cl ] = 18 10 10 from the data booklet solve: [Ag + ]*(60 M) = 18 10 10 [Ag + ] = 30 10 11 M 13 *** If they ask for K sp, don t just take the one from the booklet! *** K sp = [Ag + ][Cl ] = (34 10 6 )(50 10 5 ) = 17 10 10 14 [Cd 2+ ] = 8 10 14 M 15 omitted 16 K sp = [Pb 2+ ][IO 3 ] 2 = 37 10 13 from the data booklet [Pb 2+ ] [IO 3 ] 0035 M 33 10 6 M 83 10 8 M 21 10 3 M 34 10 2 M 33 10 6 M Key page 9

Solubility #4 (KEY) 1 dilutions: [BaCl 2 ] = 0200 M = 00800 M ; [NaCl] = 0400 M = 0240 M dissociations: BaCl 2 (aq) Ba 2+ (aq) + 2 Cl (aq) & NaCl(aq) Na + (aq) + Cl (aq) final concentrations: [Ba 2+ ] = 00800 M ; [Na + ] = 0240 M ; [Cl ] = 2(00800 M) + 0240 M = 0400 M 2 *** see #1 for sample calculations [Al 3+ ] = 00500 M ; [Ba 2+ ] = 0400 M ; [Cl ] = 3(00500 M) + 2(0400 M) = 0950 M 3 K sp = [Cu 2+ ][S 2 ] = 60 10 37 from the data booklet solve: (020 M)*[S 2 ] = 60 10 37 [S 2 ] = 30 10 36 M 4 *** see #3 ***: [F ] = 11 10 4 M 5 formula equation: Pb(CH 3 COO) 2 (aq) + Na 2 SO 4 (aq) PbSO 4 (s) + 2 NaCH 3 COO(aq) potential precipitate is: PbSO 4 (s) solubility equilibrium: PbSO 4 (s) Pb 2+ (aq) + SO 4 2 (aq) K sp = [Pb 2+ ][SO 4 2 ] = 18 10 8 from the data booklet dissociations: Pb(CH 3 COO) 2 (aq) Pb 2+ (aq) + 2 CH 3 COO (aq) Na 2 SO 4 (aq) 2 Na + (aq) + SO 4 2 (aq) dilutions: [Pb(CH 3 COO) 2 ] = 10 10 3 M = 20 10 4 M [Pb 2+ ] = 20 10 4 M [Na 2 SO 4 ] = 15 10 4 M = 12 10 4 M [SO 4 2 ] = 12 10 4 M Trial ion product: trial Ksp (or Q) = (20 10 4 )( 12 10 4 ) = 24 10 8 Since trial K sp is greater than K sp, a precipitate will form 6 dilutions: [Mg 2+ ] = 10 10 2 M = 25 10 3 M [CO 3 2 ] = 20 10 2 M = 15 10 2 M K sp = [Mg 2+ ][CO 3 2 ] = (25 10 3 )(15 10 2 ) = 38 10 5 Key page 10

7 ***see #6 ***: K sp = [Ag + ][I ] = 88 10 17 8 K sp = [Tl + ][I ] = 89 10 8 solve: (0001)*[I ] = 89 10 8 [I ] = 89 10 5 M 9 a) [Ag +! ] = = 92 6784 10 3 M (2 sf) K sp = [Ag + ][Cl ] = 18 10 10 from data booklet solve: [Cl ] = 19 10 8 M b) [Cl ] = 20 10 5 M c) [Cl ] = 50 10 2 M 10 dissociations: Na 2 S (aq) 2 Na + (aq) + S 2 (aq) Pb(CH 3 COO) 2 (aq) Pb 2+ (aq) + 2 CH 3 COO (aq) ***double dilution: [Na 2 S] = 10 10 7 M = 50 10 11 M [S 2 ] = 50 10 11 M dilution: [Pb(CH 3 COO) 2 ] = 76 10 17 M " = 68 4 10 17 M [Pb 2+ ] = 68 10 17 M potential precipitate is: PbS(s) solubility equilibrium: PbS(s) Pb 2+ (aq) + S 2 (aq) K sp = [Pb 2+ ][S 2 ] = (68 10 17 )(50 10 11 ) = 34 10 27 11 ***see #5***: Trial ion product = 67 10 5, which is greater than K sp of AgBr (54 10 13 ) So, yes, a precipitate will form 12 Trial ion product = 80 10 9, which is greater than K sp (37 10 15 ) So yes, a ppt will form 13 Trial ion product = 15 10 21, which is greater than K sp (50 10 22 ) So yes, a ppt will form 14 Trial ion product = 50 10 18, which is greater than K sp (37 10 19 ) So yes, a ppt will form 15 [Cu 2+ ] = 17 10 4 M 16 [Cl ] = 4 10 8 M (rounded from 45 10 8 M) Key page 11

Solubility #5 (KEY) 1 To increase solubility of PbCl 2 (diverse ion effect): (i) add PbCl 2 to a solution that precipitates with Pb 2+ For example: NaBr, K 2 SO 4 add PbCl 2 to a solution that precipitates with Cl For example: AgNO 3, CuNO 3 To decrease solubility of PbCl 2 (common ion effect): (i) add PbCl 2 to a solution that already contains Pb 2+ For example: Pb(NO 3 ) 2 add PbCl 2 to a solution that already contains Cl For example: NaCl 2 To increase solubility of Sr(OH) 2 (diverse ion effect): (i) add Sr(OH) 2 to a solution that precipitates with Sr 2+ For example: Na 2 SO 4 add Sr(OH) 2 to a solution that precipitates with OH For example: Pb(NO 3 ) 2 To decrease solubility of Sr(OH) 2 (common ion effect): (i) add Sr(OH) 2 to a solution that already contains Sr 2+ For example: Sr(NO 3 ) 2 add Sr(OH) 2 to a solution that already contains OH For example: NaOH 3 Solubility represents the maximum amount of solute that can exist as ions in a solution at equilibrium The solution has become saturated No more AgCl can dissolve! 4 Solubility represents the maximum amount of solute that can exist as ions in a solution at equilibrium The solution has become saturated Adding more water can dissolve more solute, but in the same proportion as dictated by the original solubility For example, if the solubility of a substance is 38 grams per litre of solution, one litre of solution can dissolve 38 grams Two litres of solution can dissolve 76 grams, but the solubility still still the same (76 grams per 20 L = 38 g/l) 5 To remove the CaCO 3 deposit on the pipes, one can increase the solubility of CaCO 3 by diverse ion effect We can rinse the pipes with a solution that precipitates with Ca 2+ ions (such as NaOH), or a solution that precipitates with CO 3 2 ions (such as Fe(NO 3 ) 2 ) 6 SrCl 2 would be most soluble in Na 2 SO 4, as SO 4 2 ions would precipitate with Sr 2+ ions SrCl 2 would be least soluble in 1M MgCl 2, since MgCl 2 contains twice as much common ions (ie Cl ) as in Sr(NO 3 ) 2 7 Most soluble ----------------------------------------------------------------------------- Least soluble 2M AgNO 3 > 1M AgNO 3 > 1M KNO 3 > 1M NaCl > 1M Na 2 SO 4 highest [Ag + ] no common or diverse ion highest [Na + ] Key page 12

Solubility #6 (KEY) 1 [Br ] = 00282 M 2 volume of I = 176 ml 3 [Br ] = 0150716 M [NaBr] = 0150716 M moles of NaBr = 0150716 M * 200 L = 0301432 mol mass of NaBr = 0301432 mol * 1029 g/mol = 310 g NaBr 4 omitted 5 986% pure Solubility #6a (KEY) 1 volume of NaCl = 160 ml 2 mass of Cl ion = 00717 g 3 [KI] = 00961 M Key page 13