Question

a) When 12.0 mL of a 6.97×10^-4 M cobalt(II) nitrate solution is combined with 25.0 mL of a 3.58×10^-4 M sodium hydroxide solution does a precipitate form?   (yes or no)

For these conditions the Reaction Quotient, Q, is equal to .

b) When 25.0 mL of a 3.03×10^-4 M nickel(II) bromide solution is combined with 18.0 mL of a 3.70×10^-5 M potassium carbonate solution does a precipitate form?   (yes or no)

For these conditions the Reaction Quotient, Q, is equal to .

c) When 15.0 mL of a 4.05×10^-4M sodium phosphate solution is combined with 25.0 mL of a 9.04×10^-4M zinc iodide solution does a precipitate form? (yes or no)

For these conditions the Reaction Quotient, Q, is equal to .

Answer 1

can gole First to determine the concentrations of coat cation and oth anion are hige enogn to allow for the formation all Salt precipitates out GOOH) in soluble of solution The dissociation of Colmaz. cobalt(") nitrate can be expressed as, Co (MO3)2 + (aq) + 200zlag cas so one mole of Co (MO) No ion mole Co2+ and a mole in solution. soles of 62+ present in 120 m2 of a 6.97 x264 in CO (NO₂), solution, (1L = loro mw) 1000 L = 8.364 X Lobo sodium hydroxide (Maolt), the can be written as, Maof (an Mat (ag tou (ar - so every mole of Mash gives one mole Nat I and one more of ion in solution. i holes of OH in present in 25.00m2 of 3.58 x 10rum solution, how = cx VMaoH = 3.588104 more IL Xisco Total volume of solutie = 8.95*10b mole Vsolution (Vo(No + MaoH) nco?? - C x VCO (NOR).- 6.97 X16 mol 12. IL Similarly , dissociation for عاه m iu ton, ( 12 + 25 m2 o now the concentration = 37 mL resulting Solution will Gott and oth in the [ Go²+] a n co2+ 8.364810 6 more solution 37x10 31 = 2 26x10 now- 8.95 xloro male and [OH-] = - -4 solution - 37x10 3 = 2.4 18x10 m

AS 5.90 X10-15) -X103 • The disso nation equilibrium of 60 (04) can be written as, Co (OH) (s) = colt cad) + 2OH(aq) i Solubility product a usp=[ Co2+] [ OH- 598x10-15 = [ Co2+] [OH-] Now reaction quotient, a =[6024] [OH-] = 12.26x104) (2.418 4104) = '1-321X10") ksp BY Ksp (solubility Precipitation products or takes place. MES! Real tion quotient s = 1.321x10 mo first to determine the con antrations of Ni2+ cation and co2- anion are high enough to allow for the formation on Mi coz insoluble talt precipitates out of solution • The dissociation of NiBro, nickel bromide canbe expressed as Nibra (6) ► Niat cap +2 Brican so, one more of Nibry gives one mole of Ni2+ ion and a more of Br non solution - moles of mist hoiat cx Vw; 2+ = 3103x10 mole 25 1_xol • The dissociation = of of ke 7.575 8156 mole can be Kalog, written potassium carbonate as, K2 coz (a ) - 24+ (ay) + coz.com So, one more of kz co 3 gives one mole of 60 and a mole of kt con in solution, i moles of coz ion in 18.0mL of 2.70x105m solution, = CXVkzcoz = 3.70 x 10 mole = 6.66x107 mole 2 is L

- 43x163L mole - 1.7616x104m 43 x103 L = 1.5488x105m Total volume in solution Vsolation = ( Uni Brg + V K2 103) ? ( 25.0 +18.0) mL now, the = 43m2 concentrations of Milt and Gogh in the resulting solution will be I Ni2+ 47.575 x 10-6 more and [co - 6.66610-7 mole - 1.5488x105m 0 The dissociation equilibrium of Nico can be dritten as Nicoz (s) F Mist (as) + 1032 (99) i solubility product, Ksp = [Milt] [co ] 9 1.42 x10-7 - ('Mi2+][co] Now the reaction quotient, a 9= Emilt] [ cog-] = (1.7616X864) (1.5488 x1015) = 2.7284 8109 << ksp s kap) 9, the precipitation ce take place. can not - Все о How o Неть ф. 212 8ч х10 Ч МУ e first to de termine the concentrations of Zn2+ ion and phosphate (po 3- ion are high enough to allow for the formation of 7m3 (po42 insoluble salt precipitates out of solution . The dissoud Hon of sodium phosphate, Nazpoy can be expressed as . Mag poy cay 3 Nat can & pou (at) so, one more of Nazpoy gives B mole of Nat and one mole of por ion in solution i moles of poy in in 15.0 m2 of 4.05 x 10tum solution, - 405x10-4 mole por 15. ex 15 - 11 xlorot = 6.075x10-6 mole

Todo Zoe Iz and znic iodide dissociates as Zn Iz cas y Zn2+ (aq) + 2 I Can gives one one mole of Zn I mole at Zn²tion and solution, two two - - in mole of mole of 'I- I- ion ion . Moles, of zo2+ ion in 25.0m2 of 9.048 104m Salution, 9.048 104 mole & 25 L IL Total volume - 2.26x9 05 mole of solution Vsolution = (V Mazpoyt Van Iz ) now, - (15 +25) ma the concentrations in solution of 22+ = 40m will be and pozion [ Zn2+] = nenet 2126x155 mol - Vsolution 40 x 10 3L [ po 3-] - np., 3 = 5.65x10-4m | VIA . . = 6 015 016 8 mek 40 x 10-3L The dissociation = 1518 x 10-4 M equilibrium of be written as, Znz (po42 can Znz (pon) ₂) 3 202 + (at) + 2 por can a solubility product, s Ksp = [224] P[ Powdy 7 9.0x1733[220]3[ pory? I Reaction quotient, 8- tzn 2033 pogoje =( 5.65 x10433 ( 1.518 X10yr = 4,1561 X 10-18 » ksp ksp, Csolubility produc. 69.0x1033) will takes place [Yes Reaction quotient, o 4.156 vol85 - As 87 kgp, (solubility produet), Precipitation