Question

Effect of common ions and pH on Kap a. Consider calcium carbonate CaCO3 (limestone) with a Kn of 4.96 x 10. What is the molar solubility of CaCO3? b. What is the molar solubility of CaCO3 at pH 5.0? What is the molar solubility of CaCO3 in a solution of 0.100 M NaCO3(aq)?

Answer 1

Answer Now a) Calog (t) = Cal* + 2032- Given Ksp = 4.968109 Solubility product Ksp - Tea?"] [203] If we call the solubility s, then s= [ Ca2+] = [ co3? -] Ksp = [cat] [6032-) 4.96x10) sxs , 4.96 X10-9 so V6.96X10-9 S, 7.04 xio's mol. Solubility - 7104 xros molila. 0 Cacoz dissociates by ca cos(s) = * (aq) + Coler) Carbonate will be distributed as co? Heo and H₂CO3

г јн, 74ka, Гно) +ко, к. ЕН,0 + H2 3 = Hзо* + + год — и )+o1] [н соз") Lus xiom 1H2 сез] " н, он не е њо* + 0,2- ka, , Інo') Гn') - 69x10-11 1603) The a expressions are в , Тв, соз) , јно)”. «І, Іно, ) , К, Ін,o') [но")"+ K4) [+ър") + Kar Ku 4 4, Гео) , Ко, и.. А Гн'], К., Г»,6 ) + Ke, Ka, | At PH - [14, 07 - 05 «, , 45х76+xy 6x - II (ro-3)+(45У 10 (5)+ 5xi': :49xto" = 1'000 Х10-? Ел

to find the molar colubility we use the le to find the amounts of solution phase table to find species the a - CaCO3(s) ca 2 (aq) & cor (94) initial solid change enge -x - tx ta final solid x a 1. The ksp equation is east in the form 1 ksp = [Ca 2+] X₂ FA Using the valuses from the table 1 ksp = d₂2 where x is equal to [Ca 2+ ] FA and is the molar for solubility of CaCO3 x= ſkape 14.96X10-9 V 199x107 A 10:0249 x=0'16 moli The molar solubility of CaCO₃ at pH 5 - is equal to 0'16 nol I.

6 Caloz is dissolved in oilm Na2CO₃ solution . Cacolo) cat (aq) + cos (aq) do is already Initial solid 0 0'l present in solwion Change -n tx ta equibrium solid x (olta) Ksp = [ca?"] [ cos] 4.96X169-12 1 (0:1 + ) 4.96X169, 011x + a xis very small the tog ñ can be neglegible 4.96X10-9 20'1x x = 4.96X1009 O x = 4.96410 moli The molar solubility of CaCO3 in a solution of oim Na₂CO3 is equal to 4.968 108 mol i