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6) A solution contains equal concentrations of Cu?", Co?, Fe?', and Mg". If we add 0.050 M sodium hydroxide (NaOH) to this solution dropwise, which ion will precipitate LAST? Compound K Cu(OH) 2.2 x 10 CO(OH), 5.92 x 10- Fe(OH) 4.87 x 10- Mg(OH) 2.06 x 10

Answer 1

In this case, the common-ion effect has a greater role. NaOH is considered strong electrolytes and therefore NaOH will suppress the dissociation of Cu(OH)2, Mg(OH)2, Co(OH)2 and Fe(OH)2 . By the addition of NaOH solubility decreases. A detailed description of the answer is attached below

It Page I. £n: Solubilily equillibrium of CaCOHA 2. Kep = [cu?+] [O-]>= $(25)3453 Adding 0.05 M Naot, newn solubility is assumed as ste Cu 2+] = st. COH-] = 0.05 +25 nalar - 20 1/ kop [cucot 77 = s' (0.05+25) 2_2.2x10² cgiven) Assume es <<< 0.05 egh becomes s'(0.05) - 2.5810351 = 2.2x10² 5 = 8.8x10-18 M -20 Similarly calculating other three (brief ) • COLOH) . [cont] [ot] 2 3 4 5 New molar solubility s Ksp - SC0.05 405')25.92 xro-15 neglecting a5 2.5410 3 = 5.92X1075 s'= 2.368x60 12M

A Page : 2. • Tecol Fe 2+ + 2600) = 45 Adeling 0.05m NagH on S'C0.054 28') ? 24,87x1067 piest 10%s' 24.83X10* s ' = 4.87X1647 - 1948X10M 7.5X103 • MgCOH, Adding 0.05M Naoh, neglecting 28 UK 2.5X10 3 3! = 2:06x100 3P IP s = 2:0661013_ 8.24 x 10 m 2.5x103 Solubility order MacOH > COCOH)2 > FeCoroa > Cu COH)2 8.24x10 " x 2.36x10-12 m > 1.94881044 3 8.8x10 As solubility increases it will least precip Date Higher soluble compound precipitate last: Mo (OH) will Precipitate last.