Question

A solution contains Al3+ and Co2+. The addition of 0.3779 L of 1.739 M NaOH results in the complete precipitation of the ions as Al(OH), and Co(OH),. The total mass of the precipitate is 23.32 g. Find the masses of A18+ and Co2+ in the solution. mass of A13+: mass of Co2+:

Answer 1

Moles of Naoh added = Molarity x volume (lit") = 0.3779 X 1.739 = 0.657 moles - Aist and co2t will react as; Apt + 3NaOH ALCOH)2 + 3 Nat ---C02+ + 2NaOH Co. (OH)2 + 2Na+ Suppase; moles of Alpt in the solution=n moes of colt am So, according to the above reaction, moles of NaOH reacted with Al3+ = 3n moles of Nach reacted with co2 = 2m but total available moles of NaOH = 0.657 13n+ 2m = 0.657 - (1) Now, moles of AI COH)bermed acceeding to stoichio - - Metry of the reaction = n similarly, moles of coloth formed = m. so mass of Al(OH), = moles x Moler mass 21x78 3.78n_gm - mass of coCOH), = mx g 3 = 93m gm but, given the total moss of ppt = 23.32gm 178 +93 m = 23.32 -(2) multiplying the eq? (12 by 26 and substracted from (2) So, 30,

. 78 n +93 m = 23.32 789 752m = 17:08 41m = 6.24 = [m = 0.152) C10 - 3n+2(0.152) = 0.657 om = 0.1181 from mass of AMF = moles of A1Bf(n) x Molare mass - 0.118 * 2 7 = 3.186 gm. ang = 0.152 X 59 = 8.968 gm