Question

A solution contains Al3+ and Co2+. The addition of 0.3561 L of 1.650 M NaOH results in the complete precipitation of the ions as Al(OH)3 and Co(OH)2. The total mass of the precipitate is 21.11 g. Find the masses of Al3 and Co2+ in the solution.

Answer 1

035612 , 1.650 in Nash moles = molarity & volume a moles of Naott - 0.3561 X 1650 = 0.588 mol 1. Naoll - Natton a moles of off = moles of NaOH = 0.588 mal à moles of OH = 0.588 mof - i) out of alllam precipitate, Let 2 gm is Allot) 3 and (21-11-22) gm is CoCopla 1 molar mass of Allolla = 78 glmal a moles of Alloh!3 = To Al (H)2 = Al3+ + 304" a moles of om = 34 mm - 3 de mel - je molar mass of ColoHl2 = 93 simul a moles of Co Co M12 = (21:11–2) mel 93 CO (OH)2 + CO2+ + 2014 a moles of CoH) = 2x (21-11-2) = 42.2223 ml - forom each & 0.588 = 3x + 4222-22 S3 a e' = 788 con mass of AlloH)3 = r = 7.88 gm mass of co COH)2 = 21.11-= 13.23 gm a moles of Allott3 = 7,88 = 0.101 mel 70 moles of ColoHl2 = 13:23 = 0.142 may CScanned with CamScanner 93

Allohn - Alt + 3011 moles of Al3+ = 0.101 mal molan mass of Al3+ = argimel a s (mass of Al3+ = 0101427 = 2.73 gmy answer, Co (0412 G Co2+ + 2OH = moles of co2t = 0-142 mal molar mass of colt = sg almol n (mass of Co2+ = 0.142X 59 =8-39 gm answer CScanned with CamScanner