Question

A solution contains A1²+ and Co2+. The addition of 0.3814 L of 1.662 M NaOH results in the complete precipitation of the ions as Al(OH), and Co(OH),. The total mass of the precipitate is 23.01 g. Find the masses of Alº+ and Co2+ in the solution. mass of Al3+: g mass of Co2+: mass of CoP+: [ 60

Answer 1

According to the condition given in question we have moles of NOOH = MXV n = 1.662 x 0.3814 ne 0.6338 mol Solution contains Al +3 and cote which on addition of Naolt gives Al+3+306' -> Alloh)3 co +2 + 204 → ColoH) the total mass of precipitate = 23.019 mass of Al 10H)3 = X9 mass of CO (OH) = (23.01- * ) moles of AI COHg = gooms of Allontz molad mass.

3x 789 = 23.01-2 moles of colom), ११.१५ Now total mol t= 0.6338 - + 2 (23.01 -1) - -0.6338 ११.१५ 0.038461 + 2x23.0/ 2x ११.१५ 20.6338 ११.१५ .038464 +0,495 -0.215d 20.6338 0.01696 1 = 0.1388 ____L = 8.18 grams

So mass of AI COHg = 8118 9 moles of AI (OH)3 = 8.18 = 0.10 49 78 morers of Altse 0.1049 moles. m moles of Alt3 - mass of Alt3 molar mass 0.1049 = WCAL+3) 26198 w(Al+3) = 0.1049 x 26.98 w (Al +3) = 2.83 g

Mass of 10 (OH)ą = 63.01-9) 14.019 moles of colo), = 14001 92.94 : 0.1507 w (60+2) = 0.1507 x 58.93 Wat Crot?) = 888 9 Mass of A4+3 = 2.83], Mags of co+= 1888],

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