Question

1. (a) 0.05 mole of MgBr2 is dissolved in 1 liter of solution and the following species form: Mg2+, Br and MgOH' (a soluble hydroxy complex of Mg?) write a mass balance on the Mg species. Write a mass balance relating the molar concentration of Br to that of Mg species (b) write a charge balance for the solution in Part a.

please explain. thanks

Answer 1

According to the condition given in question Mass of Mg BQ = 005 x 124 +80x2) = 9.29. 2 mg Bog & H40 - Mgt2 + 4 B8 + mg (0H) + mass of my in Mg Bog = Mass of Myth - + mass of my in mg (OH)t Mobs of mg+2 = 0,05 = moles of mg oht

mass of Mg +2 = 0.05 x 24 Mass of mg in mg (oh) = 0,05 (24) Total mass of mg in product = 10:05 05 + 0.05 ) 24. 0.05 = 0.65 x 24 moles of my in MgBrg = 0:05 Mass of my Pn reactant = 0.05x24. Mass in reactant = mass in product [er'] = 0.05x2=0:1 . [mgt?} = 0.05 [mg (oh) 3 = 0,05 [B0-] = 0.1*80 = 8 [mgt2 ] = ,05 x 2430.06. [Mg Com)+] = 0.05 x 24 30.06 2 So tatal mass = Initial mass of MgBog = 9.29

b) The charge balance is given by 2 mg B6, + H20 --> Mg+2+480 + Mg low) H+ Emg +2 + 4 B8 T HGO Mg +²4 435 + Mg lotit overall charge as you can see from both ends are "zero"