how do you find the Ca2+ concentration. with charge balance and mass balance. th

Question

how do you find the Ca2+ concentration. with charge balance and mass balance. thank you. please show all work.

Write the charge and mass balance for aqueous Cay(PO4)2 if the present species are Ca^2+, CaOH^+, CaPO4^-, PO4^3-, HPO4^2-, H2PO4^-, and H3PO4. Find the concentration of Ca2^+ in a saturated solution of CaSO4 Including the change balance and mass balance equation in your answer, and explain All approximation made. (You can stop your work after the first cycle calculation of both activity coefficients and Ca^2+ concentrations). Given:

Explanation / Answer

M2+ + X! "MX+ K1 = [MX+ ] [M2+ ][X! ] MX+ + X! "MX 2 (aq) K2 = [MX 2(aq ) ] [MX+ ][X! ] [MX+ ] = K1[M2+ ][X! ] [MX 2(aq ) ] = K2[MX+ ][X! ] = K2K1[M2+ ][X! ] 2 Mass balance equation: MX 2 = 0.10 mol 1L = 0.10 M = [MX 2(aq )]+ [MX+ ]+ [M2+ ] = = K2K1[M2+ ][X! ] 2 + K1[M2+ ][X! ]+ [M2+ ] Charge balance equation: 2 [M2+ ]+ [MX+ ] = [X! ] By substituting of [M2+ ] and [MX- ] we have an expression for the [X- ] 2 [M2+ ]+ K1[M2+ ][X! ] = [X! ] By solving this equation for [X- ], we have: [X! ] = 2[M2+ ] 1! K1[M2+ ] Now we put expression for [X- ] into the mass balance equation: 0.10 M = K2K1[M2+ ] 2[M2+ ] 1! K1[M2+ ] " # $ % & ' 2 + K1[M2+ ] 2[M2+ ] 1! K1[M2+ ] + [M2+ ] 3. When ammonium sulfate dissolves, both the anion and cation have acid-base reactions: (NH 4 )2SO4 !2NH 4 + +SO4 2" Ksp = 276 NH 4 + !NH 3 (aq) + H+ Ka = 5.7"10#10 SO4 2! + H 2 O "HSO4 ! + OH! Kb = 9.8#10!13 a. write a charge balance for this system [NH4 + ] + [H+ ] = 2 [SO4 2- ] + [HSO4 - ] + [OH- ] b. write a mass balance for this system [NH3] + [NH4 + ] = 2 {[SO4 2- ] + [HSO4 - ]} c. find the concentration of the NH3(aq) if the pH is 9.25 [H+ ] = 10-9.25 and [OH-] = 10-4.75 [NH 3 ][H+ ] [NH 4 + ] = 5.7!10"10 = [NH 3]!10"9.25 [NH 4 + ] # [NH 3] [NH 4 + ] =10"0.75 =1.014 Therefore, [NH3] = 1.014 [NH4 + ] From another hand, [HSO4 ! ][OH! ] [SO4 2! ] = 9.8"10!13 = [HSO4 ! ]10!4.75 [SO4 2! ] # [HSO4 ! ] [SO4 2! ] = 9.8"10!8.25 = 5.51"10!8 Therefore, [HSO4 ! ] = 5.51"10!8 [SO4 2! ] Putting these values of [NH3] and [HSO4 - ] into the mass balance gives: 1.014 ![NH 4 + ]+ [NH 4 + ] = 2 [SO4 2" ]+ 5.51!10"8 [SO4 2" ( ]) 1.007![NH 4 + ] = [SO4 2" ] Now an expression for the Ksp is used: Ksp = [NH 4 + ] 2 [SO4 2! ] =1.007 [NH 4 + ] 3 " [NH 4 + ] = 276 1.007 3 = 6.5M [NH 3 ] =1.014[NH 4 + ] =1.014 ! 6.5 = 6.59M

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