Question

1. Finish the mass-balance expressions for a 0.0100 M solution of HCl that is in equilibrium with an excess of solid BaSO4? The three pertinent equilibria in the solution are: BaSO4(s) = Ba2+ + 502- S02- + H20+ = HSOZ + H2O 2H20 = H30+ + OH- [Ba2+] =

Answer 1

There are three equilibria in this solution! Basou (s) - Ba2+ + sou S02 ++30° = Hsou +H20 -------(2) 2H20 - H₂O + + OH annnnn.. (3) As the one and only source of two sulphate species is the dissolved Basoy, the concentration of Ba 2+ must be equal to the total concentration of sulphate containing species, so we can write the first mass-balance equation – [Ba2+] . [s0,2-] + [#504] According to the equation (2) hydronium ions in the solution are either free H₂ot or they react with sulphate ion to form bisulphate ion CHSOuZ so, we can write. [H30') - [13.0t7+ [hson] -------(4) where [H₂ other is the concentration of tot from all sources. and [H₂O + 7 is the free exuilibrium. concentration of hydronium ion. The protons that contribute to [H3O+] total have two sources - aqueous Hcl and the dissociation of water. Here we refer to the hydronium ion concentration from the complete dissociation of Hal. So total concentration of Hoot is total [190 *Itopad [H043n u + C 430 43.0 --- --(W)

comparing equation (1) and (5) we can white [H20" Lotal = "Hg 0] + [750] - [H204) fiel + [H30*34/20 But [H 3 Ota Ha(e), and since the only source of hydroxide is the dissociation of water. Then we can write [H3O+] = [OH- ] By substituting these two into the exquation above- (H 30 + lotal = [H 3 Ot] + [Hson] = HU(R) + (8th] and the mass-balance equation is then [113 0+7 + HSO4 = 0.0100 +1047 Here Ha(e) is concentration of the. total NO Hence [Ba2+] = [sour] + [h sou] and the finished mass. balance exuation is [H3O+] + [Hsoi] = 0.0100 + [04]