Question

Consider 1.0 L of an cu solution that contains 0.050 molsufaric acid to which 0.27 molarium nitrate is the pl the concentration of burim ions in the final solution, and the mass of solid formed ded. Assuming changes volume of the solution, determine Ka(,0) 1.2 x 10- K., (BaSO) = 1.5 x 10" (Bar) - PETER Mass of Baso, produced Try Another Version to the attempts remaining

Answer 1

Comment in case of any doubt.

Given Ksp (Basou) = 1.5X6%, the solubility equilibrium of Basoy is as follows: Basoy(s) = Böttcas) + Sos/ag), Keçi Ksp 1.5x17° Then equilibrium equo constant of reverse of above equation is given by keq = this p = TSxgE 6.67X108 (2 Bapt caz)+ souca) 2 BasOys), Keq hace 1:5869 = 6.67 X108 Eque 1.2x102 Equilibrium for Kag of Has Oy is as follows: üs HS OT109) + H2O1l Hz0+(29)+ som Cag), Kaş 1-2017 Adding is and il gives: Baztcaq) + HSOT (AL) + Hold) = BasOy(s) + H2O*(aq),kép = 6.67810* Sime kéz is very high, so above reaction 2 = 8X10C almost goes to completion and limiting reactant, which is Hsou will be fully consumed . Thus, moles of Basoy produced = moles of HSO4, which . is limiting reagent = 0.050 mal and [Ba2+] left = 0.27-0.05= 0.22 mal And tt produced in solution = ax males of so, initially as My Soy almost dissociate I son pH = -log(0:1) = 1 Mass of Rasoy produced = 0.05 mol X-233.389 = 11.679 Crazt] = 10.32]m [ 13.91 = 2X0.05 = 0.10mal