Question

Please explain solution in detail to the below problem -

Q) If you mix 48 mL of 0.00012 M BaCl₂ with 24 mL of 1.0 X 10^-6 M Na₂SO_4, Will a precipitate of BaSO₄ form?

Answer 1

no of moles of BaCl2 = molarity * volume in L

                                    = 0.00012*0.048   = 5.76*10^-6 moles

BaCl2 --------------> Ba^2+ (aq) + 2Cl^-

5.76*10^-6 moles    5.76*10^-6 moles

total volume = 48+24   = 72ml   = 0.072L

molarity of Ba^2+   = no of moles/total volume in L

                               = 5.76*10^-6/0.072   = 8*10^-5 M

no of moles of Na2So4 = molarity * volume in L

                                      = 1*10^-6 *0.024   = 2.4*10^-8 moles

total volume = 48+24   = 72ml   = 0.072L

   Na2So4 --------------------> 2Na^+ (aq) + SO4^2-

2.4*10^-8 moles                                         2.4*10^-8 moles

molarity of SO4^2-     = no of moles/total volume in L

                                  = 2.4*10^-8/0.072   = 3.33*10^-7 M

Qsp     = [Ba^2+][SO4^2-]

           = 8*10^-5 *3.33*10^-7   = 2.664*10^-11

Ksp = 1.1*10^-10

Ksp>Qsp will not form precipitate BaSo4