(a) Will Mn(OH)2 precipitate from solution if the pH of a 0.050 M solution of MnCl2 is adjusted to 8.0? (b) Will Ag2SO4 precipitate when 100 mL of 0.010 M AgNO3 is mixed with 20 mL of 5.0 x 10-2 M Na2SO4 solution?

Solubilty product of Mn(OH)2 is, ksp = 2 × 10-13Solubilty product of Ag2SO4 is, ksp = 1.5 × 10-5 To determine whether the precipitation is formed or not for the given solutions.In (a) pH of the solution is 8.0 so pOH = 14 – pH = 6. It means that the concentration of OH- is, [OH-] = 10-6 M Concentration of MnCl2 is 0.05 M so concentration of Mn2+ is also 0.05 M i.e. [Mn2+] = 0.05 M. Mn(OH)2 ↔ Mn2+ + 2OH- Ionic product of Mn(OH)2 is Q=Mn2+OH−2Q=0.0510−62Q=5×10−14 Since ionic product of Mn(OH)2 is less than its solubility product therefore no precipitation is formed.In (b) after mixing, final volume of the solution is 100 ml + 20 ml = 120 ml = 0.12 L Number of moles of AgNO3=ConcentrationVolumeNumber of moles of AgNO3=0.010 M0.1 LNumber of moles of AgNO3=0.001 mol Since AgNO3 contains 1 Ag+ therefore number of moles of Ag+ is same as that of AgNO3. Concentration of Ag+ in the solution is given by, Ag+=Moles of Ag+Volume of solutionAg+=0.0010.12Ag+=0.008333 MNumber of moles of Na2SO4=ConcentrationVolumeNumber of moles of Na2SO4=5.0×10−2 M0.02 LNumber of moles of Na2SO4=0.001 mol Since Na2SO4 contains 1 SO42- therefore number of moles of SO42- is same as that of Na2SO4. Concentration of SO42- in the solution is given by, SO42−=Moles of Ag+Volume of solutionSO42−=0.0010.12SO42−=0.008333 MAg2SO4 ↔ 2Ag1+ + SO42- Ionic product of Ag2SO4 is Q=Ag1+2SO42−Q=0.00833320.008333Q=5.79×10−7 Since ionic product of Ag2SO4 is less than its solubilty product therefore no precipitation is formed.In (a) no precipitate of Mn(OH)2 will form. In (b) no precipitate of Ag2SO4 will form.