A solution contains 0.133 g of dissolved lead. How many moles of sodium chloride must be added to the solution to completely precipitate all of the dissolved lead? What mass of sodium chloride must be added?
Here, we have to determine the moles of NaCl and mass of NaCl.
Calculation of the mass of lead which is dissolved into the solution.
Molar mass of Lead =207.2 g/mol
Mass of lead dissolved in the solution = 0.133 g
Therefore, mole required for Lead is = = 0.000642 mol
Calculation for the mole of NaCl required to dissolve the precipitation.
Pb (s) + 2 NaCl(aq)-------> PbCl2(aq)
Therefore, 2 mole of NaCl required to dissolve the ppt = 2x 0.000642 mol =0.00128 mol
Thus, NaCl required to dissolve the lead precipitation = 0.00128 mol =1.28 x 10-3 mol