posted by Abril .
A solution is prepared by mixing 0.12 L of 0.12 M sodium chloride with 0.22L of a 0.19M MgCl2 solution.What volume of a 0.22 M silver nitrate solution is required to precipitate all the Cl- ion in the solution as AgCl ?
First determine the moles chloride ion.
moles = M x L = 0.12L x 0.12 M NaCl = xx moles
moles = M x L = 0.22 L x 0.19 M MgCl2 = yy moles MgCl2. Multiply that by 2 since there are 2 moles Cl in each mole MgCl2 to obtain the moles chloride in that solution. yy moles x 2 = zz moles.
Total moles chloride = xx + zz = total.
What volume of 0.22 M AgNO3 will ppt the total moles. M = moles/ L. You know moles and you know M, calculate Liters.