posted by Jimmy on .
A 35ml 0.1M AgCl solution is mixed with 15ml 0.25M MgCl2 solution. Calculate the moles and molarities of the Ag+, NO3-,Mg2+, and Cl- ions in the solution mixture.
I think it was 35 mL of 0.1M AgNO3, not AgCl. When the two solutions are mixed the reaction is:
2AgNO3(aq) + MgCl2(aq) ---> AgCl(s) + Mg(NO3)2(aq)
Moles of AgNO3 = (0.035L)(0.1mol/L) = 0.0035 moles of AgNO3. Moles of Ag+ ion = 0.0035 also.
Likewise, moles of MgCl2 = (0.015 L)(0.25 mol/L) = 0.00375 mol MgCl2. Moles of Cl- ion = (2)(0.00375) = 0.0075
The mole ratio of AgNO3 to MgCl2 is 2/1. However, the ratio used is 0.0035/0.00375 which is much smaller. That means AgNO3 is the limiting reagent (gets used up) in forming the precipitate AgCl. Some of the chloride ion is also used:
0.0075 - 0.0035 = 0.0040 moles of Cl- left in slution.
Total volume of solution = 15 + 35 = 50 mls = 0.050 L
From what I have shown you so far,
Final concentration of Ag+ = 0
Final concentration of Cl- = 0.0040 mol/0.050 L, or [Cl-] = 0.080 mol/L
Final concentration of NO3- = 0.0035 mol/0.050L = 0.070 mol/L
The final concentration of Mg+2, I leave to you.