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 solution.
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.
To calculate the moles and molarities of the ions in the solution mixture, we need to use the following formulas:
Moles = Molarity x Volume (in liters)
Molarity = Moles / Volume (in liters)
First, let's calculate the moles and molarities of Ag+ ions:
For the AgCl solution:
Moles of AgCl = Molarity x Volume
= 0.1 M x 0.035 L
= 0.0035 moles
Molarities of Ag+ ions:
Moles of Ag+ = Moles of AgCl
= 0.0035 moles
Volume of the mixture = volume of AgCl solution + volume of MgCl2 solution
= 0.035 L + 0.015 L
= 0.05 L
Molarity of Ag+ ions:
Molarity of Ag+ = Moles of Ag+ / Volume of the mixture
= 0.0035 moles / 0.05 L
= 0.07 M
Next, let's calculate the moles and molarities of NO3- ions:
For the AgCl solution:
Moles of NO3- = Molarity x Volume
= 0.1 M x 0.035 L
= 0.0035 moles
For the MgCl2 solution:
Moles of NO3- = Molarity x Volume
= 0.25 M x 0.015 L
= 0.00375 moles
Total Moles of NO3- = Moles from AgCl solution + Moles from MgCl2 solution
= 0.0035 moles + 0.00375 moles
= 0.00725 moles
Molarities of NO3- ions:
Moles of NO3- = Total Moles of NO3-
= 0.00725 moles
Molarity of NO3- ions:
Molarity of NO3- = Moles of NO3- / Volume of the mixture
= 0.00725 moles / 0.05 L
= 0.145 M
Now, let's calculate the moles and molarities of Mg2+ ions:
For the MgCl2 solution:
Moles of Mg2+ = Molarity x Volume
= 0.25 M x 0.015 L
= 0.00375 moles
Molarities of Mg2+ ions:
Moles of Mg2+ = 0.00375 moles
Molarity of Mg2+ ions:
Molarity of Mg2+ = Moles of Mg2+ / Volume of the mixture
= 0.00375 moles / 0.05 L
= 0.075 M
Finally, let's calculate the moles and molarities of Cl- ions:
For the AgCl solution:
Moles of Cl- = Moles of AgCl
= 0.0035 moles
For the MgCl2 solution:
Moles of Cl- = Molarity x Volume
= 0.25 M x 0.015 L
= 0.00375 moles
Total Moles of Cl- = Moles from AgCl solution + Moles from MgCl2 solution
= 0.0035 moles + 0.00375 moles
= 0.00725 moles
Molarities of Cl- ions:
Moles of Cl- = Total Moles of Cl-
= 0.00725 moles
Molarity of Cl- ions:
Molarity of Cl- = Moles of Cl- / Volume of the mixture
= 0.00725 moles / 0.05 L
= 0.145 M
Summary:
Moles and Molarities of ions in the solution mixture are:
Ag+: Moles = 0.0035 moles, Molarity = 0.07 M
NO3-: Moles = 0.00725 moles, Molarity = 0.145 M
Mg2+: Moles = 0.00375 moles, Molarity = 0.075 M
Cl-: Moles = 0.00725 moles, Molarity = 0.145 M