If 50.0 mL of a 1.00 M solution of CaCl2 (solution A) is mixed with 85.0 mL of pure water to produce solution B and then 42.5 mL of solution B is mixed with 52.0 mL of solution A to give solution C, what is the concentration of CL^-(aq) in the final solution?
from A to B, it was diluted to 1M*50/(50+85)=.37M, but twice that in Cl conc.
then, mixing it with A, (.425*.37*2+.52*1*2)/(.425+.52)=1.43M in Cl
check my thinking.
To find the concentration of Cl^-(aq) in the final solution, we need to calculate the moles of Cl^- ions present in each step.
First, let's calculate the moles of Cl^- ions in solution A.
Moles of Cl^- in solution A = Volume of solution A (in liters) * Molarity of CaCl2
Given:
Volume of solution A = 50.0 mL = 50.0/1000 = 0.050 L
Molarity of CaCl2 = 1.00 mol/L
Moles of Cl^- in solution A = 0.050 L * 1.00 mol/L = 0.050 mol
Next, let's calculate the moles of Cl^- ions in solution B.
Since we add pure water to solution A, the number of moles of Cl^- ions remains the same.
So, moles of Cl^- in solution B = 0.050 mol
Now, let's calculate the moles of Cl^- ions in solution C.
Moles of Cl^- in solution C = Moles of Cl^- in solution A + Moles of Cl^- in solution B
Moles of Cl^- in solution C = 0.050 mol + 0.050 mol = 0.100 mol
Finally, to find the concentration of Cl^- ions in the final solution, we divide the moles of Cl^- ions in solution C by the total volume of solution C.
Total volume of solution C = Volume of solution A + Volume of solution B
Total volume of solution C = 50.0 mL + 85.0 mL = 135.0 mL = 135.0/1000 = 0.135 L
Concentration of Cl^- in solution C = Moles of Cl^- in solution C / Total volume of solution C
Concentration of Cl^- in solution C = 0.100 mol / 0.135 L = 0.741 M
Therefore, the concentration of Cl^- ions in the final solution (solution C) is 0.741 M.