posted by bri on .
If 125mL of a 1.10 NaCl solution is mixed with 105mL of a 0.850M CaCl2 solution but NO reaction takes place, calculate the final concentration of each of the ions in the final solution.
Please reply with the steps on how to do this question so I can learn. I can't find examples or anything related in my textbooks and I take it via correspondence, so no class instructor to ask for help. Thank you
mols Na ions from NaCl = M x L = ?
mols Cl^- from NaCl = M x L = ?
mols Ca^2+ = M x L = ?
moles Cl^- from CaCl2 = 2 x mols CaCl2.
M = moles/L.
For Na it is 0.125 x 1.10 = 0.1375 and total volume is 125 mL + 105 mL = 230 mL = 0.230L
(Na^+) = 0.597 M
Ca^2+ is done the same way.
Cl^-: Add mols Cl^- from NaCl and mols Cl^- from CaCl2 for total mols Cl^- and divide by total volume for M Cl^-.