Question

What is the concentration of a chloride ion in a solution made by mixing 110.0mL of a 2.00M barium chloride and 250mL of .450M sodium chloride?

Answer 1

BaCl2 - 110 mL and 2.00M + NaCl - 250mL 0.450M = new solution with 110+250=360mL volume
so what we have to do here is to find moles of chloride ions separately...
from Molarity = mole / volume:
for BaCl2 => 2.00M = mole / 0.110L(we have to change unit to Liters)
mole of BaCl2 = 0.220 moles: each BaCl2 contains 2 chloride ions so there must be 0.220 * 2 = 0.440 moles of Chloride ions..
for NaCl => 0.450M = mole/ 0.250L(unit conversion) => mole of NaCl = 0.113 moles of NaCl ; in each NaCl there is one chloride ion, so there must be 0.113 moles of chloride ion.
add Chloride ion amounts(moles) = 0.220+0.113 = 0.333 moles of Cl in new solution
molarity of chloride = 0.333mole/0.36L = 0.925M
(look carfully, there may be some mistakes, but I'm pretty sure)