Question

Calculate the concentration of chloride ions when 100.0 mL of 0.233 M sodium chloride is mixed with 250.0 mL of 0.150 M calcium chloride.

Answer 1

The dissociation of both salts NaCl and CaCl₂ are as follows;
NaCl --> Na⁺ + Cl⁻

CaCl₂ --> Ca²⁺ + 2Cl⁻

the molar ratio of NaCl to Cl⁻ is 1:1
therefore number of NaCl moles is equal to number of Cl⁻ ions dissociated from NaCl
then number of Cl⁻ ion moles - 0.233 mol/L x 0.1000 L = 0.0233 mol

molar ratio of CaCl₂ to Cl⁻ ions is 1:2
1 mol of CaCl₂ gives out 2 mol of Cl⁻ ions.
number of CaCl₂ moles - 0.150 mol/L x 0.2500 L  = 0.0375 mol 
then the number of Cl⁻ ion moles - 0.0375 x 2 = 0.0750 

total number of Cl⁻ ion moles = 0.0233 mol + 0.0750 mol = 0.0983 mol
volume of solution - 100.0 + 250.0 = 350.0 mL 

concentration of Cl⁻ = 0.0983 mol / 0.3500 L = 0.281 M
concentration of Cl⁻⁻ is 0.281 M