A 25.00 mL solution of 0.150 M NaCl is mixed with 10.00 mL of a 0.0750 M CaCl2 solution. Assuming the total volume is 35.00 mL, calculate the concentration of chloride ions in the resulting solution.

 To calculate the concentration of chloride ions in the resulting solution, we can follow these steps:

Step 1: Calculate the moles of chloride ions from NaCl Since there are 25.00 mL of a 0.150 M NaCl solution, which implies there is a 1:1 ratio between NaCl and Cl⁻ ions, we calculate the moles of Cl⁻ ions from NaCl like so: Moles of Cl⁻ from NaCl = Molarity (M) of NaCl × Volume (L) of NaCl solution = 0.150 mol/L × 0.02500 L = 0.00375 moles

Step 2: Calculate the moles of chloride ions from CaCl2 Since 1 mole of CaCl2 yields 2 moles of Cl⁻ ions due to the 1:2 ratio, we calculate the moles of Cl⁻ ions from CaCl2 like so: Moles of Cl⁻ from CaCl2 = 2 × (Molarity (M) of CaCl2 × Volume (L) of CaCl2 solution) = 2 × (0.0750 mol/L × 0.01000 L) = 2 × 0.000750 moles = 0.001500 moles

Step 3: Calculate the total moles of chloride ions Total moles of Cl⁻ ions = Moles of Cl⁻ from NaCl + Moles of Cl⁻ from CaCl2 = 0.00375 moles + 0.001500 moles = 0.005250 moles

Step 4: Calculate the new concentration of chloride ions Concentration (C) = Total moles of Cl⁻ ions / Total volume of the mixture (L) = 0.005250 moles / 0.03500 L = 0.150 M

So, the concentration of chloride ions in the resulting solution is 0.150 M.