Question

What is the molar concentration of chloride ions in a solution prepared by mixing 100.0 mL of 2.0 M KCl with 50.0 mL of a 1.50 M CaCl2 solution?

Answer is 2.3 mol/L please show how we got to this

Answer 1

We know that each KCl has one mole of Cl^- ions and each CaCl2 has two moles of Cl^- ions.

So we need to find out total number of moles of chloride ions in the newly made solution.

Now KCl solution:

2.0 M solution means 2.0 moles per 1000 ml of solution.

So for 100 ml of solution , number of chloride ions = 0.2 moles.

CaCl2 fraction:

1.5 M solution means 1.5 moles of CaCl2 per 1000 ml solution. Then number of caCl2 for 50 ml solution would be =

= 0.075 moles of CaCl2

and (0.075 x 2) = 0.15 moles of chloride ions.

Now total number of moles of chlorided ions = 0.2 + 0.15

= 0.35 moles

Molarity = (moles ) x (1000 / volume in ml)

Molarity = 0.35 x (1000 / (100ml + 50ml))

= 2.33 M