Calculate the chloride ion concentration of a sol. prepared by adding 8.003 g KCl(s) to 159.8 mL of a 0.496 M?

Question

Calculate the chloride ion concentration (in mol/L) in a solution prepared by adding 8.003 g KCl(s) to 159.8 mL of a 0.496 M BaCl2(aq) solution. Assume the volume of the solution isn't changed by the addition of the solid.

Calculate the chloride ion concentration (in mol/L) in a solution prepared by adding 8.003 g KCl(s) to 159.8 mL of a 0.496 M BaCl2(aq) solution. Can you show me how you got this solution?

Answer 1

Ok, the first step is to calculate the mass of the chloride ions already in solution.

It’s 0.496 molar, and a one liter one molar solution of BaCl2 will have 2*35.45g Cl so it’s 2*35.45g*.496M = (35.166 g Cl / liter) * .1598 liters = 5.62 g Cl in our current sample.

Now let’s calculate the mass of the new ions we are adding by calculating the mass ratio for KCl and scaling with our sample mass: (35.45 g Cl / (39.09 g K + 35.45 g Cl) ) * 8.003 g KCl = 3.806 g Cl

So when we add the solute the total mass of chloride ions will be 5.62 g Cl + 3.806 g Cl = 9.426 g Cl in a .1598 Liter soln.

So now we scale to a one molar solution to get our final answer:

[Chloride Ion] = (9.426 g Cl / .1598 L) * (1.000L / 35.45 g cl ) = 1.66 Molar

Answer 2

Ummm, I'd say the concentration would be about...medium.