[Violet89]

If 52.00 ml of 0.500 M CaBr2 solution is added to 128 ml of 0.320 M solution of KCl solution, what will the molarity of CaBr2 and the molarity of KCl in the final solution? (assume volumes are additive)

I don't know where to get started on this problem. Anyone know where I can find an example of this type of problem? Any help will be much appreciated!

I started off with solving for moles of solute for the first solution... 

[UG]

I started off with solving for moles of solute for the first solution... 

That is good, so then the new molarity is just (mole of solute)/(new volume of solution).

[Violet89]

I started off with solving for moles of solute for the first solution... 

That is good, so then the new molarity is just (mole of solute)/(new volume of solution).

I divided by the new volume (128 ml) and got the new molarity of CaBr2. In order to solve for the new molarity of KCl, do I use the new molarity of CaBr2?

[UG]

The volumes are additive, meaning that the new volume is 52 mL + 128 mL = 180 mL. This is the volume you use to calculate the molarity of both CaBr₂ and KCl. To solve the molarity of KCl, do the same as you did before, calculate the moles of solute and then divide by the new volume.

[Violet89]

The volumes are additive, meaning that the new volume is 52 mL + 128 mL = 180 mL. This is the volume you use to calculate the molarity of both CaBr₂ and KCl. To solve the molarity of KCl, do the same as you did before, calculate the moles of solute and then divide by the new volume.

Makes sense! I understand now. Thank you very much for your help. 

[Borek]

The volumes are additive

Too strong statement. They are not.

In most cases this is only a (good) approximation. Perfectly applicable here.

[UG]

Indeed, I was just re-stating what it said in the question