Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Violet89 on February 08, 2012, 07:58:20 PM
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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... ???
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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).
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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?
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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.
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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. ;D
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The volumes are additive
Too strong statement. They are not.
In most cases this is only a (good) approximation. Perfectly applicable here.
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Indeed, I was just re-stating what it said in the question :P