Ignoring protolysis reactions, indicate the concentration
of eac...

Question

Ignoring protolysis reactions, indicate the concentration 
of each ion or molecule present in the following solutions: 
(d) a mixture of 35.0 mL of 0.010 M Na2CO3 and 
50.0 mL of 0.200 M K2SO4. Assume the volumes are additive.

Accepted Answer

Hey everyone, we're asked to calculate the concentration of calcium ions when 50 mL of 0.40 moller calcium chloride and 80 mL of 0.15 moller calcium chloride solutions are mixed. We're told to assume volumes are additive first. Let's go ahead and calculate the moles of each of our solutions. Starting off with our 50 ml of 0.40 moller calcium chloride. We're going to convert this into leaders with our dimensional analysis and we know that we have 10-30 ml per one leader. Now we're going to multiply this by our 0.40 Molar, which is moles of calcium chloride per leaders of calcium chloride. Now we're going to look at our multiple ratios and we know that per one mole of calcium chloride, we have one mole of our calcium ions. Now to calculate this out and cancel out all of our units, we end up with 0.2 mol of calcium ions in our 0.40 moller calcium chloride solution. Now let's go ahead and look at our next solution. Starting off with our 80 mL, we're going to convert this into leaders and we know that we have 10 to the third milliliters per one leader. Now we're going to multiply this by 0.15 molar which is 0.15 mol of calcium chloride per one liter of calcium chloride. Next looking at our multiple ratios, we know that one mole of calcium chloride contains one mole of our calcium ions calculating this out and canceling out all of our units, We end up with 0.12 mol of our calcium ions in our 0.15 moller calcium chloride solution. Now to calculate the concentration of our calcium ions, we can go ahead and take the two values we calculated which is 0.2 mol Plus 0.012 mol of our calcium ions. And we're going to divide this by our leaders. And since we were told to assume volumes are additive, we're going to take our 0.050 leaders Which was our 50 ml converted into leaders plus our 0.080 L. Again, these were our milliliters that we converted into leaders calculating this out. We end up with 0.25 moles over liters of our calcium ions and we know that moles over liters is simply Moeller. So our final answer is going to be 0.25 molar of our calcium ions. Now, I hope that made sense. And let us know if you have any questions

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