Dilution: Ions in Mixed Solutions

Question

In the laboratory a student combines 42.1 mL of a 0.373 M sodium carbonate solution with 24.0 mL of a 0.311 M potassium carbonate solution.

What is the final concentration of carbonate anion ?

?M

Answer 1

Hi Meagan,

If we assume no additional ionization of carbonate occurs, then

the Na₂CO₃ solution has

42.1mL*(1 L/1000mL)*.373 mol/L = .0157033 mol of CO₃ ion.

Similarly, the K₂CO₃ solution has

24.0mL*(1 L/1000mL)*.311 mol/L = .007464 mol of CO₃ ion.

Adding the moles gives .0231673 mol of CO₃ in a total volume of

(.0240 L+ .0421 L) = .0661 L, so the molarity of CO₃ is

.0231673 mol/.0661 L = 0.350489 M of CO₃ ion.

Obeying significant digits rules makes the answer

0.350 M of CO₃ ion.

The answer is in between the two original concentrations that were mixed which is reasonable.

(My calculator batteries are weak, so you may want to double check the calculations!)

Answer 2

Molarity is moles of solute per liter of solution, so in order to find the molarity of the carbonate ion in the final solution, we need to find the total number of moles of carbonate ion and divide it by the total volume of the solution in liters.

 

There is one mole of carbonate ion (CO3) per mole of sodium carbonate (Na2CO3), so using the appropriate stoichiometry, we find that the first solution contains [(0.373 moles of sodium carbonate)/(liter)]*[(1 mole of carbonate ion)/(mole of sodium carbonate)]*[(1 liter)/(1000 mL)]*[42.1 mL] = 0.0157 moles of carbonate ion.

 

Similarly, there is one mole of carbonate ion (CO3) per mole of potassium carbonate (K2CO3), so the second solution contains [(0.311 moles of potassium carbonate)/(liter)]*[(1 mole of carbonate ion)/(mole of potassium carbonate)]*[(1 liter)/(1000 mL)]*[24.0 mL] = 0.00746 moles of carbonate ion.

 

Thus, the final solution contains 0.0157 + 0.00746 = 0.0232 moles of carbonate ion.

 

To find the volume of the final solution we simply add the volumes of the two solutions, so we get 42.1 mL + 24.0 mL = 66.1 mL. Converting this to liters gives us (66.1 mL)*[(1 L)/(1000 mL)] = 0.0661 L.

 

 

The concentration of carbonate ion in the final solution is therefore (0.0232 moles)/(0.0661 L) = 0.350 moles/L = 0.350 M.