In the laboratory you dissolve 15.4 g of potassium carbonate in a volumetric fla

Question

In the laboratory you dissolve 15.4 g of potassium carbonate in a volumetric flask and add water to a total volume of 250 mL.

What is the molarity of the solution? ______ M.

What is the concentration of the potassium cation? _______ M.

What is the concentration of the carbonate anion? ______ M.

In the same laboratory you dissolve 16.9 g of calcium acetate in a volumetric flask and add water to a total volume of 375 mL.

What is the molarity of the solution? _____ M.

What is the concentration of the calcium cation? _____ M.

What is the concentration of the acetate anion? _____ M.

Lastly, in a separate area of the laboratory you dissolve 12.4 g of zinc fluoride in a volumetric flask and add water to a total volume of 500 mL.

What is the molarity of the solution? _____ M.

What is the concentration of the zinc cation? ____ M.

What is the concentration of the fluoride anion? _____ M.

Explanation / Answer

Moles of potassium carbonate in the flask = mass/ molar mass = 15.4 g/ 138.20 gmol-1 = 0.11 mol

molarity = number of moles/ 1000ml of the solution.

So, molarity of the solution = 0.11 x 4 = 0.44 M

the molarity of the solution =0.44 M.

Potassium carbonate dissociates in solution as follows

K2CO3 ---------> 2K +   + CO32-

So, 1 mole of K2CO3 will dissociate to give 2 mol of K + and 1 mol of carbonate ion.

the concentration of the potassium cation = 0.88 M.

the concentration of the carbonate anion = 0.44 M.

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Moles of Cacium acetate in the flask = mass/ molar mass = 16.9 g/ 158.16 gmol-1 = 0.1068 mol

molarity = number of moles/ 1000ml of the solution.

So, molarity of the solution = 0.1068 x 1000/375 = 0.28 M

the molarity of the solution =0.28 M.

Calcium acetate dissociates in solution as follows

Ca(OCOCH3)2 ---------> Ca2+   + 2OCOCH3-

So, 1 mole of Ca(OCOCH3)2 will dissociate to give 1 mol of Ca2+ and 2 mol of acetate ion.

the concentration of the Calcium cation = 0.28 M.

the concentration of the acetate anion = 0.56 M.

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Moles of Zinc fluoride in the flask = mass/ molar mass = 12.4 g/ 103.37 gmol-1 = 0.119 mol = 0.12 mol

molarity = number of moles/ 1000ml of the solution.

So, molarity of the solution = 0.12 x 2 = 0.24 M

the molarity of the solution =0.24 M.

Zinc fluoride dissociates in solution as follows

ZnF ---------> Zn2+   + 2 F-

So, 1 mole of ZnF will dissociate to give 1 mol of Zn2 + and 1 mol of fluoride ion.

the concentration of the potassium cation = 0.24 M.

the concentration of the carbonate anion = 0.48 M.

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