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A 8.50 g sample of KCl is dissolved in 66.0 mL of water. The resulting solution is then added to 72.0 mL of a 0.280 M CaCl2(aq) solution. Assuming that the volumes are additive, calculate the concentrations of each ion present in the final solution.

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Answer:

For KCl

[K+]= 12.5M and [Cl-] = 12.5M.

For CaCl2

[Ca+] = 2.03M and [Cl2] = 2 * 2.03 = 4.06M because there are 2 Cls in the compound

Then, for the amount of chlorine in both compound = 12.5 + 4.06 = 16.56M

Explanation:

Mass of KCl = 8.50 g

Molecular weight of KCl = (1 atom X 39.0983 Potassium) + (1 atom X 35.453 Chlorine) = 74.55 g/mol

Volume of KCl solution = 66.0 mL = O.066L

Volume of CaCl2 = 72.0 mL

Then,

Number of moles of KCl = mass of KCl/molecular weight

= 8.50/74.55 = 0.1140 Mol's

Initial molarity of KCl = Number of solute/Volume of the solution

0.1140/66*10^-3 = 1.727M

If volumes are additives, then add the of volume of KCl and CaCl2 together,

66 + 72 = 138 ml

To get the ion concentration, divide the molarity of each by the new volume

Then,

For KCL = mole of KCl/new volume

= 1.727/138*10^-3 = 12.5M/L

For CaCl2 mole of CaCl2/new volume = 0.280/138*10^-3 = 2.03M/L