Question

Suppose

0.332g

of zinc chloride is dissolved in 50.mL

of a

48.0mM

aqueous solution of potassium carbonate.

Calculate the final molarity of chloride anion in the solution. You can assume the volume of the solution doesn't change when the zinc chloride is dissolved in it.

Be sure your answer has the correct number of significant digits.

Answer 1

Moles of zinc chloride (ZnCl₂) = mass of ZnCl₂ / molar mass of ZnCl₂ = 0.332 g/(136.29 g/mol) = 0.002436 mol

One mole zinc chloride gives two moles of chloride ions.

ZnCl₂ -------> Zn²⁺ + 2Cl^-

Moles of Cl^- = 0.002436 mol ZnCl₂ x ( 2 mol Cl^- / 1 mol ZnCl₂) = 0.004872 mool Cl^-

Volume of solution = 50. mLx(1 L/1000 mL) = 0.050 L

Chloride ion concentration , [Cl^- ] = moles Cl^- /volume of solution in L = 0.004872 mol/ 0.050 L = 0.097 mol/L = 0.097 M

So, the final molarity of chloride anion in the solution is 0.097 M.