Question

Suppose 0.829g of zinc chloride is dissolved in 100.mL of a 0.60M 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

Answer:

Molarity of  $Cl^{-}$ anion in solution is 0.122 M.

Explanation:

Chemical Reaction: $ZnCl_{2}+K_{2}CO_{3}\rightarrow ZnCO_{3}+2KCl$

As KCl is a strong electrolyte therefore it dissociates completely in solution. Hence $Cl^{-}$ present in final solution is equal to $Cl^{-}$ present in 0.829 g of  $ZnCl_{2}$.

Molar mass of $ZnCl_{2}$ = 136.286 g/mol

So, 0.829 g of $ZnCl_{2}$ = $\frac{0.829}{136.286}$ moles of $ZnCl_{2}$

1 mol of $ZnCl_{2}$ contains 2 moles of $Cl^{-}$.

Hence number of moles of $Cl^{-}$ in solution = $(2\times \frac{0.829}{136.286})$ moles

So, molarity of $Cl^{-}$ in solution = [(number of moles of $Cl^{-}$ in solution)/(volume of solution in mL)]$\times 1000$

= $\frac{(2\times \frac{0.829}{136.286})}{100.}\times 1000$ M

= 0.122 M