Question

uppose of nickel(II) iodide is dissolved in of a aqueous solution of potassium carbonate. Calculate the final molarity of iodide anion in the solution. You can assume the volume of the solution doesn't change when the nickel(II) iodide is dissolved in it. Be sure your answer has the correct number of significant digits.

Answer 1

Explanation:

Let us assume that mass of nickel(II) iodide is 2.90 g, volume is 150.ml, and concentration is 0.70 M.

When $K_{2}CO_{3}$ is added into the solution then it will not affect the $I^{-}$ concentration.

         $NiI_{2} \rightarrow Ni^{2+} + 2I^{-}$

Moles of $NiI_{2}$ = $\frac{mass}{\text{molar mass of nickel iodide}}$

                          = $\frac{2.90}{312.50}$

                          = 0.00928 mol

Moles of $I^{-}$ = $2 \times moles of NiI_{2}$

                  = $2 \times 0.00928$

                 = 0.01856 mol

Molarity of $I^{-}$ = $\frac{moles of I^{-}}{\text{volume of solution}}$

                        = $\frac{0.01856}{0.150}$

                        = 0.124 M

Thus, we can conclude that final molarity of iodide anion in the solution is 0.124 M.