Question

Suppose 3.63g of nickel(II) iodide is dissolved in 350.mL of a 75.0mM 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. (M)

Answer 1

NiI2 + K2CO3 $\rightarrow$ 2KI + NiCO3

Given : Mass of NiI2 = 3.63 g

350 ml or 0.35 L of 75mM or 0.075 M K2CO3

(1M = 1000 mM)

Molar mass of NiI2 = 312.5 g/mol

Moles of NiI2 = given mass / molar mass

= 3.63/312.5

= 0.0116 moles

Moles of K2CO3 = molarity x volume

= 0.075 x 0.35

= 0.2142 moles

So, NiI2 is the limiting reagent ( as it has least number of moles) and will determine the amount of product formed.

So, moles of KI = 2 x 0.0116 ( according to balanced reaction)

concentration of KI = moles / volume

As volume of solution is same

[KI] = (2 x 0.0116)/0.35

= 0.066 M

KI $\rightarrow$ K+ + I-

So, [I-] = 0.066M or 66.0 mM