Question

Suppose 0.591 g of zinc iodide is dissolved in 50. mL of a 57.0 m M 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 zinc iodide is dissolved in it Round your answer to 2 significant digits. M M10 ? X

Answer 1

Median un) 0.5919 of Zn I2 molar mass of 2nI₂ = 319.22g/mol moles = mass 0.5919 molar mass 319.22 g mol moles of Zn I ₂ = 1.8514x10 mol ZnI2 Zn²+ + 21- . 1 mol of Zn. I = 2mol of I- : 1.8514 mol of Zr I = 3.7028 ml of I- Reaction taking place; Zn I₂ + K[oz → ZnCoz +2KI final volume = 50mL moles of I = 3.70 28x03 mol. molarity = moles - Volume = 3.70 28X163 mol 50x103 L M = 0.074056 M 1 Molarity of final I solution will be 0.0740 56M After Rounding offi TM = 0.074 M I final molarity So, molarity of final I is 0.074M