Question

Suppose 1.38g of potassium iodide is dissolved in 300.mL of a 18.0mM aqueous solution of silver nitrate. Calculate the final molarity of iodide anion in the solution. You can assume the volume of the solution doesn't change when the potassium iodide is dissolved in it. Round your answer to 3 significant digits.

Answer 1

Answer will be as follows-

AgNay + KI Agt + KNO3 Number of moles of KI = Amount of KI (8) Numb Molarity mw of KI (mole) - 1.388/166.0039 mole = 0.008313 moles number of moles of AgNO₃ = volume in Litve & = 0.BLY 10x10²m = 0.003 moles According to balance equation i moleo of Agno reacts with mole of KI therefore Amount of unreacted KI = 0.008313 - 0.003 (moles) - 0.005313 moles Molarity of KI & Nauber of moles of unreacted KI volume of solution (2) 0.005313 moles = 0.01771M 0.3 L molarity of iodide ion = molarity of KI = 0.01771M ( ~ 1.778 10 M Huswer