Question

Suppose 1.27 g of potassium iodide is dissolved in 100. mL of a 44.0 m M 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.

This is a limiting reactants question with Stoichiometry

Answer 1

Man (9) 766 g/med g/mol 1.87g Moles of KI in 1.27g = - 0.00765 moles. Molar mass (g/mol) KI → kt + I So moles of kI = moles of iodide anion = 0.00765 moles. Moles of AgNO3 in solur = molarity (M) x vol.in litres = 44.0 x 1000 = 0.0044 moles 1000 (An Imm= 10 3M s 1 ml= 1032 ) Moles of Ay Moa = moles of Agt & moles of iodide ion will used for ion = 0.0044 moles. AS Agt + 1 - Ay I (solid precipitates) and there moles of 0 are no more available as moles as forms solid precipitater with Agt of AgNO3. So, moles of so in solur available = 0.00765 - 0.0044=0.00325 volume of solen = 100ml= 0.1002. So, molarity of fodide ion = moles - 0.00325 moles volin citres This final molarity of iodide ion = 0.0325 m =0.0325m 0.100L