Answer to Question #95918 in General Chemistry for taylor

Question

Suppose 21.4g of sodium iodide is dissolved in 350.mL of a 0.60 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 sodium iodide is dissolved in it.

Answer 1

21.4 g NaI = 21.4/149.84Mole NaI ( Sodium Iodide)

= 0.842 mole NaI

( Molecular Weight of NaI =149.84 g/Mole)

0.6 M Silver Nitrate will be precipitated out by the 0.6 M sodium iodide

Therefore, (0.842-0.6)= 0.242 mole Sodium Iodide will remain in the 350 ml solution.

Therefore, in the 350 ml solution now therr is 0.242 mole Iodide ion.

The Final molarity of Iodide ion in the sollution

= (0.242*350)/1000 M

=0.0847M