Question

What volume (in mL) of $0.01MAgN{O}_3$ would be required to precipitate all the $I^{-}$ in $200$ mL of a solution that contained $24.32$ ppt of $KI$?

Answer 1

PPT means parts per thousand.

$24.32$ ppt of $KI$ corresponds to $24.32$ g $KI$ per thousand mL solution.
$200$ mL of solution will contain $\frac{24.32}{1000}\times 200=4.864$ g or $0.038$ mol
Let $V$ ml be the volume of silver nitrate required.
The molarity of the solution is $0.001$ M.
The number of millimoles of silver nitrate is $0.001M\times V=0.001V$ millimoles $=1\times {10}^{-5}$ mol
This is equal to $0.038$ moles.
Hence, $V=\frac{0.038}{1\times {10}^{-5}}=3800$ mL
Therefore, $3800$ mL of $0.01M$ silver nitrate will be required.