Question

A solution is 0.10 M in Pb(NO3)2 and 0.10 M in AgNO3. Solid NaI is added until the second solid compound is on the verge of precipitating.
Which compound precipitates first and what is the I- concentration when the second compound begins to precipitate?Ksp for AgI is 1.5 x 10-16 Ksp for PbI2 is 8.7 x 10-9PbI_2,\:5.9\times10^{-4}\:M
PbI_2,\:4.4\times10^{-3}\:M
AgI,\:2.9\times10^{-4}\:M
AgI,\:9.3\times10^{-5}\:M
AgI,\:8.7\times10^{-8}\:M

Answer 1

Answer:

The compound which will precipitate first will be AgI with $2.9X10^{-4}$

Explanation:

In order to precipitate a salt solution the ionic product of salt must exceed the solubility product.

Given:

Ksp of AgI = $1.5X10^{-16}$

Ksp of PbI2= $1.5X10^{-9}$

As Ksp of AgI is very low it will precipitate faster than lead iodide.

Now, higher the concentration of AgI solution taken faster its precipitation.

In the given choice the highest concentration solution of AgI is $2.9X10^{-4}$