Answer:
(a) Increase the solid PbI₂ (s)
(b) Decrease
(c) Increase
Explanation:
In this question we are considering the effect of adding a common ion to an equilibrium.
The problem states that the saturated solution of PbI₂ is at equilibrium with undissolved PbI₂ (s) :
PbI₂ (s) ⇄ Pb²⁺+ 2 I⁻
If we increase the concentration of I⁻ by adding the highly soluble KI , the equilibrium will react according to LeChatelier´s principle by decreasing the concentration of Pb²⁺ precipitating some Pb²⁺ until it reaches equilibrium again. With this in mind, lets answer the question:
(a) Increase the solid PbI₂ (s)
(b) Decrease
(c) Increase
The addition of KI in the solution increases the settling of lead iodide at the bottom, decreases the lead ions and increases the iodide ions in the solution.
The saturated solution has been given as the one in which the salt has been divided into maximum concentration.
The equilibrium reaction of lead iodide has been:
[tex]\rm PbI_2\;\rightleftharpoons Pb^2^+\;+2\;I^-[/tex]
The addition of KI increases the product concentration, and the lead iodide has been found to be settles in increased concentration in the bottom.
The addition of KI decreases the rate of formation of product, thus the ion of lead formation will be decreased in the reaction.
The concentration of iodide ions in the solution increases with the addition of KI with common ion effect.
Learn more about common ion effect, here:
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