Question

If the molar concentration of $ Pb{{I}_{2}} $ is $ 1.5\times {{10}^{-3}}mol\,{{L}^{-1}} $ , then the concentration of iodide ions of g ion $ {{L}^{-1}} $ is?
A. $ 3.0\times {{10}^{-3}} $
B. $ 6.0\times {{10}^{-3}} $
C. $ 0.3\times {{10}^{-3}} $
D. $ 0.6\times {{10}^{-6}} $

Answer 1

Hint Form the equation for the dissociation of the compound and check the number of moles. Then use the concentration given in the question to find a relation and find the concentration of the iodide ions.

Complete step by step solution:
In order to answer the question, we need to learn about stoichiometry and molarity. Now, let us discuss about the mole concept. Now, every element in the earth's atmosphere is unique. So, in order to express the concentration, we use the mole concept. Matter is made up of atoms. Now, it is decided that when a substance contains approximately $ 6.022\times {{10}^{23}} $ atoms, then we call it 1 mole of the substance. Since every element is different, so the masses of every element should be also different. The number $ 6.022\times {{10}^{23}} $ is called the Avogadro’s constant. So, the mass of 1 mole of a particular element will not be equal to the mass of 1 mole of some other element and that is what makes it easier to differentiate between the compounds. Now, the mass of the element that is calculated, which contain the Avogadro’s number of atoms in it is called the molar mass of the element. Now, let us come to our question, and write the dissociation of $ Pb{{I}_{2}} $ . So, we have:
\[Pb{{I}_{2}}\rightleftharpoons P{{b}^{2+}}+2{{I}^{-}}\]
So, one mole of $ Pb{{I}_{2}} $ gives us 2 moles of iodide ions and 1 mole of lead ions. Now, we have been provided in the question that the molar concentration of $ Pb{{I}_{2}} $ is $ 1.5\times {{10}^{-3}}mol\,{{L}^{-1}} $ . But according to the equation of dissociation, we can see that the iodide ions will have a concentration that is double than that of $ Pb{{I}_{2}} $ . So, the required concentration is:
\[\begin{align}
& [{{I}^{-}}] =(2\times 1.5\times {{10}^{-3}})mol\,{{L}^{-1}} \\
& \Rightarrow [{{I}^{-}}] =3\times {{10}^{-3}}\,mol\,{{L}^{-1}} \\
\end{align}\]

So, we get the final answer option as A.

NOTE: In order to calculate the molar mass of a compound, the individual molar masses of the elements are added up, and multiplied with their atomicity.