Question

The volume (in mL) of $0.1 M$ $AgN{{O}_{3}}$ required for complete precipitation of chloride ions present in $30 mL$ of $0.01 M$ solution of $[Cr{{({{H}_{2}}O)}_{5}}Cl]C{{l}_{2}}$, as silver chloride is close to:
(A) 6
(B) 8
(C) 9
(D) 7

Answer 1

Hint: Silver nitrate when reacts with $[Cr{{({{H}_{2}}O)}_{5}}Cl]C{{l}_{2}}$, silver chloride will be precipitated. The chloride ions of $[Cr{{({{H}_{2}}O)}_{5}}Cl]C{{l}_{2}}$ are of two types. By equating the moles of silver nitrate to the moles of chloride ions we can get the volume.

Complete step by step solution:
Silver nitrate when reacts with $[Cr{{({{H}_{2}}O)}_{5}}Cl]C{{l}_{2}}$, silver chloride will be precipitated. But not all the chloride ions of $[Cr{{({{H}_{2}}O)}_{5}}Cl]C{{l}_{2}}$ are ionizable . this compound has two types of chloride ions: (i)- The chloride ion which is inside the coordination bracket i.e., directly linked to the central metal atom with coordinate bond, (ii)- The chloride ions that are present outside the bracket that can take part in a chemical reaction.
So, the compound $[Cr{{({{H}_{2}}O)}_{5}}Cl]C{{l}_{2}}$ has only 2 ionizable chloride ions.
Its concentration is given as 0.01 M and volume is given as 30 mL.
So, the number of moles of the compound will be:
$mole{{s}_{(compound)}}=30\text{ x 0}\text{.01}$
Since, there are two chloride ions present in the compound, the moles of chloride ions will be:
$mole{{s}_{(C{{l}^{-}})}}=2\text{ x 30 x 0}\text{.01}$
The reaction which takes for the formation of silver chloride is given below:
$AgN{{O}_{3}}+C{{l}^{-}}\to AgCl+NO_{3}^{-}$
So, in the equation the number of moles of silver nitrate and chloride ions is the same, so we can equate them.
The concentration of $AgN{{O}_{3}}$ is $0.1 M$, and the volume is $V$.
Now on equating, we get:
$0.1 \times V = 2 \times 0.01 \times 30$
$V=6\text{ }ml$
So, the volume of $AgN{{O}_{3}}$ is $6 ml$.

Therefore the correct answer is an option (A) 6.

Note: We can equate the moles only when the number of moles in the reaction is the same. It must be noted that in a coordination compound not all the ions are ionizable or take part in the chemical reaction.