Question
Question: The volume (in mL) of \(0.1 M\) \(AgN{{O}_{3}}\) required for complete precipitation of chloride ion...
The volume (in mL) of 0.1M AgNO3 required for complete precipitation of chloride ions present in 30mL of 0.01M solution of [Cr(H2O)5Cl]Cl2, as silver chloride is close to:
(A) 6
(B) 8
(C) 9
(D) 7
Solution
Silver nitrate when reacts with [Cr(H2O)5Cl]Cl2, silver chloride will be precipitated. The chloride ions of [Cr(H2O)5Cl]Cl2 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(H2O)5Cl]Cl2, silver chloride will be precipitated. But not all the chloride ions of [Cr(H2O)5Cl]Cl2 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(H2O)5Cl]Cl2 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:
moles(compound)=30 x 0.01
Since, there are two chloride ions present in the compound, the moles of chloride ions will be:
moles(Cl−)=2 x 30 x 0.01
The reaction which takes for the formation of silver chloride is given below:
AgNO3+Cl−→AgCl+NO3−
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 AgNO3 is 0.1M, and the volume is V.
Now on equating, we get:
0.1×V=2×0.01×30
V=6 ml
So, the volume of AgNO3 is 6ml.
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.