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Question: The volume (in mL) of \(0.1 M\) \(AgN{{O}_{3}}\) required for complete precipitation of chloride ion...

The volume (in mL) of 0.1M0.1 M AgNO3AgN{{O}_{3}} required for complete precipitation of chloride ions present in 30mL30 mL of 0.01M0.01 M solution of [Cr(H2O)5Cl]Cl2[Cr{{({{H}_{2}}O)}_{5}}Cl]C{{l}_{2}}, as silver chloride is close to:
(A) 6
(B) 8
(C) 9
(D) 7

Explanation

Solution

Silver nitrate when reacts with [Cr(H2O)5Cl]Cl2[Cr{{({{H}_{2}}O)}_{5}}Cl]C{{l}_{2}}, silver chloride will be precipitated. The chloride ions of [Cr(H2O)5Cl]Cl2[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(H2O)5Cl]Cl2[Cr{{({{H}_{2}}O)}_{5}}Cl]C{{l}_{2}}, silver chloride will be precipitated. But not all the chloride ions of [Cr(H2O)5Cl]Cl2[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(H2O)5Cl]Cl2[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:
moles(compound)=30 x 0.01mole{{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:
moles(Cl)=2 x 30 x 0.01mole{{s}_{(C{{l}^{-}})}}=2\text{ x 30 x 0}\text{.01}
The reaction which takes for the formation of silver chloride is given below:
AgNO3+ClAgCl+NO3AgN{{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 AgNO3AgN{{O}_{3}} is 0.1M0.1 M, and the volume is VV.
Now on equating, we get:
0.1×V=2×0.01×300.1 \times V = 2 \times 0.01 \times 30
V=6 mlV=6\text{ }ml
So, the volume of AgNO3AgN{{O}_{3}} is 6ml6 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.