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Question: : If excess of \({\text{AgC}}{{\text{O}}_3}\) solution is added to \(100\) ml of a \(0.024{\text{M}}...

: If excess of AgCO3{\text{AgC}}{{\text{O}}_3} solution is added to 100100 ml of a 0.024M0.024{\text{M}} solution of dichlorobis (ethylene diamine) cobalt (III) chloride. How many moles of AgCl{\text{AgCl}} will be precipitated?
A.0.0012
B.0.0016
C.0.0024
D.0.0048

Explanation

Solution

Molarity is the number of moles of solute dissolved in one litre solution. It is given as-
M = nV{\text{M = }}\dfrac{{\text{n}}}{{\text{V}}} Where M is molarity, n=number of moles and V is the volume in litres. We can find the number of moles of AgCl{\text{AgCl}} by using this formula.

Complete step by step answer:
Here the given complex is dichlorobis (ethylene diamine) cobalt (III) chloride. We have to find its formula. The coordination sphere contains cobalt ion (Co3+)\left( {{\text{C}}{{\text{o}}^{3 + }}} \right) , two chlorine ion (Cl−)\left( {{\text{C}}{{\text{l}}^ - }} \right) and 2 molecules of ethylene diamine (en)\left( {{\text{en}}} \right) so; the net charge in coordination sphere is-+3+2(−1)+2(0)=+1 + 3 + 2\left( { - 1} \right) + 2\left( 0 \right) = + 1 as (en) is neutral
So the complex can be written as [Co(en)2Cl2]+Cl−{\left[ {{\text{Co}}{{\left( {{\text{en}}} \right)}_2}{\text{C}}{{\text{l}}_2}} \right]^ + }{\text{C}}{{\text{l}}^ - } .Now we know the complex so we can dissociate to find how many chlorine ions will the complex release to form AgCl{\text{AgCl}}-
⇒[Co(en)2Cl2]+Cl−→[Co(en)2Cl2]+ + Cl−\Rightarrow {\left[ {{\text{Co}}{{\left( {{\text{en}}} \right)}_2}{\text{C}}{{\text{l}}_2}} \right]^ + }{\text{C}}{{\text{l}}^ - } \to {\left[ {{\text{Co}}{{\left( {{\text{en}}} \right)}_2}{\text{C}}{{\text{l}}_2}} \right]^ + }{\text{ + C}}{{\text{l}}^ - }
This means that only one chlorine ion (per complex molecule) is released in the solution to participate in the precipitation reaction. We will find the numbers of moles of AgCO3{\text{AgC}}{{\text{O}}_3} in complex-
Now here Molarity (M) =0.0240.024, V=100100 ml=0.10.1 L as 1000ml = 1L1000{\text{ml = 1L}}
We know the formula is-
⇒\Rightarrow M = nV{\text{M = }}\dfrac{{\text{n}}}{{\text{V}}} Where M is molarity , n=number of moles and V is the volume in litres
On putting the values we get-
⇒0.024=n0.1\Rightarrow 0.024 = \dfrac{{\text{n}}}{{0.1}}
On solving we get the number of moles of AgCo3{\text{AgC}}{{\text{o}}_3} in the complex-
⇒n=0.024×0.1=0.0024\Rightarrow {\text{n}} = 0.024 \times 0.1 = 0.0024 moles
Since AgCO3{\text{AgC}}{{\text{O}}_3} produces an equimolar amount of AgCl{\text{AgCl}}, so their moles will also be equal.
So the number of moles of AgCl{\text{AgCl}} precipitated =0.0024

Hence, the correct answer is ‘C’.

Note:
-The two chlorine ions inside the square bracket are secondary valencies and are covalently bonded to cobalt so they can’t participate in reaction
-But the chlorine ion outside the bracket is primary/ionic valency so it can participate in reaction.