Question

: If excess of ${\text{AgC}}{{\text{O}}_3}$ solution is added to $100$ ml of a $0.024{\text{M}}$ solution of dichlorobis (ethylene diamine) cobalt (III) chloride. How many moles of ${\text{AgCl}}$ will be precipitated?
A.0.0012
B.0.0016
C.0.0024
D.0.0048

Answer 1

Hint: Molarity is the number of moles of solute dissolved in one litre solution. It is given as-
${\text{M = }}\dfrac{{\text{n}}}{{\text{V}}}$ Where M is molarity, n=number of moles and V is the volume in litres. We can find the number of moles of ${\text{AgCl}}$ 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 $\left( {{\text{C}}{{\text{o}}^{3 + }}} \right)$ , two chlorine ion $\left( {{\text{C}}{{\text{l}}^ - }} \right)$ and 2 molecules of ethylene diamine $\left( {{\text{en}}} \right)$ so; the net charge in coordination sphere is-$ + 3 + 2\left( { - 1} \right) + 2\left( 0 \right) = + 1$ as (en) is neutral
So the complex can be written as ${\left[ {{\text{Co}}{{\left( {{\text{en}}} \right)}_2}{\text{C}}{{\text{l}}_2}} \right]^ + }{\text{C}}{{\text{l}}^ - }$ .Now we know the complex so we can dissociate to find how many chlorine ions will the complex release to form ${\text{AgCl}}$-
$ \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 ${\text{AgC}}{{\text{O}}_3}$ in complex-
Now here Molarity (M) =$0.024$, V=$100$ ml=$0.1$ L as $1000{\text{ml = 1L}}$
We know the formula is-
$ \Rightarrow $ ${\text{M = }}\dfrac{{\text{n}}}{{\text{V}}}$ Where M is molarity , n=number of moles and V is the volume in litres
On putting the values we get-
$ \Rightarrow 0.024 = \dfrac{{\text{n}}}{{0.1}}$
On solving we get the number of moles of ${\text{AgC}}{{\text{o}}_3}$ in the complex-
$ \Rightarrow {\text{n}} = 0.024 \times 0.1 = 0.0024$ moles
Since ${\text{AgC}}{{\text{O}}_3}$ produces an equimolar amount of ${\text{AgCl}}$, so their moles will also be equal.
So the number of moles of ${\text{AgCl}}$ 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.