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Question: A solution is a mixture of 0.06 M KCl and 0.06M KI. \({\rm{AgN}}{{\rm{O}}_{\rm{3}}}\) is being added...

A solution is a mixture of 0.06 M KCl and 0.06M KI. AgNO3{\rm{AgN}}{{\rm{O}}_{\rm{3}}} is being added drop by drop till AgCl starts precipitating (KspAgCl=1×10−10{K_{sp}}{\rm{AgCl}} = 1 \times {10^{ - 10}} and KspAgI=4×10−16{K_{sp}}{\rm{AgI}} = 4 \times {10^{ - 16}}). The concentration of iodide ion at this stage will be nearly equal to:
A.4×10−5 M4 \times {10^{ - 5}}\,{\rm{M}}
B. 2.4×10−7 M2.4 \times {10^{ - 7}}\,{\rm{M}}
C. 2.0×10−8 M2.0 \times {10^{ - 8}}\,{\rm{M}}
D. 4×10−8 M4 \times {10^{ - 8}}\,{\rm{M}}

Explanation

Solution

Here, a solution of mixture of KCl and KI is given and AgNO3{\rm{AgN}}{{\rm{O}}_{\rm{3}}}is added till the precipitation of AgCl. To calculate iodide ion, first we calculate the minimum silver ions required for the precipitation of AgCl and AgI. Then, we compare the concentration of silver ions in both the cases and then we have to identify which one undergoes precipitation first and then calculate the iodide ion concentration using suitable Ag+{\rm{A}}{{\rm{g}}^ + }ion in solubility product expression.

Complete step by step answer:
When the two solutions KCl and KI is mixed, the resulting solution contains two anions (Cl−{\rm{C}}{{\rm{l}}^ - }, I−{{\rm{I}}^ - }) which can combine with Ag+{\rm{A}}{{\rm{g}}^ + } ions to result an insoluble precipitate of AgCl{\rm{AgCl}} and AgI{\rm{AgI}}.

Now, we write solubility equilibrium for silver chloride.

AgCl→Ag++Cl−{\rm{AgCl}} \to {\rm{A}}{{\rm{g}}^ + } + {\rm{C}}{{\rm{l}}^ - }

The solubility product expression for AgCl is,

Ksp(AgCl)=[Ag+][Cl]−{K_{sp({\rm{AgCl}})}} = \left[ {{\rm{A}}{{\rm{g}}^ + }} \right]{\left[ {{\rm{Cl}}} \right]^ - } …… (1)

Now, there are two conditions that silver ions can combine with (Cl−,I−{\rm{C}}{{\rm{l}}^ - },{{\rm{I}}^ - }).

The concentration of Cl−{\rm{C}}{{\rm{l}}^ - }=0.06 and KspAgCl=1×10−10{K_{sp}}{\rm{AgCl}} = 1 \times {10^{ - 10}}(Given)

Now, we have to calculate the minimum concentration of silver ions required for precipitation of silver chloride using equation (1).

[Ag+]=Ksp(AgCl)[Cl]−\left[ {{\rm{A}}{{\rm{g}}^ + }} \right] = \dfrac{{{K_{sp({\rm{AgCl}})}}}}{{{{\left[ {{\rm{Cl}}} \right]}^ - }}}

⇒[Ag+]=1×10−100.06 \Rightarrow \left[ {{\rm{A}}{{\rm{g}}^ + }} \right] = \dfrac{{1 \times {{10}^{ - 10}}}}{{0.06}}

⇒[Ag+]=1.66×10−9 \Rightarrow \left[ {{\rm{A}}{{\rm{g}}^ + }} \right] = 1.66 \times {10^{ - 9}}

Now, for silver iodide the equilibrium is,

AgI→Ag++I−{\rm{AgI}} \to {\rm{A}}{{\rm{g}}^ + } + {{\rm{I}}^ - }

The solubility product expression for AgI is,

Ksp(AgI)=[Ag+][I]−{K_{sp({\rm{AgI}})}} = \left[ {{\rm{A}}{{\rm{g}}^ + }} \right]{\left[ {\rm{I}} \right]^ - } …………..…… (2)

The concentration of l−{{\rm{l}}^ - }=0.06 and KspAgI=4×10−16{K_{sp}}{\rm{AgI}} = 4 \times {10^{ - 16}}
(Given)

Now, we have to calculate minimum silver ions required for precipitation of AgI using equation (2).

Ksp(AgI)=[Ag+][I]−{K_{sp({\rm{AgI}})}} = \left[ {{\rm{A}}{{\rm{g}}^ + }} \right]{\left[ {\rm{I}} \right]^ - }

⇒[Ag+]=4×10−160.06 \Rightarrow \left[ {{\rm{A}}{{\rm{g}}^ + }} \right] = \dfrac{{4 \times {{10}^{ - 16}}}}{{0.06}}

⇒[Ag+]=6.67×10−15 \Rightarrow \left[ {{\rm{A}}{{\rm{g}}^ + }} \right] = 6.67 \times {10^{ - 15}}

We see that silver ions required for precipitation of AgI are very less than AgCl. This indicates that silver iodide precipitates first consuming some iodide ions from the solution. So, we need concentration of silver ions equal to start precipitating is,

⇒[Ag+]=1.66×10−9 \Rightarrow \left[ {{\rm{A}}{{\rm{g}}^ + }} \right] = 1.66 \times {10^{ - 9}}

Present in the solution. This indicates that, at this point, the corresponding concentration of iodide ions must be at maximum that corresponds to solubility equilibrium condition. Now, using equation (2),

⇒[I−]=4×10−161.66×10−9=2.4×10−7 M \Rightarrow \left[ {{{\rm{I}}^ - }} \right] = \dfrac{{4 \times {{10}^{ - 16}}}}{{1.66 \times {{10}^{ - 9}}}} = 2.4 \times {10^{ - 7}}\,{\rm{M}}

Therefore, iodide concentration at the stage when AgCl starts precipitating is 2.4×10−7 M2.4 \times {10^{ - 7}}\,{\rm{M}}.

So, the correct answer is Option B.

Note:
Remember that, solubility product (Ksp)\left( {{K_{sp}}} \right) is the product of concentration of ions of the salt in its saturated solution at a given temperature raised to the power the number of ions produced by dissociation of one mole of salt. A salt in a solution is precipitated only when the product of ionic concentration is more than the value of Ksp{K_{sp}}.