Question

When \[10{\text{ ml}}\] of \[0.2{\text{ M AgN}}{{\text{O}}_3}\] is mixed with \[10{\text{ ml}}\] of \[0.1{\text{ M NaCl}}\], then what is the concentration of \[C{l^{ - 1}}\] in the resulting solution \[\left[ {{K_{sp}}(AgCl){\text{ = 1}}{{\text{0}}^{ - 10}}{\text{ }}{{\text{M}}^2}} \right]\]
\[(i){\text{ 1}}{{\text{0}}^{ - 5}}{\text{ M}}\]
\[(ii){\text{ 1}}{{\text{0}}^{ - 9}}{\text{ M}}\]
\[(iii){\text{ 2 }} \times {\text{ 1}}{{\text{0}}^{ - 9}}{\text{ M}}\]
\[(iv){\text{ 1}}{{\text{0}}^{ - 8}}{\text{ M}}\]

Answer 1

Hint: Since the volume of both the solutions are given equal therefore the concentration of each solution will become half. Hence we will calculate the concentration of ions when they start dissociating into ions. We will find the concentration of \[C{l^{ - 1}}\] in the resulting solution by using \[{K_{sp}}\] of \[AgCl\].
Formula Used:
\[{K_{sp}}\left( {AgCl} \right){\text{ = }}\left[ {A{g^ + }} \right]{\text{ }}\left[ {C{l^{ - 1}}} \right]\]

Complete answer:
When the equal volume of \[0.2{\text{ M AgN}}{{\text{O}}_3}\] and \[0.1{\text{ M NaCl}}\] are mixed then the volume of resulting solution becomes double which means the concentration of each will be halved. Therefore the concentration of \[0.2{\text{ M AgN}}{{\text{O}}_3}\] becomes \[0.1{\text{ M AgN}}{{\text{O}}_3}\] and \[0.1{\text{ M NaCl}}\] will be halved and become \[0.05{\text{ M NaCl}}\]
The dissociation of \[0.1{\text{ M AgN}}{{\text{O}}_3}\] can be written as,
\[{\text{AgN}}{{\text{O}}_3}{\text{ }} \rightleftharpoons {\text{ A}}{{\text{g}}^ + }{\text{ + N}}{{\text{O}}_3}^{ - 1}\]
We can find the concentration of its consecutives ions at \[t{\text{ = 0 }}\] and \[t{\text{ = }}t\] as,

Time(t)	[$AgNO_3$]	[$Ag^+$]	[$NO_3^{-1}$]
t=0	0.1	0	0
t=t	0	0.1	0.1

Similarly the dissociation of \[0.05{\text{ M NaCl}}\] can be written as,
\[{\text{NaCl }} \rightleftharpoons {\text{ N}}{{\text{a}}^ + }{\text{ + C}}{{\text{l}}^{ - 1}}\]
We can find the concentration of its consecutives ions at \[t{\text{ = 0 }}\] and \[t{\text{ = }}t\] as,

Time(t)	[NaCl]	[$Na^+$]	[$Cl^{-1}$]
t=0	0.05	0	0
t=t	0	0.05	0.05

When both these solution mixed then we get \[AgCl\] whose dissociation can be represented as,
\[{\text{AgCl }} \rightleftharpoons {\text{ A}}{{\text{g}}^ + }{\text{ + C}}{{\text{l}}^{ - 1}}\]
Since the limiting reagent is \[C{l^{ - 1}}\]. Therefore for above reaction the concentration of \[\left[ {A{g^ + }} \right]\] will be equal to \[\left( {0.1{\text{ - 0}}{\text{.05}}} \right){\text{ M}}\] which is small in number so it can be ignorable. Thus \[\left[ {A{g^ + }} \right]\] will be equal to \[0.05{\text{ }}M\].
The \[{K_{sp}}\] for the reaction will be represented as,
\[{K_{sp}}{\text{ = }}\left[ {A{g^ + }} \right]{\text{ }}\left[ {C{l^{ - 1}}} \right]\]
Since the value of \[{K_{sp}}\] is given in question, thus on substituting the value we get the result as,
\[{\text{1}}{{\text{0}}^{ - 10}}{\text{ = }}\left[ {0.05} \right]{\text{ }}\left[ {C{l^{ - 1}}} \right]\]
\[\dfrac{{{\text{1}}{{\text{0}}^{ - 10}}}}{{0.05}}{\text{ = }}\left[ {C{l^{ - 1}}} \right]\]
\[\left[ {C{l^{ - 1}}} \right]{\text{ = 2 }} \times {\text{ 1}}{{\text{0}}^{ - 9}}{\text{ M}}\]
Therefore the concentration of \[C{l^{ - 1}}\] in the resulting solution will be \[{\text{2 }} \times {\text{ 1}}{{\text{0}}^{ - 9}}{\text{ M}}\].

Note:
Molarity is the ratio of number of moles and volume of solution. Therefore when volume doubles the molarity is reduced to half. Remember the concept of limiting reagent while calculating the concentration of \[\left[ {A{g^ + }} \right]\]. Concentration is similar to molarity when volume of solution is constant.