91Ó°ÊÓ

In polar liquids, the ionic material dissociates into its ions in ionic equilibrium. The ions generated in the solution are always in equilibrium with the undissociated solute.

Consider the given values,

$\begin{array}{rcl}\mathrm{V}\left({\mathrm{AgNO}}_3\right)& =& 25\mathrm{ml}\\ & =& 0.025\mathrm{l}\\ \text{c}\left({\text{AgNO}}_{\text{3}}\right)& =& 0.044\mathrm{M}\\ \mathrm{V}\left({\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3\right)& =& 25\mathrm{ml}\\ & =& 0.025\mathrm{l}\\ \mathrm{c}\left({\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3\right)& =& 0.57\mathrm{M}\end{array}$

First, we can write the forming reaction$\text{Ag}{\left({\text{S}}_{\text{2}}{\text{O}}_{\text{3}}\right)}_{\text{2}}^{\text{3 -}}$

$$\begin{gathered} {\text{Ag}}^{+}{\text{+2S}}_2{\text{O}}_3^{2-}⇌\text{Ag}{\left({\text{S}}_{\text{2}}{\text{O}}_{\text{3}}\right)}_{\text{2}}^{\text{3 -}} \\ {\text{K}}_{\text{f}}\text{=}\frac{{\displaystyle \left[\text{Ag}{\left({\text{S}}_{\text{2}}{\text{O}}_{\text{3}}\right)}_{\text{2}}^{\text{3 -}}\right]}}{{\displaystyle {\text{[Ag}}^{+}\text{±Õ×}{\left[{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}^{\text{2 -}}\right]}^{\text{2}}}} \end{gathered}$$

Now, we need to understand which is the limiting reagent among${\text{Ag}}^{\text{+}} \text{and}  {\text{S}}_{\text{2}}{{\text{O}}_{\text{3}}}^{\text{- 2}}$

${\text{Number of moles of Ag}}^{+}\text{=0.025×0.044²Ñ}\left({\mathrm{AgNO}}_3\right)\text{=0.0011mole}$

From the balanced reaction, it is evident that, 1 mole of ${\text{Ag}}^{\text{+}}$react with 2 moles of ${\text{S}}_{\text{2}}{{\text{O}}_{\text{3}}}^{\text{- 2}}$. Therefore,

$\text{Mole} \text{ratio=}\frac{\text{1} \text{mol} {\text{Ag}}^{\text{+}}}{\text{2} \text{mol} {\text{S}}_{\text{2}}{{\text{O}}_{\text{3}}}^{\text{- 2}}}$

${\text{No of moles ofS}}_2{{\text{O}}_3}^{-2}\text{reacted = 0.0011}  \text{³¾´Ç±ô×}2=0.0022 \text{mol}$

We can now compute the remaining thiosulfate concentration present in the solution.

$\begin{array}{rcl}\mathrm{Total}\mathrm{volume}& =& \text{0.025L + 0.025L = 0.05L}\\ \left[{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}^{\text{2 -}}\right]& =& \frac{{\displaystyle \left(0.01425-0.0022\right)\text{mol}}}{{\displaystyle \text{0.05L}}}\\ & =& \text{0.241M}\end{array}$

${\text{No of moles of AgS}}_2{{\text{O}}_3}^{-3}\text{formed=0.0011}  \text{mol}$

We can now compute the concentration of a generated complex ion using the following formula:

$\begin{array}{rcl}\left[\text{Ag}{\left({\text{S}}_{\text{2}}{\text{O}}_{\text{3}}\right)}_{\text{2}}^{\text{3 -}}\right]& =& \frac{0.0011\mathrm{mol}}{0.05\mathrm{l}}\\ & =& 0.022\mathrm{M}\end{array}$

And now we use the expression to determine the concentration of :

$$\begin{gathered} \left[{\text{Ag}}^{\text{+}}\right]=\frac{{\displaystyle \left[\text{Ag}{\left({\text{S}}_{\text{2}}{\text{O}}_{\text{3}}\right)}_{\text{2}}^{\text{3 -}}\right]}}{{\displaystyle {\text{K}}_{\text{f}}\text{×}{\left[{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}^{\text{2 -}}\right]}^2}} \\ =\frac{{\displaystyle 0.022 \text{M}}}{{\displaystyle {\text{4.7×10}}^{13}\text{×}{\left(0.0241 \text{M}\right)}^2}} \\ =8.1×{10}^{-15}\mathrm{M} \end{gathered}$$

Therefore, the concentration of ${\mathrm{Ag}}^{+}$is$8.1×{10}^{-15}\mathrm{M}$