Question

What is $\left[{\mathrm{Ag}}^{+}\right]$when25.0 mL each of 0.044 M ${\mathrm{AgNO}}_3$and 0.57 M ${\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3$are mixed[${\text{K}}_{\text{f}}$of $\mathrm{Ag}{\left({\mathrm{S}}_2{\mathrm{O}}_3\right)}_2^{3-}=4.7\times {10}^{13}$]?

Short answer

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

Step-by-step solution

Concept Introduction

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.

Calculation of

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$\text{Ag}{\left({\text{S}}_{\text{2}}{\text{O}}_{\text{3}}\right)}_{\text{2}}^{\text{3 -}}$

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

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

${\text{Number of moles of Ag}}^{+}\text{=0.025×0.044M}\left({\mathrm{AgNO}}_3\right)\text{=0.0011mole}$

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,

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

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

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 :

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

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