Q19.103P - Chemistry: Molecular Nature Of Matter And Change

Question

When$\mathbf{\text{0.84}}$  g of  ${\mathbf{\text{ZnCl}}}_{\mathbf{\text{2}}}$ is dissolved in 245 mL of 0.150 M  , $\mathbf{\text{NaCN}}$what are $\mathbf{\left[}{\mathbf{\text{Zn}}}^{\mathbf{\text{2+}}}\mathbf{\right]}$ , $\mathbf{\left[}{{\mathbf{\text{Zn(CN)}}}_{\mathbf{\text{4}}}}^{\mathbf{\text{2-}}}\mathbf{\right]}$ , and  [ $\mathbf{\left[}{\mathbf{\text{CN}}}^{\mathbf{\text{-}}}\mathbf{\right]}$ ${\mathbf{\text{K}}}_{\mathbf{\text{f}}}$of ${{\mathbf{\text{Zn(CN)}}}_{\mathbf{\text{4}}}}^{\mathbf{\text{2-}}}{\mathbf{\text{=4.2×10}}}^{\mathbf{\text{19}}}$]?

Step-by-Step Solution

Answer

The required values are:

$\begin{array}{c}\left[{\text{Zn(CN)}}_{\text{4}}^{\text{2-}}\right]\text{=0.02516M}\\ \left[{\text{CN}}^{\text{-}}\right]\text{=0.150M-4×0.02516M}\\ \text{=0.0494M}\\ \left[{\text{Zn}}^{\text{2+}}\right]\text{=}\frac{\left[{\text{Zn(CN)}}_{\text{4}}^{\text{2-}}\right]}{\text{Kf×}{\left[{\text{CN}}^{\text{-}}\right]}^{\text{4}}}\\ {\text{=1.01×10}}^{\text{-16}}\text{M}\end{array}$

Step 1: 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.

Step 2:Find the values of [ Zn 2+ ] [ Zn(CN) 4 2- ] [ CN - ]

Simplify the given values:

 $\begin{array}{c}\text{m}\left({\text{ZnCl}}_{\text{2}}\right)\text{=0.84 gV(NaCN)}\\ \text{=245ml}\\ \text{=0.245lc(NaCN)}\\ \text{=0.150M}\end{array}$

We can start by writing the formation equation for :${\text{Zn(CN)}}_{\text{4}}^{\text{2-}}$

 $\begin{array}{l}{\text{Zn}}^{\text{2+}}{\text{+4CN}}^{\text{-}}\rightleftharpoons {\text{Zn(CN)}}_{\text{4}}^{\text{2-}}\\ \text{Kf=}\frac{\left[{\text{Zn(CN)}}_{\text{4}}^{\text{2-}}\right]}{\left[{\text{Zn}}^{\text{2+}}\right]\text{×}{\left[{\text{CN}}^{\text{-}}\right]}^{\text{4}}}{\text{=4.2×10}}^{\text{19}}\end{array}$

Now, we can calculate$\left[{\text{Zn}}^{\text{2+}}\right]$  :

$\begin{array}{c}\text{n}\left({\text{Zn}}^{\text{2+}}\right)\text{=}\frac{{\text{0.84gZnCl}}_{\text{2}}}{{\text{136.286g/molZnCl}}_{\text{2}}}\\ \text{=0.0061 mol}\\ \left[{\text{Zn}}^{\text{2+}}\right]\text{=}\frac{\text{0.0061mol}}{\text{0.245l}}\\ \text{=0.02516M-x}\end{array}$

We put "-  x" because some of the${\text{Zn}}^{\text{2+}}$  will be consumed during the reaction

Now we can do the same for $\left[{\text{CN}}^{\text{-}}\right]$ :

 $\left[{\text{CN}}^{\text{-}}\right]\text{=0.150M-4x}$

We put " -4x " because for every${\text{Zn}}^{\text{2+}}$  we need  .

 And now we do the same for :$\left[{\text{Zn(CN)}}_{\text{4}}^{\text{2-}}\right]$

 $\left[{\text{Zn(CN)}}_{\text{4}}^{\text{2-}}\right]\text{=x}$

We must now suppose that all of the initial  $\left[{\text{Zn}}^{\text{2+}}\right]$ has completely reacted, resulting in a concentration of the produced ${\text{Zn(CN)}}_{\text{4}}^{\text{2-}}$  be $\text{0.02516M}$ and that is " x " value.

 $\left[{\text{Zn(CN)}}_{\text{4}}^{\text{2-}}\right]\text{=0.02516M}$

Now that we have our "  x " value we can calculate  :$\left[{\text{CN}}^{\text{-}}\right]$

$\begin{array}{c}\left[{\text{CN}}^{\text{-}}\right]\text{=0.150M-4×0.02516M}\\ \text{=0.0494M}\end{array}$ 

Now we have to use the Kf expression to calculate  $\left[{\text{Zn}}^{\text{2+}}\right]$ :

 $\begin{array}{c}\left[{\text{Zn}}^{\text{2+}}\right]\text{=}\frac{\left[{\text{Zn(CN)}}_{\text{4}}^{\text{2-}}\right]}{\text{Kf×}{\left[{\text{CN}}^{\text{-}}\right]}^{\text{4}}}\\ {\text{=1.01×10}}^{\text{-16}}\text{M}\end{array}$

Therefore, the values are

$\begin{array}{c}\left[{\text{Zn(CN)}}_{\text{4}}^{\text{2-}}\right]\text{=0.02516M}\\ \left[{\text{CN}}^{\text{-}}\right]\text{=0.150M-4×0.02516M}\\ \text{=0.0494M}\\ \left[{\text{Zn}}^{\text{2+}}\right]\text{=}\frac{\left[{\text{Zn(CN)}}_{\text{4}}^{\text{2-}}\right]}{\text{Kf×}{\left[{\text{CN}}^{\text{-}}\right]}^{\text{4}}}\\ {\text{=1.01×10}}^{\text{-16}}\text{M}\end{array}$