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 ${{\mathbf{\text{Zn(CN)}}}_{\mathbf{\text{4}}}}^{\mathbf{\text{2-}}}{\mathbf{\text{=4.2×10}}}^{\mathbf{\text{19}}}$]?

Short 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-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.

Step 2:Find the values of  [Zn2+][Zn(CN)42-][CN-]

Simplify the given values:

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

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$\left[{\text{Zn}}^{\text{2+}}\right]$ :

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

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

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