Question

For a solution of ${\mathrm{Ni}}^{2+}$and ethylenediamine, the following

equilibrium constants apply atrole="math" localid="1663385095626" ${\mathbf{20}}^{\mathbf{°}}$c:

role="math" localid="1663386154645" $$\begin{gathered} {\mathbf{Ni}}^{\mathbf{2}}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}{\mathbf{NCH}}_{\mathbf{2}}{\mathbf{CH}}_{\mathbf{2}}{\mathbf{NH}}_{\mathbf{2}}\mathbf{}\mathbf{}\mathbf{Ni}{\mathbf{\left(}\mathbf{cn}\mathbf{\right)}}^{\mathbf{2}\mathbf{+}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{log}\mathbf{}{\mathbf{K}}_{\mathbf{1}}\mathbf{}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{25} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Eilhylenediamine}\mathbf{}\mathbf{}\mathbf{(}\mathbf{abbreviated}\mathbf{}\mathbf{en}\mathbf{)} \\ \mathbf{Ni}{\mathbf{\left(}\mathbf{en}\mathbf{\right)}}^{\mathbf{2}}\mathbf{+}\mathbf{}\mathbf{cn}\mathbf{}\mathbf{}\mathbf{}\mathbf{Ni}{\mathbf{\left(}\mathbf{en}\mathbf{\right)}}_{\mathbf{2}}^{\mathbf{2}\mathbf{+}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{log}\mathbf{}{\mathbf{K}}_{\mathbf{2}}\mathbf{}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{32} \\ \mathbf{Ni}{\mathbf{\left(}\mathbf{en}\mathbf{\right)}}^{\mathbf{2}}\mathbf{+}\mathbf{}\mathbf{cn}\mathbf{}\mathbf{}\mathbf{}\mathbf{Ni}{\mathbf{\left(}\mathbf{en}\mathbf{\right)}}_{\mathbf{3}}^{\mathbf{2}\mathbf{+}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{log}\mathbf{}{\mathbf{K}}_{\mathbf{3}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{49}\mathbf{}\mathbf{}\mathbf{} \end{gathered}$$

Calculate the concentration of free role="math" localid="1663386355087" ${\mathbf{Ni}}^{\mathbf{2}\mathbf{+}}$ in a solution prepared by mixing

0.100 mol of en plus 1.00 mL of role="math" localid="1663386332852" $\mathbf{0}\mathbf{.}\mathbf{0100}{\mathbf{MNi}}^{\mathbf{2}\mathbf{+}}$and diluting to 1.00 L with dilute base (which keeps all the enin its unprotonated form). Assume that nearly allnickel is in the form$\mathbf{}\mathbf{Ni}{\left(\mathrm{cn}\right)}_{\mathbf{3}}^{\mathbf{2}\mathbf{+}}\mathbf{}\mathbf{}\mathbf{so}\mathbf{}\left[\mathrm{Ni}{\left(\mathrm{cn}\right)}_3^{2+}\right]\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{x}{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{M}$.Calculate the concentrations of

$\mathbf{Ni}{\mathbf{\left(}\mathbf{en}\mathbf{\right)}}^{\mathbf{2}\mathbf{+}}\mathbf{}\mathbf{and}\mathbf{}\mathbf{Ni}{\mathbf{\left(}\mathbf{en}\mathbf{\right)}}_{\mathbf{3}}^{\mathbf{2}\mathbf{+}}$to verify that they are negligible incomparison with $\mathbf{Ni}{\mathbf{\left(}\mathbf{en}\mathbf{\right)}}_{\mathbf{3}}^{\mathbf{2}\mathbf{+}}$.

Short answer

For given solution,

- Concentration of free ${\mathrm{Ni}}^{2-}\mathrm{is}4.7\mathrm{x}10{}^{21}\mathrm{M}$

- Concentration of $\mathrm{Ni}{\left(\mathrm{cn}\right)}^2\mathrm{is}1.5\mathrm{x}{10}^{-14}\mathrm{M}$and concentration of $\mathrm{Ni}{\left(\mathrm{cn}\right)}_3^{2+}\mathrm{is}1.5\mathrm{x}{10}^{-14}\mathrm{M}$On verifying, these two solutions are very less than${10}^{-21}\mathrm{M}$.

Step-by-step solution

concentration for given solution:

For given solution, - Concentration of free ${\mathrm{Ni}}^2$ must be calculated.-

Concentration of$\mathrm{Ni}{\left(\mathrm{en}\right)}^2$and$\mathrm{Ni}{\left(\mathrm{en}\right)}_3^{2I}$must be calculated to prove that theyare insignificant in comparison with$\mathrm{Ni}{\left(\mathrm{en}\right)}_3^{2_I}$.

solve for concentration:

Given

Given equilibrium is,

$$\begin{gathered} {\mathrm{Ni}}^2+{\mathrm{H}}_2{\mathrm{NCH}}_2{\mathrm{CH}}_2{\mathrm{NH}}_2\mathrm{Ni}{\left(\mathrm{cn}\right)}^{2+}\log {\mathrm{K}}_1=7.25 \\ \mathrm{Ni}{\left(\mathrm{en}\right)}^2+\mathrm{cn}\mathrm{Ni}{\left(\mathrm{en}\right)}_2^{2+}\log {\mathrm{K}}_2=6.32 \\ \mathrm{Ni}{\left(\mathrm{en}\right)}^2+\mathrm{cn}\mathrm{Ni}{\left(\mathrm{en}\right)}_3^{2+}\log {\mathrm{K}}_3=4.49 \end{gathered}$$

Assuming that all Ni in the form of $\mathrm{Ni}{\left(\mathrm{en}\right)}_3^{2_{\mathrm{I}}}$and the concentration of $\mathrm{Ni}{\left(\mathrm{en}\right)}_2^{2_{\mathrm{I}}}$is equal to $1.00\mathrm{x}{10}^{-5}\mathrm{M}$Three moles of $\mathrm{en}$leaves concentration of en at 0.100M .

Adding above three equations, we get

$$\begin{gathered} {\mathrm{Ni}}^{2-}+3\mathrm{en}\mathrm{Ni}{\left(\mathrm{en}\right)}_3^{2+} \\ \mathrm{K}={\mathrm{K}}_1{\mathrm{K}}_2{\mathrm{K}}_3=2.14\mathrm{x}{10}^{18} \\ \left[{\mathrm{Ni}}^{2+}\right]=\frac{\left[\mathrm{Ni}{\left(\mathrm{en}\right)}_3^{2-}\right]}{\mathrm{K}\left[\mathrm{en}\right]3} \\ =\frac{(1.00\mathrm{x}{10}^{-})}{\left(2.14\mathrm{x}{10}^{18}\right){\left(0.100\right)}^3} \\ =4.7\mathrm{x}10{}^{21}\mathrm{M} \end{gathered}$$

verify the concentration of solutions

Now verify that the concentration of solutions$\mathrm{Ni}{\left(\mathrm{en}\right)}^2\mathrm{and}\mathrm{Ni}{\left(\mathrm{en}\right)}_2^{2_{\mathrm{I}}}\mathrm{and}\mathrm{N}{\left(\mathrm{en}\right)}^{2-}\mathrm{and}\mathrm{Ni}{\left(\mathrm{en}\right)}_2^{2+}\underline{\mathrm{x}}{10}^{-21}\mathrm{M}$

That is,

Concentration of $\mathrm{Ni}{\left(\mathrm{en}\right)}^{2+}={\mathrm{K}}_1\left[{\mathrm{Ni}}^{2+}\right]\left(\mathrm{en}\right)$

$=1.5\mathrm{x}{10}^{-14}\mathrm{M}$

Concentration of data-custom-editor="chemistry" $\mathrm{N}{\left(\mathrm{en}\right)}_2^{2_{\mathrm{I}}}=\mathrm{K}2$

$\left[\mathrm{Ni}{\left(\mathrm{en}\right)}^2\right]\left[\mathrm{en}\right]=3.2\mathrm{x}{10}^9\mathrm{M}$