Question

Mixing the chelating reagent B with Ni(II) forms the highly colored NiB₂²⁺, whose solutions obey Beer’s law at 395 nm over a wide range. Provided the analytical concentration of the chelating reagent exceeds that of Ni(II) by a factor of 5 (or more), the cation exists, within the limits of observation, entirely in the form of the complex. Use the accompanying data to evaluate the formation constant K_f for the process

${\mathrm{Ni}}^{2+}+2\mathrm{B}\rightleftharpoons {{\mathrm{NiB}}_2}^{2+}$

Short answer

The value for formation constant for the process is:$K_f=3.43\times {10}^5$

Step-by-step solution

Given Information

Explanation

Molar absorptivity of NiB₂²⁺,

$$\begin{gathered} 0.718=\mathrm{\epsilon }\times 1.00\mathrm{cm}\times 2.00\times {10}^{-4}\mathrm{M} \\ \mathrm{\epsilon }=\frac{0.718}{1.00\mathrm{cm}\times 2.00\times {10}^{-4}\mathrm{M}} \\ \mathrm{\epsilon }=3.59\times {10}^3{\mathrm{M}}^{-1}{\mathrm{cm}}^{-1} \\ 0.276=3.59\times {10}^3{\mathrm{M}}^{-1}{\mathrm{cm}}^{-1}\times \left[{{\mathrm{NiB}}_2}^{2+}\right]\times 1.00\mathrm{cm} \\ \left[{{\mathrm{NiB}}_2}^{2+}\right]=\frac{0.276}{3.59\times {10}^3{\mathrm{M}}^{-1}{\mathrm{cm}}^{-1}\times 1.00\mathrm{cm}} \\ \left[{{\mathrm{NiB}}_2}^{2+}\right]=7.69\times {10}^{-5}\mathrm{M} \\ \left[{\mathrm{Ni}}^{2+}\right]\mathrm{left}=2.00\times {10}^{-4}-7.69\times {10}^{-5}=1.231\times {10}^{-4}\mathrm{M} \\ \left[\mathrm{B}\right]\mathrm{left}=1.5\times {10}^{-3}-(2\times 7.69\times {10}^{-5})=1.35\times {10}^{-3}\mathrm{M} \\ {\mathrm{K}}_{\mathrm{f}}=\frac{\left[{{\mathrm{NiB}}_2}^{2+}\right]}{\left[{\mathrm{Ni}}^{2+}\right]{\left[\mathrm{B}\right]}^2} \\ {\mathrm{K}}_{\mathrm{f}}=\frac{7.69\times {10}^{-5}}{1.231\times {10}^{-4}\times {(1.35\times {10}^{-3})}^2} \\ {\mathrm{K}}_{\mathrm{f}}=3.43\times {10}^5 \end{gathered}$$