Question

A simultaneous determination for cobalt and nickel can be based on absorption by their respective 8-hydroxyquinolinol complexes. Molar absorptivities corresponding to their absorption maxima are given below:-

Calculate the molar concentration of nickel and cobalt in each of the following solutions using the following data: Absorbance, A (1.00-cm cells) Solution 365 nm 700 nm (a) 0.349 0.022 (b) 0.792 0.081

Short answer

Part(a)$$\begin{gathered} c_{\text{Ni}}=5.34\times {10}^{-5}M \\ {c}_{\text{Co}}=5\times {10}^{-5}M \end{gathered}$$

Part (b)$$\begin{gathered} c_{\text{Ni}}=1.88\times {10}^{-4}M \\ {c}_{\text{Co}}=3.99\times {10}^{-5}M \end{gathered}$$

Step-by-step solution

Part (a) Step1. Given information

For cobalt

$$\begin{gathered} \text{Molarabsorptivityat365nm}=3529 \\ A=0.349 \\ l\left(A\right)=1\mathrm{cm} \end{gathered}$$

For Nickel:-

$$\begin{gathered} \text{Molarabsorptivityat365nm}=3228 \\ A=0.349 \\ l\left(A\right)=1\mathrm{cm} \end{gathered}$$

Part (a) Step2. Calculate the concentration of cobalt 

$$\begin{gathered} A_{365}=0.349=3529\times 1.00\times c_{\text{Co}}+3228\times 1\times c_{\text{Ni}} \\ 0.349=3529\times c_{\text{A}}+3228\times c_{\text{Ni}}.............\left(1\right) \\ {A}_{700}=0.022=428.9\times 1.00\times c_{\text{Co}}+10.2\times 1\times c_{\text{Ni}} \\ 0.022=428.9\times c_{\text{A}}+10.2\times c_{\text{Ni}}............\left(2\right) \end{gathered}$$

$\left(1\right)\times 10.2-\left(2\right)\times 3228$

$$\begin{gathered} 3.5598-71.016=35995.8C_{\text{Co}}-1384489.2C_{\text{Co}} \\ 1348493.4C_{\text{Co}}=67.4562 \\ {C}_{\text{Co}}=5.00\times {10}^{-5}\text{mol/L} \end{gathered}$$

Part (a) Step3. Calculate the concentration of nickel 

Substitute the value of the concentration of cobalt in equation (1):-

$$\begin{gathered} 0.349=3529\times 5.00\times {10}^{-5}+3228\times c_{\text{Ni}} \\ 0.349=0.17645+3228c_{\text{Ni}} \\ {c}_{\text{Ni}}=5.35\times {10}^{-5}{\text{molL}}^{-1} \end{gathered}$$

Part (b) Step1. Calculate the concentration of cobalt.

$$\begin{gathered} A_{365}=0.792=3529\times 1.00\times c_{\text{Co}}+3228\times 1\times c_{\text{Ni}} \\ 0.792=3529\times c_{\text{A}}+3228\times c_{\text{Ni}}.............\left(1\right) \\ {A}_{700}=0.081=428.9\times 1.00\times c_{\text{Co}}+10.2\times 1\times c_{\text{Ni}} \\ 0.081=428.9\times c_{\text{A}}+10.2\times c_{\text{Ni}}............\left(2\right) \end{gathered}$$

$\left(1\right)\times 10.2-\left(2\right)\times 3228$

$$\begin{gathered} 8.0784-261.468=3599.5C_{\text{Co}}-1384489.2C_{\text{Co}} \\ 1348493.4C_{\text{Co}}=253.3896 \\ {C}_{\text{Co}}=1.88\times {10}^{-4}\text{mol/L} \end{gathered}$$

Part (b) Step2. Calculate the concentration of nickel.

Substitute the value of the concentration of cobalt in the equation (1) as follows:-

$$\begin{gathered} 0.792=3529\times 1.88\times {10}^{-4}+3228C_{\text{Ni}} \\ {C}_{\text{Ni}}=3.98\times {10}^{-5}\text{mol/L} \end{gathered}$$