Question

The sodium salt of 2-quinizarinsulfonic acid (NaQ) forms a complex with Al31 that absorbs radiation strongly at 560 nm.32 (a) Use the data from Owens and Yoe’s paper to find the formula of the complex. In all solutions, c_Al = 3.7 x 10^-5 M, and all measurements were made in 1.00-cm cells. (b) Find the molar absorptivity of the complex.

Short answer

a) The formula of the complex is AlQ²⁺.

b)${\mathrm{\epsilon }}_{{\mathrm{AlQ}}^{2+}}=1.35\times {10}^4{\mathrm{cm}}^{-1}{\mathrm{M}}^{-1}$.

Step-by-step solution

Part (a) Step 1: Given Information

c_Al = 3.7 x 10^-5 M,

Part (a) Step 2: Explanation

The plot between concentration and absorbance is:

The point of intersection of the two linear portions will give the concentration of NaQ needed to react with all Al³⁺ ions present in the solution.

At the point of intersection y and x values are equal

$$\begin{gathered} 447.41\mathrm{x}+0.4609=13250\mathrm{x}-0.001 \\ 12802.59\mathrm{x}=0.4619 \\ \mathrm{x}=3.61\times {10}^{-5}\mathrm{mol}/\mathrm{L} \\ {\mathrm{C}}_{\mathrm{Q}}=3.61\times {10}^{-5}\mathrm{mol}/\mathrm{L} \\ {\mathrm{C}}_{\mathrm{Al}}=3.7\times {10}^{-5}\mathrm{mol}/\mathrm{L} \\ {\mathrm{C}}_{\mathrm{Q}}:{\mathrm{C}}_{\mathrm{Al}}=3.61\times {10}^{-5}:3.7\times {10}^{-5} \\ {\mathrm{C}}_{\mathrm{Q}}:{\mathrm{C}}_{\mathrm{Al}}=1:1 \\ \therefore {\mathrm{Q}}^{-}:{\mathrm{Al}}^{3+}=1:1 \end{gathered}$$

Therefore, the formula of the complex is AlQ²⁺.

Part (b) Step 1: Given Information

The Beer-Lambert Law is:

$\mathrm{A}=\mathrm{\epsilon lc}$

where,

A = absorbance

$\epsilon$- molar absorptivity

l – length of the solution light passes through (cm)

c – concentration of solution (mol/L)

Part (b) Step 2: Explanation

Using Beer-Lambert Law:

$\mathrm{A}=\mathrm{\epsilon lc}$

Substituting the values:

$$\begin{gathered} 0.500={\mathrm{\epsilon }}_{{\mathrm{AlQ}}^{2+}}\times 1.00\times 3.7\times {10}^{-5}\mathrm{M} \\ {\mathrm{\epsilon }}_{{\mathrm{AlQ}}^{2+}}=\frac{0.500}{1.00\mathrm{cm}\times 3.7\times {10}^{-5}\mathrm{M}} \\ {\mathrm{\epsilon }}_{{\mathrm{AlQ}}^{2+}}=1.35\times {10}^4{\mathrm{cm}}^{-1}{\mathrm{M}}^{-1} \end{gathered}$$