Question

An unknown sample of Cu2+gave an absorbance of 0.262 in an atomic absorption analysis. Then 1.00 mL of solution containing 100.0 ppm (=g/mL)Cu2+was mixed with 95.0mL of unknown, and the mixture was diluted to 100.0mL in a volumetric flask. The absorbance of the new solution was 0.500.

(a) Denoting the initial, unknown concentration as [Cu2+]i, write an expression for the final concentration,Cu2+f, after dilution. Units of concentration are ppm.

(b) In a similar manner, write the final concentration of added standard Cu2+, designated as [S]f.

(c) Find [Cu2+]i in the unknown.

Short answer

$$\begin{gathered} \mathrm{a}){\left[{\mathrm{Cu}}^{2+}\right]}_{\mathrm{final}}=0.95{\left[{\mathrm{Cu}}^{2+}\right]}_{\mathrm{initial}} \\ \mathrm{b}){\left[\mathrm{S}\right]}_{\mathrm{final}}=1.00{\mathrm{ppmCu}}^{2+} \\ \mathrm{c}){\left[{\mathrm{Cu}}^{2+}\right]}_{\mathrm{initial}}=1.04{\mathrm{ppmCu}}^{2+} \end{gathered}$$

Step-by-step solution

Absorbance definition

The amount of light absorbed by a solution is measured by absorbance (A), also known as optical density (OD).

Given data

We are given the following:

absorbance unknown=0.262

absorbanceunknown+ spike =0.500

$$\begin{gathered} {\left[\mathrm{S}\right]}_{\mathrm{initial}}=100.0{\mathrm{ppmCu}}^{2+} \\ {\mathrm{V}}_{\mathrm{initial}}=95.0\mathrm{mL} \\ {\mathrm{V}}_{\mathrm{final}}=100.0\mathrm{mL} \\ {{\mathrm{V}}_{\mathrm{spike}}}_{,\mathrm{initial}}=1.00\mathrm{mL} \\ {\mathrm{V}}_{\mathrm{spike},\mathrm{final}}=100.0\mathrm{mL} \end{gathered}$$

Solving for part (a)

The expression for dilution is${\mathrm{C}}_1{\mathrm{C}}_2={\mathrm{V}}_1{\mathrm{V}}_2$. Following the same expression, we can denote that final concentration Cu2+can be expressed as:

$$\begin{gathered} {\left[{\mathrm{Cu}}^{2+}\right]}_{\mathrm{initial}}\times {\mathrm{V}}_{\mathrm{initial}}={\left[{\mathrm{Cu}}^{2+}\right]}_{\mathrm{final}}\times {\mathrm{V}}_{\mathrm{final}} \\ {\left[{\mathrm{Cu}}^{2+}\right]}_{\mathrm{final}}=\frac{{\left[{\mathrm{Cu}}^2\right]}_{\mathrm{initial}}\times {\mathrm{V}}_{\mathrm{initial}}}{{\mathrm{V}}_{\mathrm{final}}} \\ {\left[{\mathrm{Cu}}^{2+}\right]}_{\mathrm{final}}=\frac{{\left[{\mathrm{Cu}}^2\right]}_{\mathrm{initial}}\times 95}{100} \\ {\left[{\mathrm{Cu}}^{2+}\right]}_{\mathrm{final}}=0.95{\left[{\mathrm{Cu}}^{2+}\right]}_{\mathrm{initial}} \end{gathered}$$

Solving for part (b)

b) The final concentration of the additional standard may be calculated in the same way as the previous expression:

$$\begin{gathered} {\left[\mathrm{S}\right]}_{\mathrm{initial}}\times {\mathrm{V}}_{\mathrm{spike},\mathrm{initial}}={\left[\mathrm{S}\right]}_{\mathrm{final}}\times {\mathrm{V}}_{\mathrm{spike},\mathrm{final}} \\ {\left[\mathrm{S}\right]}_{\mathrm{final}}=\frac{{\left[\mathrm{S}\right]}_{\mathrm{initial}}\times {\mathrm{V}}_{\mathrm{spike},\mathrm{initial}}}{{\mathrm{V}}_{\mathrm{spike},\mathrm{final}}} \\ {\left[\mathrm{S}\right]}_{\mathrm{final}}=100.0{\mathrm{ppmCu}}^{2+}\times \frac{1.00\mathrm{mL}}{100.0\mathrm{mL}} \\ {\left[\mathrm{S}\right]}_{\mathrm{final}}=1.00{\mathrm{ppmCu}}^{2+} \end{gathered}$$

Solving for part (c)

We are asked to solve for ${\left[{\mathrm{Cu}}^{2-}\right]}_{\mathrm{initial}}$ present in the unknown. But before that, we need to find the expression for the concentration of ${\mathrm{Cu}}^{2+}$after spiking. The equation for solving role="math" localid="1663396450456" ${{\mathrm{Cu}}^{2+}}_{\mathrm{after}\mathrm{spike}}$is given by:

$$\begin{gathered} {\left[{\mathrm{Cu}}^{2+}\right]}_{\mathrm{after}\mathrm{spike}}={\left[{\mathrm{Cu}}^{2+}\right]}_{\mathrm{initial}}\times \frac{\mathrm{V}\mathrm{initial}}{{\mathrm{V}}_{\mathrm{final}}}+{\left[{\mathrm{Cu}}^{2+}\right]}_{\mathrm{spike}}\times \frac{{\mathrm{V}}_{\mathrm{spike},\mathrm{initial}}}{{\mathrm{V}}_{\mathrm{spike},\mathrm{final}}} \\ {\left[{\mathrm{Cu}}^{2+}\right]}_{\mathrm{after}\mathrm{spike}}={\left[{\mathrm{Cu}}^{2+}\right]}_{\mathrm{initial}}\times \frac{95.0\mathrm{mL}}{100.0\mathrm{mL}}+100.0{\mathrm{ppmCu}}^2\times \frac{1.00\mathrm{mL}}{100.0\mathrm{mL}} \\ {\left[{\mathrm{Cu}}^{2+}\right]}_{\mathrm{after}\mathrm{spike}}=0.95{\left[{\mathrm{Cu}}^{2+}\right]}_{\mathrm{initial}}+1.00{\mathrm{ppmmCu}}^{2+} \end{gathered}$$

Now that we have the expression for the concentration of ${\mathrm{Cu}}^{2+}$after spike, we can plug it into the equation and solve for role="math" localid="1663396479141" ${\left[{\mathrm{Cu}}^{2+}\right]}_{\mathrm{initial}}$ $$\begin{gathered} {\left[{\mathrm{Cu}}^{2+}\right]}_{\mathrm{initial}}=\frac{0.262}{0.5}\times \left(0.95{\left[{\mathrm{Cu}}^{2+}\right]}_{\mathrm{initial}}+1.00{\mathrm{ppmCu}}^{2+}\right) \\ 0.5022{\left[{\mathrm{Cu}}^{2+}\right]}_{\mathrm{initial}}=0.524{\mathrm{ppmCu}}^{2+} \\ {\left[{\mathrm{Cu}}^{2+}\right]}_{\mathrm{initial}}=1.04{\mathrm{ppmCu}}^{2+} \end{gathered}$$