Question

It took 2.30 min with a current of 2.00 A to plate out all the silver from 0.250 L of a solution containing Ag⁺. What was the original concentration of Ag⁺ in the solution?

Short answer

Theconcentration of Ag⁺ is 0.0114 M.

Step-by-step solution

Calculating mass of electron

The reduction reaction can be denoted as,

${\mathrm{Ag}}^{+}+{\mathrm{e}}^{-}\to \mathrm{Ag}$

The mole number of electron can be denoted as,

$\begin{array}{rcl}\mathrm{Moles} \mathrm{of} \mathrm{electron}& =& (2.3 \min )\left(\frac{60 \mathrm{s}}{1 \min }\right)\left(\frac{2\mathrm{C}}{1 \mathrm{s}}\right)\left(\frac{1 \mathrm{mol}}{96485 \mathrm{C}}\right)\\ & =& 0.00286 \mathrm{mol}\end{array}$

Finding the concentration

The concentration can be found as,

$$\begin{gathered} \mathrm{Concentration}=\frac{\mathrm{Moles} \mathrm{of} {\mathrm{Ag}}^{+}}{\mathrm{Volume} \mathrm{of} \mathrm{solution}} \\ \mathrm{Concentration}=\frac{0.00286 \mathrm{mol}}{0.25 \mathrm{L}} \\ \mathrm{Concentration}=0.0114 \mathrm{M} \end{gathered}$$$$\begin{gathered} \mathrm{Concentration}=\frac{\mathrm{Moles} \mathrm{of} {\mathrm{Ag}}^{+}}{\mathrm{Volume} \mathrm{of} \mathrm{solution}} \\ \mathrm{Concentration}=\frac{0.00286 \mathrm{mol}}{0.25 \mathrm{L}} \\ \mathrm{Concentration}=0.0114 \mathrm{M} \end{gathered}$$