Question

(a) Suppose that the reservoir in Figure 26-7 contains 1.5 L of 2.0 M ${\mathbf{K}}_{\mathbf{2}}{\mathbf{PO}}_{\mathbf{4}}$. For how many hours can the reservoir provide 20mM KOH at a flow rate of 1.0 mL/min if 75% consumption ofin the reservoir is feasible?

(b) What starting and ending current would be required to produce a gradient from 5.0 mM KOH to 0.10 M KOH at 1.0 mL/min flow rate?

Short answer

The solutions to the above questions are

$\left(a\right)3.8\times {10}^{2}h$

(b).0.16 A.

Step-by-step solution

Calculating the moles of K+ in reservoir

$$\begin{gathered} \mathrm{n}\left({\mathrm{K}}^{+}\right)=0.75\times 1.5\mathrm{L}\times \left(2\times \frac{\mathrm{n}\left({\mathrm{K}}_2{\mathrm{PO}}_4\right)}{1\mathrm{L}}\right)\left(2\times \frac{{\mathrm{n}}_1\left({\mathrm{K}}^{+}\right)}{\mathrm{n}\left({\mathrm{K}}_2{\mathrm{PO}}_4\right)}\right) \\ =4.5\mathrm{mol} \end{gathered}$$

Step 2: Calculation of the flow rate

$$\begin{gathered} \mathrm{v}=(20\times {10}^{-3}\mathrm{mol}/\mathrm{L}(\mathrm{KOH})\times (0.001\mathrm{L}/\min ) \\ =2\times {10}^{-5}\mathrm{mol}/\min \mathrm{of}\mathrm{KOH} \end{gathered}$$

The time required would be

$$\begin{gathered} \mathrm{t}=\frac{\mathrm{n}\left({\mathrm{K}}^{+}\right)}{\mathrm{v}} \\ \mathrm{t}=\frac{4.5\mathrm{mol}}{2\times {10}^{-5}\mathrm{mol}/\mathrm{minofKOH}} \\ \mathrm{t}=2.25\times {10}^5\min =3.8\times {10}^2\mathrm{h} \end{gathered}$$

Calculation of the moles of KOH per minute (flow rate)

$$\begin{gathered} \mathrm{v}=(20\times {10}^{-3}\mathrm{mol}/\mathrm{L}(\mathrm{KOH})\times (0.001\mathrm{L}/\min ) \\ =5\times {10}^{-6}\mathrm{mol}/\mathrm{minKOH} \end{gathered}$$

which is

$8.33\times {10}^{-8}\mathrm{mol}/\mathrm{sKOH}$

Conversion of moles into coulombs

$8.33\times {10}^{-8}\mathrm{mol}/\mathrm{sKOH}$X Faraday’s constant 8mA

$=8.33\times {10}^{-8}\mathrm{mol}/\mathrm{sKOH}\times (9.6485\times {10}^4\mathrm{C}/\mathrm{mol})=8\mathrm{mA}$

Calculation of final flow rate

A gradient from 5 mM KOH to 0.1 M KOH with a 1.0 mL/min flow rate requires 20 times more current so gradient would be

$$\begin{gathered} 20\times 8\mathrm{mA}=160\mathrm{mA} \\ =0.16\mathrm{A} \end{gathered}$$