Chapter 26: Q13 P (page 747)

(a) Suppose that the reservoir in Figure 26-7 contains 1.5 L of 2.0 M $K_{2} {PO}_{4}$ . 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?

Expert verified

The solutions to the above questions are

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

(b).0.16 A.

Step by step solution

$n (K^{+}) = 0.75 \times 1.5 L \times (2 \times \frac{n (K_{2} {PO}_{4})}{1 L}) (2 \times \frac{n_{1} (K^{+})}{n (K_{2} {PO}_{4})}) = 4.5 mol$

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

The time required would be

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

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

which is

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

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

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

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

$20 \times 8 mA = 160 mA = 0.16 A$

Unlock Step-by-Step Solutions & Ace Your Exams!

Over 30 million students worldwide already upgrade their learning with Vaia!

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.