Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Material balance. If you intend to measure all the anions and cations in an unknown, one sanity check on your results is that the total positive charge should equal the total negative charge. The table lists concentrations of anions and cations measured in pond water in an undergraduate experiment and expressed in mg/mL.

Find the total concentration of negative and positive charge (mol/L) to assess the quality of the analysis. What do you conclude about this analysis?

Short Answer

Expert verified

The total concentration of negative and positive charge (mol/L) is as follows:

Total cation charge: 0.00202M

Total anion charge: -0.00159M

Step by step solution

01

Calculate the concentration by using equation:

c(species;mol/L)=-valuefromspreadsheetmolar-mass¯(of)anion/cation×10-6/10-3c(totalconcentration)=(species;mol/L)

02

Calculation of values

From the above equation we conclude that

Total cation charge: 0.00202M

Total anion charge: -0.00159M

03

Analysis

From this we can conclude that some concentrations can be inaccurate, or some ionic species were not detected.

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

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

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

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Figure 26-23 shows the separation of substituted benzoates. There is a peak of unknown identity at 86.0 seconds. (a) Is the unknown a cation, neutral, or an anion? (b) Find the apparent mobility and electrophoretic mobility of the unknown peak.

(a) A long thin molecule has a greater friction coefficient than a short fat molecule. Predict whether fumarate or maleate will have greater electrophoretic mobility.

(b) Electrophoresis is run with the injection end positive and the detection end negative. At pH 8.5, both anions have a charge of 22.The electroosmotic fl ow from the positive terminal to the negative terminal is greater than the electrophoretic flow, so these two anions have a net migration from the positive to the negative end of the capillary in electrophoresis. From your answer to part (a), predict the order of elution of these two species. (c)At pH 4.0, both anions have a charge close to 21, and the electroosmotic fl ow is weak. Therefore electrophoresis is run with the injection end negative and the detection end positive. The anions migrate from the negative end of the capillary to the positive end.Predict the order of elution.

What is the effect on injection time if you decrease the applied voltage by a factor of 2?

d)Li+ has lower mobility than Na+. Explain why lithium phosphate can be used at higher electric field strength than sodium phosphate to generate the same current. What is the advantage of higher field strength for this separation?

A polystyrene resin molecular exclusion HPLC column has a diameter of 7.8 mm and a length of 30 cm. The solid portions of the gel particles occupy 20% of the volume, the pores occupy 40%, and the volume between particles occupies 40%. (a) At what volume would totally excluded molecules be expected to emerge? (b) At what volume would the smallest molecules be expected? (c) A mixture of polyethylene glycols of various molecular masses is eluted between 23 and 27 ml. What does this imply about the retention mechanism for these solutes on the column?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free