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

How does a blank titration reduce titration error?

Short Answer

Expert verified

There are no other compounds interfering with the solvent, blank titration decreases titration error.

This is done by titrating the standardised titrant against the solvent without any analyte present.

Step by step solution

01

Define the Titration error.

The endpoint determination is the most common systematic error. The titration error is the difference between the equivalency point and the measured end point. Because the colour change must be seen by eye, a visual end point is always little beyond the equivalence point.

02

Explain the blank titration.

  • Titrating a known concentration of titrant into a solvent without any traces of analyte is known as blank titration.
  • This is a very useful feature to add in the early procedures of the study in order to identify potential sources of mistake if the result fluctuates.
  • Blank titration can also be used to confirm that your solvent is free of any foreign chemicals before doing the main experiment.
  • This way, you can avoid having to repeat your analysis.

By guaranteeing that there are no other compounds interfering with the solvent, blank titration decreases titration error.

This is done by titrating the standardised titrant against the solvent without any analyte present.

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

Find [Hg22+]at 34.50 and 36.5 mL

A procedure for determining halogens in organic compounds

uses an argentometric titration. To 50 mL of anhydrous ether is added

a carefully weighed sample (10โ€“100 mg) of unknown, plus 2 mL of

sodium dispersion and 1 mL of methanol. (Sodium dispersion is finely

divided solid sodium suspended in oil. With methanol, it makes

sodium methoxide,CH3O-Na+, which attacks the organic compound,

liberating halides.) Excess sodium is destroyed by slow addition of

2-propanol, after which 100 mL of water are added. (Sodium should

not be treated directly with water, because the H2produced can

explode in the presence of O2:2Na+H2Oโ†’2NaOH+H2.)This

procedure gives a two-phase mixture, with an ether layer floating on

top of the aqueous layer that contains the halide salts. The aqueous

layer is adjusted to pH 4 and titrated with , using the electrodes in

Figure 7-5. How much 0.025 70 M AgNO3solution will be required to

reach each equivalence point when 82.67 mg of 1-bromo-4-chlorobutane(BrCH2CH2CH2Cl;FM171.46)are analysed ?

How many milliliters of 0.100MKIare needed to react with 40.0mLof 0.400MHg2(NO3)if the reaction isrole="math" localid="1654768697091" Hg22+2l-โ†’Hg2I2(s) ?

: Describe the chemistry that occurs in each of the following

regions in curve (a) in Figure 7-4: (i) before the first equivalence

point; (ii) at the first equivalence point; (iii) between the first and

second equivalence points; (iv) at the second equivalence point; and

(v) past the second equivalence point. For each region except (ii),

write the equation that you would use to calculate [Ag+].

Calculate pAg+at the following points in titration (a) in Figure 7-4: (a) 10.00mL; (b) 20.00mL; (c) 30.00mL; (d) second equivalence point; (e) 50.00mL.

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