/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q3P Why does an acid-base titration ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Why does an acid-base titration curve (pH versus volume of titrant) have an abrupt change at the equivalence point?

Short Answer

Expert verified

In an acid-base titration curve, there is an abrupt change at the equivalence point due to the absence of base and a small amount of excess acid.

Step by step solution

01

Define equivalent point.

The stage of titration where the concentrations of titrate and titrant are chemically equivalent is known as the equivalence point. The equivalence point occurs a few milliseconds before or virtually simultaneously with the endpoint.

02

Reason for the abrupt change of titration curve at the equivalence point.

  • The number of moles of titrant added so far as an equivalency point correlates to the number of moles of the chemical in question titrated according to the stoichiometry of the reaction.
  • Because the pH scale is logarithmic, the pH rises slowly at initially.
  • The initial addition of the titrant to the acid (in the burette) does not result in significant modifications.
  • A buffering action occurs in this relatively flat section of the pH curve.
  • As the titration progresses and more base is introduced, some of the acid is reacted with the new base, but some excess acid will remain anywhere before the equivalence point, keeping the pH low.
  • With the addition of a few more drops around the equivalence point, a tiny excess of acid becomes a small excess of base, and the pH quickly shifts.

Hence the pH changes abruptly at the equivalence point because the base in the solution has been titrated completely.

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 91Ó°ÊÓ!

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

The dibasic compound B (pKb1 = 4.00, pKb2 = 8.00) was titrated with 1.00 M HCl. The initial solution of B was 0.100 M and had a volume of 100.0 mL. Find the pH at the following volumes of acid added and make a graph of pH versus Va: Va = 0, 1, 5, 9, 10, 11, 15, 19, 20, and 22 mL

Consider the titration of100.0mlof 0.100MNaOHwith 1.00MHBr. Find the pH at the following volumes of acid added and make a graph of pH versusVa:Va=0,1,5,9,9,9,10,10.1 androle="math" localid="1654940659782" 12ml .

11-20. The graph shows the titration curve for a protein containingamino acids with 16 basic and acidic substituents. The curve is smooth without clear breaks because 29 groups are titrated in thepH interval shown. The 29 endpoints are so close together that a nearly uniform rise results. The isoionic point is thepHof the pure protein with no ions present exceptH+ and OH-. The isoelectric point is the pHat which the average charge on the protein is zero. Is the average charge of the protein-positive, negative, or neutral at its isoionic point? How do you know?

Acid-base titration of the protein ribonuclease. [Data from C. T. Tanford and J. D. Hauenstein, "Hydrogen Ion Equilibria of Ribonuclease," J. Am. Chem. Soc. 1956, 78, 5287.]

Spectrophotometric properties of a particular indicator are given below:

A solution with a volume of 20.0 mL containing 1.40 ×10-5 M indicator plus 0.050 0 M benzene-1,2,3-tricarboxylic acid was treated with 20.0 mL of aqueous KOH. The resulting solution had an absorbance at 604 nm of 0.118 in a 1.00-cm cell. Calculate the molarity of the KOH solution.

A solution was prepared from 1.023 g of the primary standard tris (Table 11-4) plus 99.367 g of water; 4.963 g of the solution were titrated with 5.262 g of aqueousHNO3to reach the methyl red endpoint. Calculate the concentration of the HNO3(expressed as mol HNO3/kg solution)

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.