A student titrated 30.00 mL of a weak base solution, 0.150M CH:NH2 (Kb = 4.40 × 10 4), with 0.100 M HCI solution.Calculate the pH at the equivalence point.half-equivalence pointbuffering regionequivalence pointpH <7Volume of HCI addedHd

Answered: A student titrated 30.00 mL of a weak…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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A student titrated 30.00 mL of a weak base solution, 0.150
M CH:NH2 (Kb = 4.40 × 10 4), with 0.100 M HCI solution.
Calculate the pH at the equivalence point.
half-equivalence point
buffering region
equivalence point
pH <7
Volume of HCI added
Hd

Expert Solution

Step 1- Introduction

Titration is an experimental method of analysis where a solution of known concentration is titrated with a solution of unknown concentration. Acid -Base titration involves determining the unknown concentration of an acid or a base by neutralizing it with another base or acid of known concentration.

The titration of a weak base and a strong acid forms a salt. Now in case of salt formation the pOH is calculated by using the following equation :

${pOH}_{} = \frac{p_{Kw}}{2} - \frac{\log (C)}{2} - \frac{p_{K b}}{2} (i) where : K_{w} = Ionic product of water = 10^{- 14} C = Concentration of salt K_{b} = Dissociation comstant of Base Now, p_{K_{b}} = - \log (K_{b}) p_{K_{w}} = - \log (K_{w})$

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