A student titrated 30.00 mL of a weak base solution, 0.150 M CH&NH2 (Kb = 4.40 × 104), with 0.100 M HCI solution. Calculate the pH at the equivalence point.

Chemistry: An Atoms First Approach
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Chapter14: Acid- Base Equilibria
Section: Chapter Questions
Problem 110CP: A 0.400-M solution of ammonia was titrated with hydrochloric acid to the equivalence point, where...
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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
Transcribed Image Text: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 =pKw2 - logC2 - pKb 2                          (i)where : Kw = Ionic product of water =10-14C   = Concentration of salt Kb  = Dissociation comstant of BaseNow, pKb = -logKb         pKw    = -logKw

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