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Chapter 11: Q1TY (page 240)

Find the pH when Va = 14.63 mL.

Short Answer

Expert verified

The pH is 4.73

Step by step solution

01

Information given

The following information was obtained from the example to solve this problem

Titration of pyridine using HCl

Amount of pyridine=25.00 mL of 0.08364M pyridine

Concentration of HCl=0.1067 M

Equivalence point occurs at 19.60 mL

pKa =5.20

Kb = 1.59×10-9

Ka=(Kw/Kb)=6.31×10-6

02

Reaction

The titration reaction is as follows

In the mixture of pyridine and pyridinium ion, a part of pyridine will be neutralized.

03

Calculation

The fraction of pyridine titrated is (14.63/19.60)=0.746

Therefore, the remaining fraction will be

1-0.746=0.254

The pH will be

pH=pKa+logBBH+=5.20+log0.2540.746=4.73

Therefore, the pH is 4.73

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Most popular questions from this chapter

Why is it not practical to titrate an acid or base that is too weak or too dilute?

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

Calculate the pH at 10.0-mL intervals (from 0 to 100 mL) in the titration of 40.0 mL of 0.100 M piperazine with 0.100 M HCl. Make a graph of pH versus Va.

Calculate the pH at each of the following points in the titration of pH50.00mLof 0.0100MNaOH with 0.100MHCI . Volume of acid added: 0.00,1.00,2.00,3.00,4.00,4.50,4.90,4.99,5.00,5.01,5.10,5.50,6.00,8.00,and 10.00mL. Make a graph of pH versus volume of HCl added.

A 100.0-mL aliquot of 0.100 M diprotic acid H2A (pK1 = 4.00, pK2 = 8.00) was titrated with 1.00 M NaOH. Find the pH at the following volumes of base added and make a graph of pH versus Vb: Vb = 0, 1, 5, 9, 10, 11, 15, 19, 20, and 22 mL.
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