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Strike441 [17]
3 years ago
7

A chemist titrates of a aniline solution with HCl solution at . Calculate the pH at equivalence. The of aniline is . Round your

answer to decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of solution added.
Chemistry
1 answer:
Alexus [3.1K]3 years ago
7 0

Answer:

<u>The pH at equivalence is 5.24 .</u>

Explanation:

Let us calculate -

millimoles of aniline = 160 x 0.3403 = 54.448

To reach equivalence , 54.448 millimoles HNO3 must be added,

54.448 = V \times0.0501

V = 1086.79 mL HNO3 must be added

total volume = 1086.79 + 160 = 1246.79 mL

Concentration of [salt] = \frac{54.448}{1246.79} = 0.044 M

Now, at the equivalence or equivalent point ,

p(OH)=\frac{1}{2} [pKw+pKb+logC]

p(OH)=\frac{1}{2} [14+4.87+log0.044]

p(OH)=8.76

pH= 14-8.76

By solving the above equation , we get -

pH = 5.24

<u>Hence , the answer is 5.24 </u>

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