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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.

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Answer:

The pH at equivalence is 5.24 .

Explanation:

Let us calculate -

millimoles of aniline = 160 x 0.3403 = 54.448

To reach equivalence , 54.448 millimoles HNO3 must be added,

[tex]54.448 = V \times0.0501[/tex]

V = 1086.79 mL HNO3 must be added

total volume = 1086.79 + 160 = 1246.79 mL

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

Now, at the equivalence or equivalent point ,

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

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

[tex]p(OH)=8.76[/tex]

[tex]pH= 14-8.76[/tex]

By solving the above equation , we get -

pH = 5.24

Hence , the answer is 5.24

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