A 25.0-mL sample of 0.125 M pyridine is titrated with
0.100 M HCl...

Question

A 25.0-mL sample of 0.125 M pyridine is titrated with
0.100 M HCl. Calculate the pH at each volume of added acid: 
one-half equivalence.

Accepted Answer

Hi everyone here we have a question asking us at the half equivalence point, what is the ph of the thai tradition of 15 mL of 0.183 moller aniline with 0.15 moller nitric acid. So because this is a week based strong acid titillation, the ph at one half the equivalence point is the P. K. A. Of the conjugate acid of the week based solution. The ph equals P K. A. Equals negative log of K A. Equals negative log. The kb of an ailing Equals 3. Times 10 to the -10. The K. A equals the kw, divided by the kb. K. A Equals one times 10 To the -14, divided by 3.8 times To the negative Equals 2.631, 6 times 10 To the -5. So our P. K. A equals negative log of 2. times 10 To the negative 5th Equals 4.58. And that is our final answer. Thank you for watching. Bye.

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