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Ch.17 - Aqueous Ionic Equilibrium

Chapter 17, Problem 74b

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: 10 mL.

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Hi everyone here, we have a question asking what is the ph of the solution of 15.0 mL of 0.113 Mohler analyst I traded with 5.0 mL of 0.156 molar nitric acid. So the KB Equals 3. times 10 To the -10th power. So we have 15 ml and we're going to multiply that by one leader Divided by 1000 ml. And we're going to multiply that by 0. moles Divided by one leader. So our male leaders are canceling out here and our leaders are canceling out here, giving us 1.69, 5 times 10 to the -3. Next we're gonna do the same with our five ml. So we have five ml times one leader Over 1000 ml. And we're going to multiply that by 0.156 Molds per leader. So our middle leaders are canceling out and our leaders are canceling out. And that gives us 7.8 Times 10 to the -4. And now we need to do an ice chart. So we're gonna let h be equal aniline and be equal our conjugate acid. So we have H B plus H plus goes back and forth from B plus plus water. And our water here isn't gonna matter because it is a liquid and not acquis. So we're going to have our initial our change and our end For channeling that is 1.695 times 10 to the -3 and then hydrogen 7.8 times 10 to the negative four. And we're starting out with zero of our conjugate acid. And then our change is going to be negative 7. times 10 to the negative four. And it will be the same for hydrogen. And for our contract it base it will be plus 7.8 times 10 to the -4. And now for our end that is 9.59. Times 10 to the -4, zero And 7. times 10 to the -4. And now that we have our ice chart done, the ph equals the P K. A. Plus the log times the concentration of the base divided by the concentration of the acid. So our P. K. A. Equals the negative log of K. A. Equals the negative log of kW, divided by K. B. Equals the negative log of one times 10 to the -14, divided by 3.8 times 10 to the -10. And that equals 4.5798. So our ph Is going to equal 4. plus the log of 9.1. 5 Times 10 to the -4, divided by 7. Times 10 to the -4. And that equals 4.65. So our final answer is 4.65. Thank you for watching. Bye