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: 40 mL.

Video transcript

Hi everyone here, we have a question asking what is the ph of the solution of 15.0 mL of 0.113 moller tri methyl amine tie traded with 20.0 mL of 0.156 Mueller nitric acid. So we're going to start off with our middle leaders and we're going to convert that to Leaders. So we're going to multiply by one leader Over 1000 ml And then we're gonna multiply that by 0.113 moles over one liter. And our middle leaders are canceling out and our leaders are canceling out And that gives us 1.69, 5 times 10 to the negative third. Now we're going to do the same for our 20 ml. So we have 20 ml and we need to change that to leaders. So we're gonna multiply by one leader Over ml. And we're going to multiply that by 0.156 moles, her one liter. And our male leaders are going to cancel out and our leaders are going to cancel out. And that gives us 3. times 10 to the negative third. And now we need to make an ice chart and we're gonna let h be equal tri methyl amine and be equal our conjugate acid of tri methyl amine. So we have H B plus hydrogen goes back and forth from B plus water. And our water doesn't matter because it's a liquid not acquis and we're going to have our initial our change and our ending. So initially our tri methyl amine is going to be 1.695 times 10 to the -3. Our hydrogen is going to be 3.12 times to the -3. And our conjugate acid is going to be zero. Our change is going to be negative 1.695 times 10 to the negative three for tri methyl amine. And for our hydrogen. And it's going to be plus 1.695 times 10 to the negative three for our conjugate acid. So our ending is going to be zero for our tri methyl amine, 1.4 to 5 times 10 to the negative three for our hydrogen And 1.695 times 10 to the -3 for our conjugate acid and nitric acid is a strong acid. So our concentration of hydrogen is going to equal our concentration of nitric acid. Our total volume is 15 plus 20, Which equals 35 ml. And we're gonna multiply that by one l Over 1000 ml. So our male leaders are going to cancel out and that's going to give us 0.035 L. Our concentration of hydrogen Equals 1.4, 25 Times 10 to the -3, Divided by 0.035 L Which equals 0. 40714. Our hydrogen concentration R. P H is going to equal the negative log of our hydrogen concentration equals the negative log of 0. 40714. That equals 1.39. And that is our final answer. Thank you for watching. Bye.