Consider the titration of a 25.0-mL sample of 0.175 M CH3NH2
with...

Question

Consider the titration of a 25.0-mL sample of 0.175 M CH3NH2
with 0.150 M HBr. Determine each quantity.
c. the pH at 5.0 mL of added acid

Accepted Answer

Hi everyone here we have a question asking what is the ph of the solution of 15. mL of 0.113 moller tri methyl amine tight traded with 2.0 mL of 0.156 molar nitric acid. So our KB Equals 6. Times 10 to the -5. So now we have to calculate our moles. So we're gonna start with our 15 ml and we want to change that to leaders. So we're going to multiply by one leader Over 1000 ml Times 0.113 moles over one liter. And our male leaders are going to cancel out and our leaders are going to cancel out and that is going to give us 1. Times 10 to the -3. Next we're gonna do the same with our two ml. So we have two ml times one liter Over ml Times 0.156 moles over one liter. And our middle leaders are canceling out and our leaders are canceling out. And that gives us 3. times 10 To the -4. And now we need to make an ice chart. So we have our initial our change and our end and we're going to let h be equal tri methyl amine and be equal its conjugate acid. So we have H. B plus hydrogen goes back and forth from its conjugate acid plus water. So our initial for try methamphetamine is 1. times 10 to the negative third. Our initial for hydrogen is 3. Times 10 to the -4. Our initial for our conjugate acid is zero and our water doesn't matter because it's a liquid not acquis. So then we are going to move on to our change. So our change for tri methyl amine is going to be negative 3.12 times 10 to the negative fourth. Our change in our hydrogen is going to be negative 3.12 times 10 to the negative fourth. Our change in our conjugate acid is going to be plus 3.12 times 10 to the negative fourth. So our end is going to be one 383 Times 10 to the -3, zero and 3. times 10 to the -4. So now we can calculate our ph so r P H. Is going to equal P. K. A. Plus the log of the concentration of the base divided by the concentration of the acid. So our P. K. A. Equals negative log of K. A. Equals the negative log of Kw. Overcame me equals the negative log of one times 10 to the -14, Divided by 6.5 times 10 to the negative 5th equals nine point 8129. So our ph Equals 9. plus the log of 1.38. 3 Times 10 to the - divided by 3. Times 10 to the -4, Which equals 10.46. And that is our final answer. Thank you for watching. Bye.

Consider the titration of a 25.0-mL sample of 0.175 M CH3NH2 with 0.150 M HBr. Determine each quantity. c. the pH at 5.0 mL of added acid

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