Mathematically prove that the pH at the halfway point of
a titration of a weak acid with a strong base (where the volume
of added base is half of that needed to reach the equivalence
point) is equal to pKa for the acid.

Question

Mathematically prove that the pH at the halfway point of 
a titration of a weak acid with a strong base (where the volume 
of added base is half of that needed to reach the equivalence 
point) is equal to pKa for the acid.

Accepted Answer

Hello. Everyone in this video. We're trying to see if the statement is true or false using mathematical proof and the statement reads at the half equivalence point. Administration of a weak acid with a strong base, the page is equal to the P. K. For the acid. If you remember this statement as a matter of fact true. But we have to go ahead and use mathematical proof to back our statement up. So let's recall that in an asset based hydration, usually base is added to our acid. So at our equivalence point, at the equivalence point hepatitis E. P. Our moles of acid equal to our moles of base. So basically what that is is that the more clarity multiply per volume of our acid is also equal to the more clarity and multiplied by our volume of our base is going to equal to each other and our more clarity times our volume of our base will equal to our moles of base of course which is equal to the moles of acid. His initial amount. All right. And then at our half Echo Van's point our moles of base is equal to half the moles of acid from the initial or of initial. So basically half of the weak acid reacted is converted to its conjugate base. So here we're just gonna go ahead and assume that exactly half of the weak acid reacts. So we have that the moles a weak acid. I'll just put that W. A. For short equals to the moles our conjugate base of label the SCB at our equivalence point or half equivalence point. And since we have a buffer we can go ahead and use utilize the Henderson Hasselbach equation. And let's just go ahead and write this out. So R P H. Or I'm writing the Henderson Hasselbach equation. So we have our ph is equal to the P K. A. Plus the log of the con uh situation of our conjugate base over the concentration of our weak acid. Okay, we can go ahead and put our other writing or other way of writing a conjugate base and the weak acid. We're just going to be a minus and H A. For a weak acid. So let's go ahead and scroll down a little bit. So we have more space. So the concentration of our a minus is going to equal To the concentration of our age of eight. So if we put that on top of each other because that's a portion of our equation, we have a minus over concentration of H. A. And if the equal to each other they were equal to one. So putting this into my equation of ph equaling to the P K A plus log of one. Now one we know that the log of one is equal to zero. So we will now have that the P H is equal to P K. A. So you can say that therefore just those three dots bye mathematical proof this statement that P H. Is equal to P K A at the half equivalence point. This true and this right here is going to be my final answer for this problem. Thank you all so much for watching.

Mathematically prove that the pH at the halfway point of a titration of a weak acid with a strong base (where the volume of added base is half of that needed to reach the equivalence point) is equal to pKa for the acid.

Verified Solution

Key Concepts

Titration

pH and pKa

Henderson-Hasselbalch Equation