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Ch.17 - Additional Aspects of Aqueous Equilibria

Chapter 17, Problem 111b

(b) As the acid is titrated, the pH of the solution after the addition of 11.05 mL of the base is 4.89. What is the Ka for the acid?

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Hello. Everyone in this video. We are going to determine the K. A. For the asset of a patrician. So from the problem we can go ahead and realize that at the end point 25.3 ml was used. So then at the half equivalence point of this situation 12.65 ml was used and how we got that was well first we want to confirm what this number was. Even for I divide this number by two and we got this value which is same as here. So again, an equivalence point, the P. H. Or the half a quiz. Um equivalence point is when ph is equal to R P. K. A. Value. And the problem we are given our ph value which is the same as a P. K. A. According to this right here. So R P K. A. Is equal to 4.20. So we have our P. K. A. To solve for R. K. Value. That is very simple. We simply do 10 to the power of our negative P. K. A. So again peek A is equal to or K. Is equal to 10 to the negative 4.20. Then putting that into my calculator. I get that my K. A value for the acid is 6. times 10 to the negative five. And this right here is going to be my final answer for this problem. Thank you all so much for watching
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