The following plot shows two pH titration curves, each representing the titration of 50.0 mL of 0.100 M acid with 0.100 M NaOH:

. (b) What is the approximate pH at the equivalence point for each of the acids?

Video transcript

Hello. And this problem we are shown below the titillation curve of two acids. The concentration of 0.5 Mueller were told that a 40 million sample of each acid was then tie traded with 400.5 Mueller potassium hydroxide. Were asked to estimate the ph at the equivalence point for each acid. Let's begin by calling our two acids H. A. And H. B. And generically will refer to them as hx Where X. 10 is equal to a. or b. So our acid well then react with potassium hydroxide to form water and assault. So the volume of testing my drug side required to completely neutralize our acid you can find. Then by beginning with the volume of our acid, we'll convert that to leaders. Our concentration is 0.05 zero moles of acid. For leader Based on the story Geometry. one mole of our acid reacts with one mole of our base Base concentration is 0. no per liter. And we'll convert our volume based middle leaders. One leader is equal to 1000 ml. And so our volume of acid cancels in male leaders and leaders are moles of acid cancels our molds of base cancels and our volume of basin leaders cancels and we're left with a volume in milliliters. This works out to 40 mL of potassium hydroxide. This is the volume of potassium hydroxide required to reach the equivalence point that is to react with all of the assets that is present. So this is our equivalence point. If we begin with H. A. This is a strong acid then it will react with our base form which is a strong base to form water. And the salt assault then as potassium mines which will result in a neutral solution. And we have the conjugate base of a strong acid which will also result in a neutral solution because it will not undergo hydraulic sis to go back to form our strong acid. So the P. H. will be equal to seven at the equivalence point. For the reaction between our strong acid and strong base, consider our other acid to be a weak acid again reacting with a strong base. We will form water and the salt in this case. Then our salt as potassium mines which will result in a neutral solution. But now we have the conjugate base a weak acid which will undergo hydraulic sis. So it will react with water reform, H. B. And O H minus the sense this is the base that accepts protons from water which functions as an acid. And so the presence of hydroxide ions indicates that the ph Will be greater than seven. So we'll have a basic solution. So if we look at the answers we were provided, we are told in a that we have a strong acid and the ph is eight. So the strong acid. The ph of the equivalence point will be equal to seven. So that's incorrect. We're also told and see that the ph of the strong acid at the equivalence point will be H. And that is again incorrect. If we look at answer D says the ph of a strong acid will be equal to seven at the equivalence point. And that of the weak acid will also be equal to seven at the equivalence point. That is incorrect. So the ph at the equivalence point for weak acid will be greater than that. So, if you look at answer B, you see that we're told that the ph for a strong acid at the equivalence point will be seven and that for weak acid, the ph will be nine, which is greater than seven, so it's in the basic region. And so answer B. Then is our correct answer. Thanks for watching. Hope This helped.