A sample of 0.1687 g of an unknown monoprotic acid was dissolved in 25.0 mL of water and titrated with 0.1150 M NaOH. The acid required 15.5 mL of base to reach the equivalence point. (b) After 7.25 mL of base had been added in the titration, the pH was found to be 2.85. What is the Ka for the unknown acid?

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Hello everyone today. We have the following problem. A 0.1164 g sample of an unknown mono protic acid was dissolved in 35 mL of water a 350.106 Mueller potassium hydroxide solution was used to titrate the resulting solution. A total of 17.8 mL of potassium hydroxide was used to reach the equivalence point. The ph of the solution is 9.15. When eight ml of the base was added, calculate the K. A. of the acid. So the first thing you want to make note of is that we are working with a mono product acid. And so this just means that we have the typical formula of H. A. And so when we react this mono product acid with potassium hydroxide, we get our potassium with the acid and water. It was noted that there was an equivalence point that was reached and the equivalence point of E. P. Is essentially when the moles of the base equal the moles of the acid. And so it's important to find the moles of both of the acid and the base in this question. So we'll start with the moles of our base or potassium hydroxide. We're going to start off with what we're given of the base, which is the volume 17.8. Middle leaders of potassium hydroxide, We want to convert this into leaders with the conversion factor that one middle leader is equal to 10 to the negative third. Leaders. Anything that's multiplied by the polarity and we have the clarity here and minorities in units of moles per liter. So we can just say this is 0.106 moles over one liter units of liter and milliliters will cancel out. And we will end up with 1.89 times 10 to the negative third moles and we'll keep this number for later. Next we need to find our moles of our acid and so we're essentially going to take our moles of our Base 1.89 times 10 to the negative 3rd moles. And that's what will be equal to. Since we have reached an equivalence point where the most of our base equal the most of our acid. And so now we must consider the moles of potassium hydroxide at our ph Of 9.15. And so we have the dissolution or the dissociation of potassium hydroxide meaning it would associate into potassium ions and hydroxide ions. And so we know that potassium is neutral but our hydroxide is considered our base. And so we must find the number of moles of our hydroxide. So our fourth step is to find the moles of our hydroxide And we can do this by taking what we have what we're given first for hydroxide which is our volume eight. Middle leaders of potassium hydroxide that was added. Of course we're going to convert this into leaders by using a conversion factor that one. Middle leader is equal to 10 to the negative third. Leaders are units for leaders with middle leaders will cancel out. And we want to multiply by the polarity, normal clarity before 0.106 moles per one leader arguments of leader were canceled out. It will be left with 8. Times 10 to the -4 malls of hydroxide that was added. And so fifth we're going to construct our I. C. F. Table so that we can find out what changed in our reaction. So essentially that's going to look like our acid reacting with our base or the hydroxide part to form our acid ion and our water. And then we have our I. C. F. On the side here. Our initial concentration for the acid we calculated was 1.89 times 10 to the negative third moles. And for hydroxide we calculated that to be 8.48 times 10 to negative fourth of course the initial concentration for our ion will be zero and water will get no concentrations because it is a constant. And so since we're tight trading with our potassium hydroxide, we're gonna subtract 8.48 times 10 to the negative fourth from both react ints Giving us zero for hydroxide And a final concentration of 1.039 times 10 to the negative third for our acid. And then we're actually going to add that 8.48 times 10 to the - to our product side. And so that final concentration will be the same. Our 6th step is going to be evaluating our acid. So before we did our base and now we do our acid. So for acid we have our acid of course reacting with our water to form that conjugate base and our hydro ni um ions. And from this we can actually find our K. A. Expression and the K. Is just the concentration of our products. Overreacting. And so that's going to be our conjugate base there with our hydro ni um there and we do not include water because that is a constant. So we just have our acid in the denominator. Let's go ahead and find our final volume as well. Our final volume, It's going to be the 35 ml of the water plus the eight Middle leaders that we had from the base to give us 43 ml. Which will actually come out to 0.043 leaders since we have to convert to leaders and now we are ready to assess for our different different concentrations. So for our step eight we're going to have the concentration of our acid Which is going to be our one 0.389 that we solved in our I. C. F. Table time tends to negative three over The volume that we just found, which is .043 L. That will give us 2.4 Times 10 to the negative to moller we're gonna have a concentration of our conjugate base which will be our 8. times 10 to the negative four moles over the volume that we just solved. Four .043 L. And that will give us 1.97 Times 10 to the negative to Mueller. And last but at least we will have our concentration of hydro ni um which will be Using the Formula 10 to the Negative p since we have the ph here Of 9.15, this rounds out to 7.08 times 10 to the -10 Moller. So we're gonna keep track of all of these numbers that we solved for For our step nine. We're going to have yet another I cf table and this is going to include our acid. So we're gonna have our acid react with water. This is going to form our conjugate base and our hydro ni um ions. As calculated our concentration of our asset is going to be 2.4. Our initial concentration is gonna be 2.4 times 10 to the negative to water. Does not get any concentrations because it is a constant. What we calculated for our conjugate base is 1. Times 10 to the -2 And we do not have an initial concentration for our hydro knee. Um that is going to be our change our change with the hydro me um is going to be seven 08 times 10 to the - Moller. And because we added on the right we have to subtract that number from our left hand side as usual. Water does not get anything. And then we have to add on this product side as well 7.08 times 10 to -10. And then you're essentially going to add these two numbers up an hour, final column, our final role. And to save on space, we will simply just refer to this as I plus C, I plus C and I plus C. That's the initial and the change. And lastly we can now solve for R K. A value by using that equation that we used earlier and plugging in the values for it. So our final concentrations of our conjugate base Is 1.9 Times 10 to the - Plus 7.08 times 10 to the -10. And then you multiply that by our 7.8 times 10 to the negative 10th from a concentration of our hydro liam. And then we put all of that over arc concentration for our acid, which is going to be our 2.4 times 10 to the - -7.8 times 10 to the -10. And when we do that we end up with a final K. A value of 5.8 Times 10 to the -10 as our final answer. Overall, I hope this helped. And until next time.

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Key Concepts

Monoprotic Acids

Titration and Equivalence Point

Acid Dissociation Constant (Ka)