The titration of 0.02500 L of a diprotic acid solution with 0.1000 M NaOH requires 34.72 mL of titrant to reach the second equivalence point. The pH is 3.95 at the first equiva-lence point and 9.27 at the second equivalence point. If the acid solution contained 0.2015 g of the acid, what is the molar mass, pKa1, and pKa2 of the acid?

Video transcript

Hey everyone, we're told that when a 15 millimeter sample of 5.45 g per liter solution of a di protic acid was tight treated against 0.15 molar potassium hydroxide, it required 24.50 mL of the base to reach the endpoint. The ph values at the 1st and 2nd equivalent points were 5.75 and 10.75 respectively, calculate the P K one P K two and the molecular weight of the acid First. Let's go ahead and start by calculating the number of moles of the acid. So we'll need to write out our reaction. We were told that we had a die protic acid so we can label this as H2 a plus potassium hydroxide. Now, when these two react, we're going to get K two A plus water. Now to balance out this equation, we need to add a coefficient of two prior to our water and a coefficient of two prior to potassium hydroxide. Now let's go ahead and calculate the mole of our di protic acid. So taking that 24.50 mL of potassium hydroxide, We first want to convert this into leaders and we know that we have 10 to the third milliliters per one leader. Now we're going to use our 0.15 molar of potassium hydroxide. Next looking at our multiple ratios between our potassium hydroxide and our dai protic acid. We can see that we have a 1-2 ratio. So we have one mole of our die protic acid per two mol of our potassium hydroxide. So when we calculate this out and cancel out our units, we end up with 0.0018375 Mol of our die protic acid. Now let's go ahead and calculate The g of sample of our acid. And we can do so by taking that 0.0156, which was the 15 ml. And we can multiply this by 5.45 g/l. Now, when we calculate this out, we end up with 0.08175g of sample of acid. Now we can use these values to determine our molar mass of our acid. So taking those grams of 0.08175 g. We can go ahead and divide this By our mole which was 0.0018375 mol of our die protic acid. This gets us to a molar mass of 44.5 g per mole. And this is going to be the molar mass of our die protic acid 1st. Let's go ahead and calculate the polarity of our an A. On. And we can do so by taking the mole of our die protic acid which was 0. mol. And dividing this by our total leaders Which was 0.0150 L plus 0.02450 L. This gets us to a polarity of 0.04652 moller. Now that we have our polarity, let's go ahead and write out our reaction. So we have our an ion and this reacts with water. When these two react, we get our acid plus our hydroxide ion. Now creating this ice chart, we know that we started with 0. molar of our an ion. And we had zero of our products formed initially we disregard our water since it is a liquid and our change is going to be a minus X. For an ion and a plus X for our product. Since we're gaining products and losing reactant, bringing this down, we have 0.4652 minus X. And an X. And an X. In our products. We can go ahead and assume that our X. Is negligible since it is a weak acid. So first let's go ahead and calculate for our KB one and this is going to be the concentration of our acid times the concentration of our hydroxide ion divided by the concentration of our an ion. So essentially this is going to be X squared divided by our concentration of our an ion. Again, the X is negligible since it is a weak acid. Now we were told that the ph at the second equivalence point is 10.75. So this means our p O. H. Is equal to 14 minus R P H. Which is 14 minus 10.75 which gets us to a p O. H of 3.25. Now we can go ahead and use this P O. H. In order to determine the polarity of our hydroxide ion as we know X. Is equivalent to our hydroxide ion. So if we take 10 to the negative P. O. H, we can go ahead and solve for our concentration. So plugging in our value, we get 10 to the negative 3.25 Which gets us to a concentration of 0.00056234 moller. Using this value to solve for RK. B one, We're going to take that value of 0. and square it. Next we're going to divide this by the value of our an ion which was 0.4652. This gets us to a kB one of 6.79764 times 10 to the negative six Mohler. So as we can see right here, our KB one is very small. So our estimation that X is negligible was correct. Now we can go ahead and use the following formula K W equals KB one times K A. To in order to solve for R p K A. Two. So plugging in the values, we know R K W is one point oh times 10 to the negative 14. And this is going to be equal to r k B one which we calculated to be 6.79764 times 10 to the negative six. And we're going to multiply this by our K A two, calculating for R K A two, we're going to divide both sides by 6.79764 times 10 to the negative six. This gets us to aka two Of 1.47109, 9 times 10 to the -9. Now, to solve for RPKA two, we're going to take the negative log of our K A two. So, plugging in these values, we get the negative log of 1.471099 times 10 to the negative nine. Which gets us to a p k A two of 8.83, which is going to be one of our answers. Now for a week. Di protic acid at the first equivalence point, we can simply take our ph and make it equal to our P K one plus R p k A two And divide this by two. So, plugging in our values, we know our ph was 4.35 and this is going to be equal to our p K one Plus our PK two which was 8.83. And we're going to divide this value by two, Solving for APKA one. We end up with a value of 2.67, which is also going to be our final answer. Now, I hope this made sense and let us know if you have any questions.