Calculate the pH at the equivalence point for titrating 0.200 M solutions of each of the following bases with 0.200 M HBr: (b) hydroxylamine 1NH2OH2.

Video transcript

Hello everyone today we are being asked what is the ph at the equivalent when 0.150 molar of pirate line is tight traded with 0.1 molar hydrochloric solution. So the first thing you do is you want to write out what our reaction local would look like. So we have our period in here and it reacts with our hydrochloric solution and it is going to form our pure iodine ion as well as our chloride ion. The equivalence point. So the equivalence point is essentially just a point where our molds of our acid equal the moles of our base. And so in this equation here, since we have a one coefficient in front of the purity and the hydrochloric solution, those two are completely consumed. Next we want to utilize our dilution formula that says that the hilarity of our base is equal to the volume times the volume of the base is equal to the parity of a, times the volume of A as well. And so by default here we can note that are pretty of, our base is equal to the majority of our acid and thus the volume of our base must be equal to the volume of our acid. And so the total volume. So the total volume must therefore equal to V. Be since it is essentially just adding or multiplying to V. A. S. Or to VBS. And so the next thing you wanna do is you want to find the concentration of our period in ion here. And so to do that we have to essentially use our initial equation that we had. So we have the polarity for our paradigm which is point 15 4.150 Molar. We then have to multiply by our volume of our base. We're gonna put this over our two V. B. Our volume of the bases will cancel each other out. And so will essentially just have .150 divided by two, which will give us 0.075 molar concentration. And then we simply have to find the K. A. Value. So according to a textbook reference, our paradigm here has a K. B value of 1.7 times 10 to the negative ninth. If we use the equation that R K W. R constant is equal to the K. A. Times the KB. We can rearrange this as K. A. Is equal to R K W. Over R K B. In a reference text. RKW is always gonna be one times 10 to the -14. Then we have our KB value which is 1.7 times 10 to the negative night. That's gonna give us a K. A value of 5.88 times to the negative six. Our next step is to make an I. C. F. Table. So we're going to make an I. C. F. Table that is going to be used with our reaction. So we have our period ayn on that reformed, it's going to react with water and that's gonna give us our period ng and our hydroxide ions. Then we'll have the initial concentration, the change in the concentration and our final concentration, Our initial concentration we calculated for our period in ion is 0.075 molar. For our water it's going to be zero because water is a constant throughout for our period. I know it's gonna be zero and hydrating for zero. Of course we are then going to be tight trading this so therefore we're going to be taken away from the eye on itself. And so that's why this gets a negative X. As our change. And so our products are going to get a positive X. We then note that when we calculate this or combine the initial and the change we get 0.75 minus X. And then we get X. And X. For our products. Let us write our K. A expression. Okay expression is going to be based on the concentration of our products. So in this case that is our Carradine and our hydro knee um over the concentration of our reactant, we're just going to be our period in ion. And as before we do not utilize, we do not utilize the water because it is a constant. And so one thing to note about this K. A expression is that it is going to be larger than 500. Which is that constantly referring to. And there were four. Therefore we can ignore the value of X when we go to calculate for it. And so we put this together as part of Step nine. When we re number these We get the K. A. value being 5.88 times 10 to the -6. Equalling X squared from the two X. Of r I C. F. Table over Just our 0.075. When we solve for X, we get that X is equal to 6.6 times 10 to the negative 4th moller. X is also going to be equal to our hydro nian concentration. Our very last step is going to be to calculate for ph and the calculation for ph is as follows negative log of our hydro um concentration. So we're gonna plug in negative log of our high joining concentration which was our six 0. times 10 to the negative fourth moller. And this is gonna give us a ph of 3.173 point 178 as our final answer. Overall. I hope that this helped. And until next time