What is the pH at the equivalence point for the titration of 0.10 M solutions of the following acids and bases, and which of the indicators in Figure 17.5 would be suitable for each titration? (c) CH3NH2 (methylamine) and HCl

Video transcript

Hello. Everyone in this video trying to calculate for the ph at the equivalence point of a 0.0.125 molar solution of dimethyl amine tie traded with 0.125 molars of hcl. So we have the volume of hcl that's being told that we have that equal to the dimethyl amine. It was used to achieve equivalence And we want to determine the suitable indicator for the said situation. Were given the K. of this ion here which is 1.86 times 10 to the -11. So our solution of dimethyl amine tight treated with Hcl. Let's go ahead and write the chemical equation occurs at equivalence point. So we just read that at equivalence point we have the Bethel means. So we have c. H. three two and H. Which is and then we're gonna add that with H plus which of course is also Aquarius plus, R. C. L minus ions. And that reacts to give already a different line below our products include this ion that were given the K. Value for as well as our chlorine and ions. And of course our H 20. And that's in its liquid state. So that's only the salt which are dimethyl amine chlorinate will be left in the solution. So all concentrations involved will be halved when equal volumes of acid and base react. Thus the dimethyl amine. See let's write this out. So 0.125 molars of dimethyl amine. So that's C H 32 and H is going to be 0.625. Again it's halved of our molar concentration of dimethyl amine. All right, so for our ion or ourselves in this ocean in equilibrium is actually write this out as well. We're gonna go ahead and formulate a ice table. Go scroll down for more space. So again, the solution in equilibrium, that equation is going to be our CH 32. NH two plus which is acquis reacts with H 20. Which is water to go ahead and be a grilling room with maybe I'll put this down more Our H plus which is a quiz. Plus our dimethyl means that C. H. Three to an H. Which is reformulating a ice table. So, if I for initial concentration C for change of concentration and E. For the concentration in equilibrium. So initially for our ion here we have let's do this in maybe a different color in black. We have 0.0625. And then for our products here we have zero and zero. And for HBO, since it's a liquid, we don't count that in our ice table. So our change here is always unknown or in our case is unknown. And then for our starting materials are change is going to be negative. And then for our products is always plus, since we don't know it well denoted as being the value of X. So again, for my star material is done for this ion will have minus X. For my two products here, we'll just have plus X and plus X. Well I coefficient to this unknown. So in front of our X. If we have a coefficient in our vast occasion and which of course we just have ones here. So we just won't put any values there because any number multiplied by one is just itself. So equilibrium is just basically taking the total of this, this and this for our slots here and here. So basically combining those two for our first ion for my region starting reagents just have 0.625 minus X from H. 30. Plus we have X. And then for my other product we also have X. Let's go ahead and scroll down for more space here. So then the K. A. Which are given for our ion CH three to N. H two plus is equal to 1.86 times 10 to the negative 11. We know that the formula for K is equal to we'll start off with our murder, we have a concentration of our products H. 30 plus multiplied by the concentration of CH32 and H all over the concentration of our starting materials in our case is just our ion because again is a liquid. So we have C. H. 32 and H. Two plus. Let's go ahead and scroll down and just fill in all the values that we have. So if we play all the values that we do have that's going to be equal to. She doesn't spread it in the second line because we know that K. Eight and that's 1.86 times 10 to the 9 11. So for the concentration of H. 30. Plus that is X. Concentration of my second Marie agent. There is also an X. And then for our denominator here we do have a value of some sort and that 0.625 minus X. So determine if our plus or minus X. Can be omitted from the expression. We want to find the ratio of the initial concentration and our K. A. Values. So maybe I'll do this off to the side here in a different color. I'll choose blue. So we have the concentration of our C. H. 32 And H. two initial Divided by R. K. A. value. That's equal to 0.0625 over 1.86 times 10 to the negative 11. So once I put that into my calculator I seek that I get a value of 3.36022 times to the positive nine. So since this value is much greater than 500. R. Plus it minus can be admitted from the expression. All right so we can see here that basically this X. Can be omitted. Go ahead, scroll down for more space to continue our calculation. Alright so if you combine this and simplify this then we can go ahead and get 1.1625 times 10 to the negative power equaling two X squared basically got rid of this denominator by multiplying both sides by this number of 0.625. And we just multiplied our two Xs on the numerator. And doing more of our calculations, we're isolating X. So we're gonna go ahead and take the square root on both sides. What we get is that X. Which we said to be the concentration of H. 30. Plus and also the concentration of dimethyl means. So ch 32 and H equals two. Once we take the square root of like we said the square root of both sides here and here to isolate our X. We get the value of 1. times 10 to the negative six molar. Because we're solving for concentration now that we have this concentration of H. 30. Plus. We actually saw for ph as well if you recall this equation is that ph is equal to the negative log of the concentration of H. 30. Plus which we have. So plug in this value. The negative log of the concentration is 1.78193 times 10 to the negative six. We get that the ph is equal to 5.97. So the ph of the equivalence point is 5.97. And since the closest p. K of the indicator is going to be equal to chloral final red because it has a PK value of 6. and it's going to be the most suitable indicator. So our answer, then is that the ph at the equivalence point Is equal to 5.97. An indicator that we can use is plural fennel red, which has a T. K. That's equal to 6.25. So this right here is going to be my final answer for this problem.