Calculate the percent dissociation of 0.10 M hydrazoic acid (HN3,...

Question

Calculate the percent dissociation of 0.10 M hydrazoic acid (HN3, Ka = 1.9 X 10^-5). Recalculate the percent dissociation of 0.10 M HN3 in the presence of 0.10 M HCl, and explain the change.

Accepted Answer

Hello everyone in this video want to calculate for the value of the percent ionization when 0.20 molars of hcl is attitude. The solution that would explain why the two values are different. So regarding our 0.20 molars of CH three C O H. Let's go ahead and do an ice table here. So the reaction that we're dealing with is C. H. Three C. O. H. Which is that reacts with water, which is a liquid. And this yields H30 plus, which is Aquarius as well as our an ion. So CH three C 00 minus. Alright, so of course we have our I C N. E. I stands for initial C. Stands for change and E stands for equilibrium. So these are all regarding concentrations. So for my first product here this is of course 0.20 molars for our second starting material, this is a liquid stage. So all these will actually be not appropriate for our ice table so we can go ahead and just ignore all of this for H 30 plus this is zero as well as our H three C 00 minus. All right, So for my change here, this is all unknown for my starting materials that will always be a minus sign. And then for my products, I will always have a plus. So because this unknown well denoted as X. So for our 1st 3rd material here, that is just minus X. Now, for my products, they're both of course unknown and both adding it. So we'll put as plus X and plus X. So now we're basically taking the sum here of this to give us the equilibrium concentration. So starting off with my first sort of material, this will equal to then 0.20 molars minus X. Our first product here, this is just X. And same thing for my second product. So you know that the K value is equal to the concentration of H. 30. Plus multiplied by the concentration of C. H. Three. C. 00 minus is going to be divided by the concentration of C. H. Three. C. O. H. Alright so it's actually plugging some of these values then we have 1.8 times 10 to the negative five. We're going to X multiplied by x over 0.20 minus X. So the concentration or the initial concentration of R C H three C. 00 H divided by the k value is really, really big. So X. Is is pretty small compared to our 00.20. So let's actually go ahead and simplify that. So 1.8 times 10 to the negative five equals two X squared over 0.20 minus X. Let's go ahead and isolate our X squared. Of course we're just ignoring this minus X here. So small. So then we have 3.6 times 10 to the negative six going to X squared and find taking the square root of both sides cancel. And just isolate our X Value which is what we want. We get that X. Is equal to 1.897 times 10 to the negative three. And this is going to be our concentration for H. 30. Plus. So then for my percent ionization is equal to the concentration of H. 30. Plus divided by the concentration of C. Nine H. 10 02 as initial state times 100. We'll go ahead and just plug in the values. Here we have 1.897 times 10 to the negative three over 0.20 molars times simplifying all this, we get them I percent. Ionization Is equal to 0.949%. All right now moving on for my 0.20 molars of ch three C. O. H. With our 0.2 molars of H. C. L. Alright, so for my chemical reaction here we have H. C. L. Which is a quiz that reacts with one mole of H. 20. In its liquid state. In this yield H. 30. Plus as well as C. L minus. Alright, so my concentration of H. C. L. Is equal to the concentration of H. 30. Plus. And this is equal to 0.20 molars. Now for my chemical reaction here again we have CH three C. O. H. This reacts with H. 202 yield H. 30. Plus as well as Ch three C. 00 minus. Alright, okay. We have our initial change in equilibrium. So for my initial concentration here we have course have 0.20 molars. Again this water so all of this will be neglected for H. 30. Plus we have 0.20. And then from my last product here we have zero changes are still all unknown. So we have minus X plus X. And plus X. There. From my equilibrium we have 0.20 molars minus X. 0.20 minus X. And X. All right. So then my value here 4K. 1.8 times 10 to the -5. This time it equals to 0.20 plus X times X. Over 0.20 - X. So again, X. Is small compared to our points too. So we go ahead and neglect our values of X. And here and here. So this leaves us. That X then is equal to one X. Of course is equal to 1.8 times 10 to the negative five molars. And this is equal to the concentration of H 30. Plus. Now, sorry for the percent iron nation of this here is of course equal to the concentration of my H. 30. Plus divided by my concentration of CH three C. O. H. As initial state times 100. All right, So putting in the vice. I know that's 1.8 times 10 to the negative five developed by 0. of course is all multiplied by 100. So then my percent ionization Is equal to 9.00 times 10 to the -3%. Now the present ionization decreases due to the presence of additional H. 30. Plus in the solution, which is also known as the common ion effect. So then my final answers for this problem is that let's just scroll back up over. Just go ahead and actually rewrite this. Alright, so for 0.20 molars of C. H. Three C. O. H. My percent ionization is 0.949% now four hours 0.20 molars of C. H. Three C. O. H. But adding are 0.20 molars of hcl. We get that percent organization is equal to nine zero times 10 to the negative three percent and the percent organization decreases when adding additional H 30. Plus in the solution. And this is just known to be the common ion effects. Alright, so this here is all going to be my final answer for this problem. Thank you all so much for watching

Calculate the percent dissociation of 0.10 M hydrazoic acid (HN3, Ka = 1.9 X 10^-5). Recalculate the percent dissociation of 0.10 M HN3 in the presence of 0.10 M HCl, and explain the change.

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