Strychnine 1C21H22N2O22, a deadly poison used for killing rodents, is a weak base having Kb = 1.8 * 10-6. Calculate the pH of a saturated solution of strychnine (16 mg/100 mL).

Video transcript

Hello everyone today we are the falling problem caffeine, a stimulant found in coffee, has a KB. Or a base constant of 4.1 times 10 to the negative fourth what is the ph of a 100.35 molar solution of caffeine. So the first thing I wanna do you wanna draw caffeine out when it reacts with water. So the formula for caffeine is c. eight h. 10 In 402. Of course this is going to be a quiz and it's going to react with water or liquid water. Excuse me. And we show that this reaction is reversible with the double headed arrow. And that's going to result in the following. We're going to have a hydrogen that is now attached to this caffeine molecule and of course you will have hydroxide ions. And so what we can do now is draw out our ice table in this and so that's just going to be our initial concentration in one row. The change and then the equilibrium or the final concentration. So we started out with 0.35 molar solution of caffeine, water does not get a concentration because it is a constant of course for our products we did not start off with anything since we are reacting to caffeine we are going to have X here as our variable to represent caffeine. We're also gonna draw minus X. To show that we are slowly getting rid of it treating it. And then for our products this means that we're going to have a positive X. For each of our products. And so if we were to add up our initial and our change or INRC we would get 0.35 -1. And then for our products we would simply just get X and X. And so now we can construct what are KB expression would look like. So our KB expression is gonna be the concentration of our products over are reacting. So that that includes the concentration of our hydroxide, times the concentration of our caffeine molecule that has that extra hydrogen attached to it. That is going to be divided by our concentration of our caffeine. Since we already have our KB value which is 4.1 times 10 to negative fourth. We can go in and plug in the rest of what we have. So this is going to equal. We set our concentrations of our products were X. So this is going to be X times X. We're going to divide that by the concentration of our reactant which was 0. minus X. Something special that we can make note of. Is that this expression here? That would be .35 moller divided by our kB expression. We're essentially cross multiplying these two variables here. This is going to be greater than 500. And so since this is a case we can omit the minus X portion of this problem. So in actuality this turns into 4.1 times 10 to the negative fourth is equal to X squared over 0.35. This gives us a value that X. is equal to 0. moller. So we're gonna hang on to this value here. The last stuff that we need to do is we need to calculate for our P. O. H. And that is equal to the negative log of our hydroxide ions. So when we plug in that value we get negative log times 0.01198 moller. And that was going to give us a value of 1.92. However, this is P O. H. And we don't want P. O. H. We want the P. H. And so to simply find the ph we can take the value that the ph is equal to 14 minus the P O. H, which was 91.92. And this gives us the final value of 12 point oh eight as our final answer. And so with that we have answered the question overall. I hope that this helped. And until next time