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in the last video we learned that if I had some let's say I have some weak acid so it's hydrogen plus some the rest of whatever the molecule is called we'll just call it a and then I think this is dense tends to be the standard convention for the rest of the acid that it can disassociate or it's in equilibrium because it's a weak acid so it can be equilibrium with since it's a weak acid it's going to produce some hydrogen proton and then the rest of the molecule is going to keep the electron so it's going to be plus all this is in an aqueous solution let me do that aqueous it's an aqueous solution and then you're going to have the rest of the acid whatever it might be a minus and that's also going to be an aqueous solution and that's the general pattern we've seen the case where a is you know a could be a could be equal to it could be an NH it could be an nh3 right if a is an nh3 then when you have this you have an NH 4 plus so this would be ammonia and this is just nh3 a could be just a it could be a fluorine it could be a fluorine molecule right there because then this would be hydrogen hydrogen fluoride or hydrofluoric acid and this would just be the - the the negative cat a negative ion of fluorine or fluorine with an extra electron so it could be a bunch of stuff we've done it you know it could be you could just throw throw in anything there and it'll work especially for the weak acids so we learned in the last video that this if this is the acid so this is the acid acid then this is the conjugate base conjugate base conjugate base and we could write the same reaction essentially it's kind of more of a basic reaction so we could say if I start with a minus that that's in an aqueous solution that's in equilibrium with this thing could grab a hydrogen from the surrounding water and become neutral then it's still in an aqueous solution and then one of those water molecules that it plucked that hydrogen off of is now going to be a hydroxy molecule right because this hydrogen remember whenever I say pluck the hydrogen pluck not just the proton not the electron for the hydrogen so the electron stays on that water molecule so has a negative charge it's in an aqueous solution so we could write the same reaction both ways and we can write equilibrium react we could write equilibrium constants for both of these reactions so let's do that let me erase this because I I can erase this stuff right there you use that space so an equilibrium reaction for this first one I could call this the D the K sub a because the equilibrium reaction for an acid and so this is going to be equal to its products so the concentration of my hydrogen times my concentration of whatever my conjugate base was divided by my concentration of my original acid my weak acid so this would be the concentration of H a H a fair enough I could also write an equilibrium constant for this basic reaction let me do it right down here so I'll call that my K sub B for this is a base equilibrium and so this is equal to the concentration of the products becoming tedious to keep switching colors actually I'll do it because it makes it easier to look at at least for me H a times the concentration of my hydroxide ions divided by my concentration of my weak base a - remember this can only be true of a weak base or weak acid if we were dealing with a strong acid or a strong base this would not be an equilibrium reaction it would only go in one direction and it only goes in one direction writing this type of equilibrium reaction makes no sense or equilibrium constant because it's not an equilibrium it only goes in one direction if this was if a was chlorine if this was hydrochloric acid you couldn't do this you would just say look if you have a mole of this you're just dumping a mole of hydrogen protons and that's all and then a bunch of chlorine anions we're not going to do anything even though they are the conjugate base they wouldn't do anything so you can only do this remember for weak acids and bases so with that said let's see if we can find a relationship between ka and Kb so let's let's do some let's divide let's see what do we have here we have an a-minus on both sides of this so maybe if we if we let's see if we did we have h over o H over a minus let's see if we divide boats let's solve for a minus right if we make if we divide if we divide multiply both sides of this equation by H a over h plus we get on the left hand side we get ka ka times the inverse of this so you have your H a over h plus h plus is equal to is equal to your concentration of your conjugate base a minus and let's do the same thing here to solve for a minus so to solve for a minus here we might have to do two steps so if we take the inverse of both sides you get 1 over KB is equal to I'll just stick to well a minus a minus over let me go down a little bit lower over H the concentration of my conjugate acid times the concentration of hydroxide multiply both sides by this and I get a minus is equal to is equal to my concentration my conjugate acid times concentration of hydroxide all of that over my base equilibrium constant now these are the same reactions right and either in either reaction for given concentrations I'm going to end up with the same concentration this is going to equal that right these are two different ways of writing the exact same reaction so let's set them equal to each other so let me copy and paste it actually so I'm saying that this thing coppy this thing is equal to this thing right here so this is equal to this is equal to let me copy and paste this that that's equal to that so let's see if we can find a relationship between ka and Kb let's see if we well well one thing we can do we can divide both sides by H a right so if we divide both sides by H a actually I could have probably done that earlier on and the whole thing yeah well anyway I could have done that earlier on but let's say if we have if we ignore this part right here this is equal to that so we could write let me erase all of this oh I'm using the wrong tool so we could say that they both equal the concentration of a minus so that's equal to that we can divide both sides by H a so if we divide so this will cancel this over here will cancel with this over here and we're getting pretty close to a neat relationship and so we get ka over our hydrogen concentration our hydrogen proton concentration is equal to our hydroxide concentration divided by Kb we can just cross multiply this so we get K a our acidic equilibrium concentration times KB is equal to our hydrogen concentration times our hydroxide concentration remember this is all in the aqueous solution what do we know about this what do we know about our our our hydrogen times our hydroxide concentration in an aqueous solution we know let me let me let me so we know that for example let me go review just to make sure I'm jogging your memory properly we could have h2o it can auto ionize into h plus plus Oh H - and this has an equilibrium K sub W it's equal to you don't you you just put the products so the concentration of the hydrogen protons times the concentration of the hydroxide ions and you don't divide by this because it's the solvent and we already figured out what this was if we have just completely neutral water this is ten to the minus seven and this is ten to the minus seven so this is equal to ten to the minus fourteen now these two things could change I could add more hydrogen I could add more hydroxide and everything we've talked about so far that's what we've been doing that's what acids and bases do they either increase this or they increase that but the fact that this is an equilibrium constant means that look I don't care what you do to this at the end of the day this will adjust for your new reality of hydrogen protons and this will always be a constant as long as we're in an aqueous solution a solution of water where water is a solvent at 25 degrees so we're always so this no matter what we do I mean in just pure water is ten to the minus seven but no matter what we do to this and this in an aqueous solution the product is always going to be ten to the minus 14th power so that's the answer to this question this is always going to be ten to the minus fourteen if you multiply hydrogen concentration times Oh H concentration now they won't each be ten to the minus seven anymore because we're dealing with a weak acid or a weak base so that they're actually going to change these things but when you multiply them you're still going to get ten to the minus fourteen and then if you take let's just take the minus log of both sides of that let me erase all of this stuff I did down here I'll need the space so let's take out let's say we take the minus logs of both sides of this equation so you get the - let me do a different color - log of course it's base ten of ka let me let me do it in the colors ka times K B is going to be equal to the minus log of this is equal to the minus log of ten to the minus fourteen so what is this equal to the log of 10 to the minus 14 is minus 14 because 10 to the minus 14th power is equal to 10 to the minus 14 you take the negative of that so this becomes 14 so the right-hand side of your equation just becomes 14 and this one we could use log properties this is the same thing as the minus log of ka we use the colors ka plus plus the minus log of KB of KB or though you could think of this this is your PKA this is your P KB so you could say this is P ka plus p kb oh I wanted to use blue + p KB and all of that is going to be equal to 14 now why is why is this useful well if you know the pKa for a weak acid for example like let's say we have nh3 NH 4 plus this is a weak acid right it can donate an H but it's not a irreversible reaction that H can be gained back so this is a weak acid if you look it up on Wikipedia you'll see it'll tell you it'll say hey it's Pete the pKa of NH 4 is equal to is equal to nine point two five right so this is nine point two five for NH 4 for ammonium so what is going to be the P KB for ammonia right let's let me write that reaction down we could write so this is NH four is in equilibrium it can this is plus it can get rid of it one of its hydrogen protons and you're just left with ammonia or you could that's so this is the acidic reaction so this is what the pKa is associated with so the equilibrium constant for this reaction is the negative log of the equilibrium constant is equal to nine point two five and if I had the reverse reaction the basic the conjugate base reaction so ammonia converts to ammonium plus it grabbed that hydrogen from a water molecule that hydrogen proton from water molecule if I wanted to figure out the PK B or the equilibrium constant or the negative log of the equilibrium concept for this reaction what is it well this one's nine point two five and nine point two five plus this PK B have to be equal to 14 so what's 14 minus nine point two five it's what four point seven five so we immediately know the equilibrium constant for the conjugate base reaction so it's a useful thing to know that the the pKa plus the PK B is equal to 14 and always remember you see these PKA PK B and say what is that well if you see a P it's a negative log of something and in this case it's a negative log of the equilibrium constant for an acidic reaction plus the negative log of the equilibrium basic reaction where this is the conjugate base of this acid that's always going to be equal to 14 if we're dealing with an aqueous solution at 25 degrees Celsius which is essentially room temperature which is usually going to be the case

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