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Video transcript

let's look at this acid-base reaction so water is going to function as a base that's going to take a proton off of a generic acid H a so lone pair of electrons on the oxygen pick up this proton leaving these electrons behind on the a so oxygen oxygen is now bonded to three hydrogen's right so it picked up a proton that's going to give this oxygen a plus one formal charge and we can follow those electrons so these two electrons in red here are going to pick up this proton forming this bond so we make hydronium h3o plus and these electrons in green right here are going to come off onto the a to make a minus let's go ahead and draw that in so we're gonna we're going to make a minus let me draw these electrons in green and give this a negative charge like that so let's analyze what happened H a donated approach on so this is our bronsted-lowry acid once H a donates a proton right we're left with the conjugate base which is a minus water h2o accepted a proton so this is our bronsted-lowry base and then once h2o accepts a proton we turn into hydronium h3o plus so this is the conjugate acid so h3o plus the conjugate acid and then a minus would be a base so if you think about the reverse reaction h3o plus donating a proton to a minus then you would get back h2o and h a once once this reaction reaches equilibrium we can write an equilibrium expression and we're going to we're going to consider we're going to consider the stuff on the left to be the reactant so we're going to let the Ford reaction and the stuff on the right to be the products so let's write our equilibrium expression and so we write our equilibrium constant and now we're going to write ka which we call the acid the acid ionization constant so this is the acid ionization constant or you might hear acid dissociation constant so acid dissociation so either one is fine alright and we know when we're writing an equilibria when we're writing an equilibrium expression we're going to put the concentration of products over the concentration of reactants so over here for our products we have h3o plus let's write the concentration of hydronium h3o plus times the concentration of a minus so times the concentration of a minus all over the concentration of our reactants so we have H a over here so we have H a so write that in and then for water we leave water out of our equilibrium expression it's a pure liquid its concentration doesn't change and so we leave we leave h2o out of our equilibrium expression all right so let's let's use this idea of of writing and ionization constant and let's apply this to a strong acid HCl is going to function as a bronsted-lowry acid and donate a proton to water which is going to be our bronsted-lowry base and so we could think about a lone pair of electrons on the oxygen taking our proton right leaving those electrons behind and so the oxygen is now bonded to three hydrogen's because it picked up a proton giving this a +1 charge and so once again let's follow those electrons in red so this electron pair picks up this proton to form this bond so we form h3o plus or hydronium and these electrons in green move off onto the chlorine so let's show that so we form the chloride anion so let me go ahead and draw in the electrons in green and I'm going to write a negative one charge here like that so another way to represent this acid-base reaction would just be to write out h2o plus HCl gives us h3o plus plus CL minus so this is just a faster way of doing it and HCl is a strong acid strong acids donate protons very easily and so we can say this this process occurs a hundred percent so we get 100 percent ionization the equilibrium is so far to the right that I just drew this one arrow down over here so we get approximately 100 percent ionization so everything turns into our products here and let's go ahead and write our equilibrium expression so ka is equal to concentration of h3o plus right so concentration of our products times concentration of cl minus all over right we have HCL and we leave out water so if we think about approximately 100 percent ionization we have all products here so we have a very very large number in the numerator and extremely small number in the denominator so if you think about what that does for your ka it's going to give you an extremely high value for your ka right so ka is much much much greater than one here and so that's how we recognize strong acids Oh an acid ionization constant that's much much greater than one now let's think about the conjugate base all right so let's go back up here so we had HCl and Cl minus as our conjugate acid-base pair and the stronger the acid the weaker the conjugate base right so HCl is a strong acid so CL minus is a weak conjugate base so let me write that here so the stronger alright the stronger the acid so stronger the acid weaker weaker the conjugate weaker the conjugate base and one way to think about that is if I look at this reaction we can think about competing base strength right so here we have bronsted-lowry base water is acting as a bronsted-lowry base and accepting a proton and over here if you think about the reverse reaction the chloride anion right would be trying to pick up a proton from hydronium for the reverse reaction here but since HCl is so good at donating protons that means that the chloride anion is not very good at accepting them so the stronger the acid the weaker the conjugate base water is a much stronger base than the chloride anion finally let's look at acetic acid so acetic acid is going to be our bronsted-lowry acid and this is going to be the acidic proton so water is going to function as a bronsted-lowry base and a lone pair of electrons and the auction is going to take this acidic proton leaving these electrons behind on the oxygen so let's go ahead and draw our products so we would form the acetate anions let me go ahead and draw in the acetate anion so negative one charge on the oxygen and let's show those electrons these electrons in green move off onto the oxygen right here giving it a negative charge we're also going to form a hydronium alright so h3o plus so let me go ahead and draw in hydronium so a plus one formal charge on the oxygen and let's show those electrons in red right so this electron pair picks up the acidic proton right form this bond and we get h3o plus so another way to write this acid-base reaction would be just to write acetic acid ch3cooh plus h2o gives us the acetate anion ch3coo minus plus h3o plus now acetic acid is a weak acid and weak acids don't donate protons very well and so acetic acid is going to stay mostly protonated so when you think about this reaction coming to an equilibrium you're going to have a relatively high concentration of your reactants here so when we write the equilibrium expression write ka is equal to the concentration of your products so ch3coo minus times the concentration of h3o plus right all over the concentration of acetic acid because we leave water out so all over the concentration of acetic acid all right the equilibrium lies to the left because acetic acid is not good at donating this proton and so we're going to get a very large number for the denominator right for this concentration so this is a very large number and a very small number for the numerator all right so this is a very small number so if you think about what that does the KA write a very small number divided by a very large number this gives you a KA value an ionization constant much less than one alright so this this value is going to be much less than one and that's how we recognize it's one way to recognize a weak acid look at the KA value
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