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buffer solutions resist changes in pH and so let's think about a solution of a weak acid and its conjugate base so here we have H a which is our generic weak acid and so the conjugate base would be a minus and a buffer solution needs to have substantial amounts of both present and that's what I'm trying to represent over here so we have a beaker that has in this case equal amounts of H a and a minus and to our buffer solution we're going to add some strong acid so a source of protons so I'm going to I'm going to draw a proton in here so H+ I could have written h3o plus because h+ and h2o give you h3o plus I'm just writing H+ to make it a little bit easier to think about so if you're increasing the concentration of h+ ions or increase the concentration of hydronium ions in solution you would think that would decrease the pH dramatically and that would happen if you didn't have something to react with the protons in solution but we do in our buffer solution right we have a minus we have our base the base is going to pick up this proton and h+ + a- are going to form h a so we're going to make more of our weak acid here so let's see we started out with some weak assets let me go ahead and draw that in here so some H a and then we're going to make more because h plus and a minus give us H so we're making more of our weak acid here what about a - alright well we're going to use one of these up - to make H a right so now we have only two of these left so we're decreasing the concentration of a - but what we've done is we've effectively removed protons from solution and that's how that's how a buffer is able to resist a change in pH if you add acid what about if you add base so next let's think about our buffer solution and this time we're going to add a strong base so we're going to increase the concentration of hydroxide ions in Ellucian so if we increase the concentration of hydroxide ions in solution you would think that would increase the pH right and it would except for the fact that we have an acid that can react with our base right so we have some weak acid present here and the hydroxide ion is going to pick up a proton from H a right so we have H+ here so - and H+ give us h2o so we're going to make we're going to make water here right so we're going to lose some of our weak acid so we started with three and now we have only two left here and if oh - if hydroxide takes a proton away from H a then we're left with a - so we're going to increase the amount of a - that we have right in solutions let me go ahead and draw that in here so once again we've we've effectively buffered against the change in pH because the hydroxide ions that we added right were reacted with the acid that was present so we've effectively removed hydroxide ions from solution alright so that's the idea of a buffer solution let's see if we can derive an equation that will allow us to do some calculations and so next let's look at what we have down here all right so we know that we have in a buffer we have we have substantial amounts of H a and a - present and so we have this reaction at equilibrium so for our equilibrium expression right concentration of products over concentration of reactions so here we have our concentration of products over our concentration of reactants once again leaving water out let's take the negative log of both sides of this so we're going to take the negative log of ka so that would be negative log of ka here and the negative log of all of this which would be equal to the negative log of all of that and I'm going to write it a little bit differently I'm going to put the concentration of hydronium ions out front here and then have concentration of a minus over concentration of H a all right I wrote it like this because it makes it a little bit easier to see a property of a logarithm let me just go ahead and write down the property that I'm referring to if you have log of a times B that's the same thing as log of a plus log of B and here we have the negative log so the negative log of a b is equal to the negative log of a plus the negative log of b so i'm just write minus the log of b here and in this case right concentration of h3o plus that would be a alright so this would be a and then this over here would be B all right so let's think about what that would give us now so let's get some more space and let's think about our log property so on the left side we have negative log of K a and the right side we would have negative log of a so that would be negative log of concentration of h3o plus all right minus minus the log of B here so that would be minus the log and B was this right here minus the log of a minus over H a alright so we can keep going here because we know the negative log of K a right the negative log of ka is equal to the pka so this is equal to the pka and then over here we have the negative log of the concentration of hydronium ions that's equal to the pH all right so negative log of concentration of h3o plus is equal to the pH and then we have minus log of the concentration of a minus over the concentration of H a and so we can rewrite this in for our final equation so organiz can rewrite this solve for pH pH is equal to PKA plus the log of a minus over over the concentration of H a here this is called the henderson hasselbalch equation so right here is the henderson hasselbalch equation it's very useful when you're doing buffer calculations all right well look at examples of this and then video