If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains ***.kastatic.org** and ***.kasandbox.org** are unblocked.

Main content

Current time:0:00Total duration:11:20

we've already seen that nh4 plus and nh3 are a conjugate acid-base pair and so let's look at NH 4 plus the ammonium ion would function as an acid and donate a proton to water to form h3o plus and if NH 4 plus donates a proton you're left with nh3 the KA for this reaction is five point six times ten to the negative ten now let's look at nh3 which we know is a weak base and it's going to take a proton from water therefore forming NH 4 plus and if we take a proton from water we're left with Oh H minus since we talked about a base here we're going to use Kb and KB for this reaction is 1.8 times 10 to the negative 5 what would happen if we add these two reactions together so we have two water molecules for our reactants so let me go ahead and write I'll write h2o plus h2o here and what about what about ammonium well we have ammonium on up on the left side for reactant we also have ammonium over here for a product and so that cancels out same thing happens with ammonia nh3 we have nh3 on the Left we have nh3 on the right so we have nh3 as a reactant nh3 as a product we can cancel those out too and so our only reactants would be 2 h2o and for our products we would get h3o plus and Oh H - so h3o plus hydronium and hydroxide and this reaction should sound familiar to you this net reaction is the autoionization of water where one water molecule acts as an acid one water molecule acts as a base and we get h3o plus and Oh H - so the equilibrium constant for the autoionization of water we've already seen that KW is equal to 1.0 times 10 to the negative 14 so we added these two reactions together and we got this for our net reaction all right what would we with ka and Kb to get kW well it turns out that you multiply them so ka times KB for a conjugate acid-base pair is equal to kW so let's do that math ka is 5.6 times 10 to the negative 10 so five point six times 10 to the negative 10 KB is equal to one point eight times 10 to the negative five one point eight times 10 to the negative five and let's get out the calculator and let's go ahead and and do that math so we have five point six times 10 to the negative 10 and we're going to multiply that by one point eight times 10 to the negative five and we get one point zero times 10 to the negative 14 so this is equal to one point zero times ten to the negative 14 which is our value for kW so when you add reactions together to get a net reaction you multiply the equilibrium constants to get the equilibrium constant for the net reaction which in this case is kW for the autoionization of water alright let's go ahead and go in even more detail here so ka right that's your that's your products over your reactants that'd be h3o plus the concentration of h3o plus times the concentration of ammonia let's go ahead and do that so the concentration of h3o plus times the concentration of ammonia and that's all over the concentration of ammonium so this is ka so this is all over the concentration of ammonium so this represents let me go ahead and highlight this here so this represents ka all right next let's think about Kb all right so over here is Kb that'd be the concentration of your products so NH 4 plus x h minus let's go ahead and do that so let's put this in parenthesis here so we have the concentration of ammonium nh4 plus all right times the concentration of O h- and that's over the concentration of Na three that's over the concentration of NH 3 here so what do we get well the ammonium would cancel out right so that cancels here and then the NH 3 cancels out and we're left with h3o plus times Oh H - all right which we know is equal to 1.0 times 10 to the negative 14 so just another way to think about this so this can be this can be important relating ka and KB to kW so if you know one you can figure out the other let's think about think about a strong acid for a second here so let's think about HCl the conjugate base to HCl would be CL - right the chloride anion here let's think about what this equation means so HCl is a strong acid which means a very high value for ka right so an extremely extremely high value for ka so what does that say about KB for the conjugate base alright the conjugate base here is the chloride anion well if ka is very large then KB must be very small for this to be equal to kW so KB is extremely small here so very small value for K B so this mathematically describes what we talked about earlier the stronger the acid the weaker the conjugate base HCl is a very strong acid has a very very high value for ka and the conjugate base is the chloride anion and it must have a very very very low value for KB which means it's an extremely weak base and so this is mathematically how to think about that relationship all right next let's look at a problem where we're calculating one of those values methylamine is a weak base and the KB for methylamine is three point seven times ten to the negative four our problem asks us to calculate the KA value for the methyl ammonium ion which is ch3 nh3 plus so we're talking about a conjugate acid-base pair there's one proton difference between those and so therefore we can use our equation ka times KB is equal to kW so we can plug in KB here and so we now we have ka times three point seven times ten to the negative four is equal to kW which is one point zero times ten to the negative 14 so let's do the math and solve for K a so one times 10 to the negative 14 we need to divide that by three point seven times ten to the negative four and so ka is equal to two point seven times ten to the negative 11 so K a is equal to two point seven times 10 to the negative 11 and so we're done that's our answer 2.7 times 10 to the negative 11 is the KA value for the methyl ammonium ion let's go a little further let's let's take our equation here ka times KB is equal to kW let's take the log of both sides alright so that would be the log of K a times K B is equal to the log of K W the log of ka times KB is the same thing as the log of ka plus the log of K be equal to the log of K W if we take the negative of everything all right let's go ahead and do that so the negative of everything negative log of ka I'll put that in parentheses plus the negative log of K B is equal to the negative log of K W and the negative log of K a we know that this is equal to the pka all right so the negative log of ka was our definition for PKA and the negative log of KB was our definition for pk B so PKA plus P KB is equal to finally the negative log of K aw that would give you 14 right--so 14.00 is the negative log of 1.0 times 10 to the negative 14 is 14 point zero zero so now we have something else that we can work with so let me go ahead and box this right here and let's let's take the KA value that we just found let's find the pKa all right so the pKa would be equal to the negative log of two point seven times ten to the negative 11 so let's do that on our calculator here so let's get some room so the negative log of two point seven times 10 to the negative 11 gives us ten point five seven right we have to round that 10 point five seven these are our two significant figures because we have two significant figures here let's go up back up to our problem here so that's the pKa for the methyl ammonium ion so let's say you're given the pKa for the methyl ammonium ion and ask for the PK be for methylamine all right so what is the PK be for methylamine all we have to do is plug into our equation alright so ten point five seven all right so ten point five seven was our PKA value so let's go ahead and write that in here so we have ten point five seven plus P KB is equal to 14 so when we solve for the PK B that would give us three point four three so the P KB is equal to three point four three all right we could double-check that let's go back up here we could double-check that because if we took the negative log of this number that's what we should get alright so let's go ahead and do that let's take the negative log the negative log of three point seven times ten to the negative four and we get three point four three so that's what we just calculated down here all right so the PK B is equal to three one two four three alright so that's that's the relationship between ka and KB and you can also talk about the relationship between PKA and pkp