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:11:22

Video transcript

ammonia is a weak base and if ammonia reacts with water water is going to function as a bronsted-lowry acid so water is going to donate a proton to ammonia and ammonia is going to accept a proton so it's going to be a bronsted-lowry base so a lone pair of electrons in the nitrogen pick up this proton which leaves these electrons behind on the oxygen and if you protonate ammonia you form ammonium the ammonium ion which is NH 4 plus so let's show those electrons so these electrons is lone pair of electrons here on the nitrogen pick up this proton and form this bond right here so we form NH 4 plus and these electrons in blue come off on to the oxygen so the oxygen had two lone pairs of electrons around it and the electrons in blue are now around it which gives it a negative 1 formal charge here so we form the ammonium ion nh4 plus and the hydroxide ion o h - so if ammonia functioned as a bronsted-lowry base over here would be the conjugate acid right so ammonium is the conjugate acid to nh3 and water functioned as a bronsted-lowry acid so over here would be the conjugate base the hydroxide anion so instead of using ammonia let's just do a generic base so here I have be written so some generic base reacts with water takes a proton from water to form B H+ and if you take a proton away from h2o you would form au H minus so once this comes to equilibrium we could write an equilibrium expression so we're going to write KB here so instead of ka we're now writing Kb because we're talking about a base and KB can be called you can call this the base ionization constant so let me write base ionization constant here or you can call this the base dissociation constant so base dissociation constant and when you're writing an equilibrium expression remember it's the concentration of your products over your reactants so over here we have the concentration of eh plus times the concentration of O H - and that's all going to be over the concentration of your reactants and we leave out water so we have only the concentration of our generic base B so you can think about Kb the same way we thought about K a all right the higher the value for K be the stronger the base because the more the more of your products you're going to make here all right so let me go ahead and write that so the stronger the stronger the base the larger the value for K be so let's talk about two weak bases here so we've already mentioned ammonia and we're going to compare ammonia to annalen so nh3 is ammonia and the KB for ammonia is 1.8 times 10 to the negative 5 and aniline is c6h5 NH 2 and notice it has a lower value for the KB for point 3 times 10 to the negative 10 is a much smaller number much smaller value than 1.8 times 10 to the negative 5 so ammonia has a higher value for the KB and therefore it's a stronger base than annalen so once again both of them are weak bases but ammonia is the stronger of the two remember when we talked about ka we also talked about PKA and so we're going to the same thing here we've talked about Kb and so now let's talk about calculating the p kb so the P KB is equal to the negative log of the KB so let's say we want to find the P KB for ammonia all we'd have to do is plug in the KB value and so the p kb would be equal to the negative log of 1.8 times 10 to the negative 5 and so let's get out the calculator here and do the problem negative log of 1.8 times 10 to the negative 5 and so we get 4 point 7 4 here so the P KB for ammonia is four point seven four and you can do the same calculation for aniline and plug in this cave and you would get a PK B of nine point three seven so comparing to week basis ammonia and annalen ammonia is the stronger of the two because it has a higher value for the KB and notice that the P KB is a lower value so once again analogous to what we talked about for PKA so next we'll do a calculation for a for a solution of ammonia our problem asks us to calculate the pH of a point five zero zero molar solution of aqueous ammonia so we have ammonia in water so we have nh3 plus h2o and ammonia is going to accept a proton from water and turn into NH four plus ammonium and if water loses a proton h2o turns into a minus so let's go ahead and write our initial concentration here so for ammonia we have 0.5 0 0 molar so we write 0.5 0 0 molar here and we're pretending like nothing has happened so the concentration of our products is zero so nothing's happened yet so the concentration of ammonium is zero and the concentration of hydroxide is also zero next let's think about the change all right so since ammonia right since nh3 turns into NH 4 plus the concentration of ammonia that's lost is the same concentration as ammonium that is gained and if we represent that concentration using X alright so if we lose a certain concentration of ammonia that's the same concentration of ammonium that we gained alright so this would be plus X over here for ammonium and it'd be the same thing for hydroxide so this would be a plus X here for the hydroxide anion therefore at equilibrium alright for ammonia the concentration of ammonia at equilibrium we would have 0.5 0 0 minus X for ammonium we would have X and for hydroxide we would also have X so next we write our equilibrium expression so we KB here is equal to the concentration of our products so that's NH four plus so I write NH four plus here times the concentration of hydroxide that's all over the concentration of ammonia right we leave water out so this is all over the concentration of ammonia here so let's plug in what we know so at equilibrium the concentration of ammonium is X alright so let's write in an X here for the concentration of ammonium and it's the same thing for the concentration of hydroxide so for hydroxide the the concentration at equilibrium is also X so we write an X right here this is all over the concentration of ammonia and that would be that would be the concentration of ammonia at equilibrium is 0.5 0 zero minus X so we put in point five zero zero minus X here this is all equal to the base ionization constant for ammonia and let's go back up to our table above to remind ourselves of what it is here right so ammonia has a KB of 1.8 times 10 to the negative five so we're going to plug that value in for KB here so this is equal to 1.8 times 10 to the negative 5 and once again we're going to assume we're going to assume that X is much much smaller than 0.5 0 0 molar because that makes our life easy for the for the math we don't have to do the quadratic formula if we do that and if we make this assumption right if we make this assumption then 0.5 0 0 minus X X is extremely small this is pretty much the same thing as 0.5 0 0 so we're going to we're going to leave that out let's rewrite our let's rewrite what we have here we have 1.8 times 10 to the negative 5 is equal to this would be x squared over zero point five zero zero all right so all we have to do now is solve for X and so we would have x squared is equal to let's see 1.8 times 10 to the negative 5 we have to multiply that by point five and we would get 9.0 times ten to the negative six so next we just take the square root of nine point zero times ten to the negative six to solve for X and let's go ahead and do that so let's get out the calculator here and let's take the square root of nine point zero times ten to the negative six and so we get point zero zero three so let's go ahead and write X is equal to X is equal to zero point zero zero three zero and remember what X refers to let's go back up here let's think about what X is talking about X is talking about a concentration and notice we have an X here for our concentration of hydroxide anions and that's what's going to allow us to eventually get the pH which is what the question asked for so X is equal to the concentration of hydroxide anions so this is Bmore all right this is our concentration of hydroxide anions and our goal is once again to find the pH of our solution so first we could find the Poh so that's one way to do it the Poh is equal to the negative log of the concentration of hydroxide ions so we go ahead and take the negative log a point zero zero three zero to calculate the Poh of our solution so let's do that on our calculator so negative log of point zero zero three zero is equal to two point five two so the Poh is equal to two point five two and then to find the pH all we have to do is subtract from 14 because pH plus Poh is equal to 14 and I believe in a previous video I forgot to put these zeros here if you're concerned about significant figures all right so you plug in your Poh into here so your pH plus two point five two is equal to 14 point zero zero and so you just solve for pH so pH is equal to 14 minus 2.5 2 which is 11.4 H so finally we've calculated the pH of our solution of ammonia