[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:00.00,0:00:01.09,Default,,0000,0000,0000,,
Dialogue: 0,0:00:01.09,0:00:04.30,Default,,0000,0000,0000,,We've dealt with the weak acid,\Nso let's try an example
Dialogue: 0,0:00:04.30,0:00:05.34,Default,,0000,0000,0000,,with the weak base.
Dialogue: 0,0:00:05.34,0:00:06.59,Default,,0000,0000,0000,,Let's say we had ammonia.
Dialogue: 0,0:00:06.59,0:00:09.73,Default,,0000,0000,0000,,
Dialogue: 0,0:00:09.73,0:00:12.35,Default,,0000,0000,0000,,That's nitrogen with\Nthree hydrogens.
Dialogue: 0,0:00:12.35,0:00:16.37,Default,,0000,0000,0000,,And it's a weak base because\Nit likes to accept hydrogen
Dialogue: 0,0:00:16.37,0:00:18.75,Default,,0000,0000,0000,,from water, leaving the water\Nwith just a hydroxide.
Dialogue: 0,0:00:18.75,0:00:21.19,Default,,0000,0000,0000,,So it increases the hydroxide\Nconcentration.
Dialogue: 0,0:00:21.19,0:00:23.94,Default,,0000,0000,0000,,So if you have some ammonia\Nin an aqueous
Dialogue: 0,0:00:23.94,0:00:28.15,Default,,0000,0000,0000,,solution, plus water.
Dialogue: 0,0:00:28.15,0:00:29.43,Default,,0000,0000,0000,,I'll throw the water in there.
Dialogue: 0,0:00:29.43,0:00:35.47,Default,,0000,0000,0000,,Plus water in an aqueous\Nsolution.
Dialogue: 0,0:00:35.47,0:00:36.94,Default,,0000,0000,0000,,It's a weak base.
Dialogue: 0,0:00:36.94,0:00:40.53,Default,,0000,0000,0000,,So this reaction doesn't go\Nin just one direction.
Dialogue: 0,0:00:40.53,0:00:41.89,Default,,0000,0000,0000,,It's an equilibrium reaction.
Dialogue: 0,0:00:41.89,0:00:45.67,Default,,0000,0000,0000,,
Dialogue: 0,0:00:45.67,0:00:49.98,Default,,0000,0000,0000,,And since this is a weak base,\Nit-- and this is where the
Dialogue: 0,0:00:49.98,0:00:53.14,Default,,0000,0000,0000,,Bronsted-Lowry definition\Nreally kind of pops out.
Dialogue: 0,0:00:53.14,0:00:57.04,Default,,0000,0000,0000,,Is that it's a proton acceptor\Ninstead of a donor.
Dialogue: 0,0:00:57.04,0:01:03.96,Default,,0000,0000,0000,,So it turns into ammonium,\Nor an ammonia cation.
Dialogue: 0,0:01:03.96,0:01:06.97,Default,,0000,0000,0000,,Ammonium has another hydrogen\Non it, so now
Dialogue: 0,0:01:06.97,0:01:08.92,Default,,0000,0000,0000,,it has another proton.
Dialogue: 0,0:01:08.92,0:01:11.20,Default,,0000,0000,0000,,So it's the plus charge.
Dialogue: 0,0:01:11.20,0:01:12.45,Default,,0000,0000,0000,,An it's an aqueous.
Dialogue: 0,0:01:12.45,0:01:15.25,Default,,0000,0000,0000,,And it took that hydrogen\Nfrom the water.
Dialogue: 0,0:01:15.25,0:01:20.05,Default,,0000,0000,0000,,So plus OH minus aqueous.
Dialogue: 0,0:01:20.05,0:01:22.43,Default,,0000,0000,0000,,And remember, if you look at\Nit from the Bronsted-Lowry
Dialogue: 0,0:01:22.43,0:01:25.11,Default,,0000,0000,0000,,definition , it was\Na proton acceptor.
Dialogue: 0,0:01:25.11,0:01:26.60,Default,,0000,0000,0000,,So that made it a base.
Dialogue: 0,0:01:26.60,0:01:28.21,Default,,0000,0000,0000,,Or if you look at the Arrhenius\Ndefinition, it
Dialogue: 0,0:01:28.21,0:01:32.01,Default,,0000,0000,0000,,increased the concentration of\NOH in the solution, so that
Dialogue: 0,0:01:32.01,0:01:33.86,Default,,0000,0000,0000,,makes it an Arrhenius base.
Dialogue: 0,0:01:33.86,0:01:37.06,Default,,0000,0000,0000,,But anyway, given that\Nwe have-- let me
Dialogue: 0,0:01:37.06,0:01:37.97,Default,,0000,0000,0000,,pick a random number.
Dialogue: 0,0:01:37.97,0:01:48.89,Default,,0000,0000,0000,,Let's say we have 0.2\Nmolar of NH3.
Dialogue: 0,0:01:48.89,0:01:51.95,Default,,0000,0000,0000,,What is going to be the pH?
Dialogue: 0,0:01:51.95,0:01:55.71,Default,,0000,0000,0000,,So what's going to be our pH of\Nthe solution, considering
Dialogue: 0,0:01:55.71,0:01:58.26,Default,,0000,0000,0000,,that it's 0.2 molar of NH3.
Dialogue: 0,0:01:58.26,0:01:59.35,Default,,0000,0000,0000,,So the first thing\Nwe need to do.
Dialogue: 0,0:01:59.35,0:02:01.93,Default,,0000,0000,0000,,We need to figure out the\Nequilibrium constant for this
Dialogue: 0,0:02:01.93,0:02:03.22,Default,,0000,0000,0000,,base reaction.
Dialogue: 0,0:02:03.22,0:02:10.82,Default,,0000,0000,0000,,And I just went to Wikipedia--\NI wanted to say liquidpedia,
Dialogue: 0,0:02:10.82,0:02:13.37,Default,,0000,0000,0000,,I'm talking about\Nliquids so much.
Dialogue: 0,0:02:13.37,0:02:14.13,Default,,0000,0000,0000,,And equilibrium.
Dialogue: 0,0:02:14.13,0:02:15.17,Default,,0000,0000,0000,,Equipedia.
Dialogue: 0,0:02:15.17,0:02:19.07,Default,,0000,0000,0000,,But I went to Wikipedia, and\Nthey have a little chart for
Dialogue: 0,0:02:19.07,0:02:20.63,Default,,0000,0000,0000,,almost any compound\Nyou look for.
Dialogue: 0,0:02:20.63,0:02:21.88,Default,,0000,0000,0000,,And they give you pKb.
Dialogue: 0,0:02:21.88,0:02:26.12,Default,,0000,0000,0000,,
Dialogue: 0,0:02:26.12,0:02:27.76,Default,,0000,0000,0000,,Which is, you see\Nthat p there.
Dialogue: 0,0:02:27.76,0:02:32.09,Default,,0000,0000,0000,,That just means it's the\Nminus log base 10 of
Dialogue: 0,0:02:32.09,0:02:33.56,Default,,0000,0000,0000,,the equilibrium constant.
Dialogue: 0,0:02:33.56,0:02:38.87,Default,,0000,0000,0000,,
Dialogue: 0,0:02:38.87,0:02:42.70,Default,,0000,0000,0000,,And they give that\Nas being 4.75.
Dialogue: 0,0:02:42.70,0:02:45.33,Default,,0000,0000,0000,,So we can just do a little bit\Nof math here to solve for the
Dialogue: 0,0:02:45.33,0:02:47.00,Default,,0000,0000,0000,,equilibrium constant.
Dialogue: 0,0:02:47.00,0:02:47.62,Default,,0000,0000,0000,,So let's see.
Dialogue: 0,0:02:47.62,0:02:53.22,Default,,0000,0000,0000,,If we multiply both sides by\Nnegative, you get log base 10
Dialogue: 0,0:02:53.22,0:02:57.46,Default,,0000,0000,0000,,of our equilibrium constant\Nfor this base reaction.
Dialogue: 0,0:02:57.46,0:02:58.57,Default,,0000,0000,0000,,That's why the b is there.
Dialogue: 0,0:02:58.57,0:03:03.03,Default,,0000,0000,0000,,Is equal to minus 4.75,\Nor 10 to the minus
Dialogue: 0,0:03:03.03,0:03:05.40,Default,,0000,0000,0000,,4.75 should be Kb.
Dialogue: 0,0:03:05.40,0:03:11.54,Default,,0000,0000,0000,,So Kb is equal to 10\Nto the minus 4.75.
Dialogue: 0,0:03:11.54,0:03:14.04,Default,,0000,0000,0000,,That's not an easy exponent to\Nfigure out in your head, so
Dialogue: 0,0:03:14.04,0:03:16.53,Default,,0000,0000,0000,,I'll bring out the calculator\Nfor that.
Dialogue: 0,0:03:16.53,0:03:27.85,Default,,0000,0000,0000,,So if we take 10 to the 4.75\Nminus, it equals, let's just
Dialogue: 0,0:03:27.85,0:03:31.46,Default,,0000,0000,0000,,say 1.8 times 10 to\Nthe negative 5.
Dialogue: 0,0:03:31.46,0:03:37.32,Default,,0000,0000,0000,,This is equal to 1.8 times\N10 to the minus 5.
Dialogue: 0,0:03:37.32,0:03:40.09,Default,,0000,0000,0000,,So now we can use this\Ninformation and we can do a
Dialogue: 0,0:03:40.09,0:03:42.26,Default,,0000,0000,0000,,mathematical thing very\Nsimilar to we
Dialogue: 0,0:03:42.26,0:03:45.01,Default,,0000,0000,0000,,did in the last video.
Dialogue: 0,0:03:45.01,0:03:46.48,Default,,0000,0000,0000,,It's going to be hard to\Nfigure out the hydrogen
Dialogue: 0,0:03:46.48,0:03:47.76,Default,,0000,0000,0000,,concentration directly, right?
Dialogue: 0,0:03:47.76,0:03:50.52,Default,,0000,0000,0000,,Because our equilibrium reaction\Nonly has hydroxide.
Dialogue: 0,0:03:50.52,0:03:53.50,Default,,0000,0000,0000,,But if we know the hydroxide\Nconcentration, then we can
Dialogue: 0,0:03:53.50,0:03:55.77,Default,,0000,0000,0000,,back into the hydrogen\Nconcentration, knowing that
Dialogue: 0,0:03:55.77,0:04:01.51,Default,,0000,0000,0000,,this plus the hydrogen\Nconcentration has to equal 10
Dialogue: 0,0:04:01.51,0:04:02.44,Default,,0000,0000,0000,,to the minus 14.
Dialogue: 0,0:04:02.44,0:04:06.62,Default,,0000,0000,0000,,Or if you figure out the pOH,\Nthat plus the pH has to be 14.
Dialogue: 0,0:04:06.62,0:04:09.00,Default,,0000,0000,0000,,And we did that a couple\Nof videos ago.
Dialogue: 0,0:04:09.00,0:04:14.06,Default,,0000,0000,0000,,So this equilibrium constant\Nor this formula
Dialogue: 0,0:04:14.06,0:04:14.88,Default,,0000,0000,0000,,would look like this.
Dialogue: 0,0:04:14.88,0:04:24.94,Default,,0000,0000,0000,,1.8 times 10 to the minus 5\Nwill be equal to-- in the
Dialogue: 0,0:04:24.94,0:04:27.88,Default,,0000,0000,0000,,denominator, we have our\Nconcentration of reactants.
Dialogue: 0,0:04:27.88,0:04:30.06,Default,,0000,0000,0000,,And remember, you don't\Ninclude the solvent.
Dialogue: 0,0:04:30.06,0:04:31.98,Default,,0000,0000,0000,,So you only include the NH3.
Dialogue: 0,0:04:31.98,0:04:35.48,Default,,0000,0000,0000,,We have 0.2 molars is what we\Nput in, but some of it, let's
Dialogue: 0,0:04:35.48,0:04:37.98,Default,,0000,0000,0000,,say X of it, is going to be\Nconverted into this stuff on
Dialogue: 0,0:04:37.98,0:04:39.67,Default,,0000,0000,0000,,the right-hand side.
Dialogue: 0,0:04:39.67,0:04:43.38,Default,,0000,0000,0000,,So in the denominator, we're\Ngoing to have 0.2 minus
Dialogue: 0,0:04:43.38,0:04:45.87,Default,,0000,0000,0000,,whatever gets converted into\Nthe right-hand side.
Dialogue: 0,0:04:45.87,0:04:49.51,Default,,0000,0000,0000,,And so then in the right-hand\Nside, we're going to have x of
Dialogue: 0,0:04:49.51,0:04:51.47,Default,,0000,0000,0000,,NH4 and x of OH.
Dialogue: 0,0:04:51.47,0:04:54.30,Default,,0000,0000,0000,,
Dialogue: 0,0:04:54.30,0:04:58.14,Default,,0000,0000,0000,,This is the concentration\Nof ammonia.
Dialogue: 0,0:04:58.14,0:05:00.30,Default,,0000,0000,0000,,And then we have x times x.
Dialogue: 0,0:05:00.30,0:05:07.83,Default,,0000,0000,0000,,This is the concentration of\NNH4 plus-- that's a 4.
Dialogue: 0,0:05:07.83,0:05:10.32,Default,,0000,0000,0000,,And then this is the\Nconcentration,
Dialogue: 0,0:05:10.32,0:05:15.33,Default,,0000,0000,0000,,right here, of OH minus.
Dialogue: 0,0:05:15.33,0:05:15.87,Default,,0000,0000,0000,,Right?
Dialogue: 0,0:05:15.87,0:05:17.86,Default,,0000,0000,0000,,And we just solve for x.
Dialogue: 0,0:05:17.86,0:05:19.85,Default,,0000,0000,0000,,Let's do that.
Dialogue: 0,0:05:19.85,0:05:21.08,Default,,0000,0000,0000,,Solve for x.
Dialogue: 0,0:05:21.08,0:05:23.39,Default,,0000,0000,0000,,And once we have x, we know\Nthe concentration of OH.
Dialogue: 0,0:05:23.39,0:05:25.31,Default,,0000,0000,0000,,We'll be able to figure out\Nthe pOH, and then we'll be
Dialogue: 0,0:05:25.31,0:05:27.90,Default,,0000,0000,0000,,able to figure out the pH.
Dialogue: 0,0:05:27.90,0:05:28.75,Default,,0000,0000,0000,,OK.
Dialogue: 0,0:05:28.75,0:05:32.30,Default,,0000,0000,0000,,Multiply this times both\Nsides of this equation.
Dialogue: 0,0:05:32.30,0:05:35.06,Default,,0000,0000,0000,,And just so you know, that same\Nsimplification step that
Dialogue: 0,0:05:35.06,0:05:36.38,Default,,0000,0000,0000,,we did in the previous thing.
Dialogue: 0,0:05:36.38,0:05:44.26,Default,,0000,0000,0000,,When this is several orders of\Nmagnitude smaller than this
Dialogue: 0,0:05:44.26,0:05:49.83,Default,,0000,0000,0000,,number right here-- I want to\Ngive you-- heuristics are just
Dialogue: 0,0:05:49.83,0:05:51.29,Default,,0000,0000,0000,,kind of rules of thumb\Nthat sometimes work.
Dialogue: 0,0:05:51.29,0:05:53.17,Default,,0000,0000,0000,,Let's just do the quadratic\Nequation.
Dialogue: 0,0:05:53.17,0:05:56.15,Default,,0000,0000,0000,,But you can kind of think about\Nsometimes when you can
Dialogue: 0,0:05:56.15,0:05:57.00,Default,,0000,0000,0000,,get rid of that middle term.
Dialogue: 0,0:05:57.00,0:05:57.76,Default,,0000,0000,0000,,But let's just multiply it.
Dialogue: 0,0:05:57.76,0:06:03.45,Default,,0000,0000,0000,,0.2 two times 1.8 is 0.36.
Dialogue: 0,0:06:03.45,0:06:07.98,Default,,0000,0000,0000,,0.36 times 10 to the\Nminus 5, right?
Dialogue: 0,0:06:07.98,0:06:11.65,Default,,0000,0000,0000,,2 times 1.8 would be\N3.6, this is 0.36.
Dialogue: 0,0:06:11.65,0:06:20.66,Default,,0000,0000,0000,,Minus 1.8 times 10 to the\Nminus 5 x, right?
Dialogue: 0,0:06:20.66,0:06:22.94,Default,,0000,0000,0000,,Is equal to that.
Dialogue: 0,0:06:22.94,0:06:24.90,Default,,0000,0000,0000,,x squared.
Dialogue: 0,0:06:24.90,0:06:29.26,Default,,0000,0000,0000,,Let's put everything on the\Nsame side of the equation.
Dialogue: 0,0:06:29.26,0:06:31.22,Default,,0000,0000,0000,,I'm going to move all of these\Nthe right-hand side, so you
Dialogue: 0,0:06:31.22,0:06:35.09,Default,,0000,0000,0000,,get 0 is equal to x squared.
Dialogue: 0,0:06:35.09,0:06:37.99,Default,,0000,0000,0000,,Add this to both sides\Nof the equation.
Dialogue: 0,0:06:37.99,0:06:44.52,Default,,0000,0000,0000,,Plus 1.8 times 10 to\Nthe minus 5 x.
Dialogue: 0,0:06:44.52,0:06:48.87,Default,,0000,0000,0000,,1.8 times 10 to the minus 5.
Dialogue: 0,0:06:48.87,0:06:50.94,Default,,0000,0000,0000,,Just so you can see the\Ncoefficients separate from the
Dialogue: 0,0:06:50.94,0:06:53.01,Default,,0000,0000,0000,,x terms.
Dialogue: 0,0:06:53.01,0:06:59.86,Default,,0000,0000,0000,,Minus 0.36 times 10\Nto the minus 5.
Dialogue: 0,0:06:59.86,0:07:00.99,Default,,0000,0000,0000,,So let's solve this.
Dialogue: 0,0:07:00.99,0:07:04.61,Default,,0000,0000,0000,,And once again, if you wanted\Nto kind of do it, you could
Dialogue: 0,0:07:04.61,0:07:06.72,Default,,0000,0000,0000,,eliminate this term and then\Njust figure out the straight
Dialogue: 0,0:07:06.72,0:07:07.54,Default,,0000,0000,0000,,up square root.
Dialogue: 0,0:07:07.54,0:07:08.36,Default,,0000,0000,0000,,But we won't do that.
Dialogue: 0,0:07:08.36,0:07:09.89,Default,,0000,0000,0000,,We'll actually use a\Nquadratic equation.
Dialogue: 0,0:07:09.89,0:07:12.61,Default,,0000,0000,0000,,So a is 1.
Dialogue: 0,0:07:12.61,0:07:13.95,Default,,0000,0000,0000,,b is this.
Dialogue: 0,0:07:13.95,0:07:14.53,Default,,0000,0000,0000,,That's b.
Dialogue: 0,0:07:14.53,0:07:15.20,Default,,0000,0000,0000,,And this is c.
Dialogue: 0,0:07:15.20,0:07:17.34,Default,,0000,0000,0000,,And you just supply than in\Nthe quadratic equation.
Dialogue: 0,0:07:17.34,0:07:20.50,Default,,0000,0000,0000,,So you get minus b.
Dialogue: 0,0:07:20.50,0:07:26.23,Default,,0000,0000,0000,,So you minus 1.8 times 10\Nto the minus 5 power.
Dialogue: 0,0:07:26.23,0:07:27.06,Default,,0000,0000,0000,,Plus or minus.
Dialogue: 0,0:07:27.06,0:07:28.79,Default,,0000,0000,0000,,We'll only have to do the plus\Nbecause if we do the minus,
Dialogue: 0,0:07:28.79,0:07:30.25,Default,,0000,0000,0000,,we'll end up with a negative\Nconcentration.
Dialogue: 0,0:07:30.25,0:07:34.44,Default,,0000,0000,0000,,So plus, the square root-- we\Nhave to do a lot of math
Dialogue: 0,0:07:34.44,0:07:36.97,Default,,0000,0000,0000,,here-- b squared.
Dialogue: 0,0:07:36.97,0:07:39.22,Default,,0000,0000,0000,,So it's 1.8 times 10\Nto the negative 5.
Dialogue: 0,0:07:39.22,0:07:41.55,Default,,0000,0000,0000,,So it's 1.8.
Dialogue: 0,0:07:41.55,0:07:45.21,Default,,0000,0000,0000,,If you square it, it's 3.24.
Dialogue: 0,0:07:45.21,0:07:51.15,Default,,0000,0000,0000,,So it's 3.24 times-- if you\Nsquare 10 to the minus 5-- 10
Dialogue: 0,0:07:51.15,0:07:57.63,Default,,0000,0000,0000,,to the minus 10 minus 4\Ntimes a, which is 1,
Dialogue: 0,0:07:57.63,0:07:59.38,Default,,0000,0000,0000,,times c, which is minus.
Dialogue: 0,0:07:59.38,0:08:04.86,Default,,0000,0000,0000,,So it's 4 times-- the minuses\Ncancel out-- times 0.36 times
Dialogue: 0,0:08:04.86,0:08:08.22,Default,,0000,0000,0000,,10 to the minus 5.
Dialogue: 0,0:08:08.22,0:08:20.14,Default,,0000,0000,0000,,Which is 4 times 0.36\Nis equal to 1.44.
Dialogue: 0,0:08:20.14,0:08:21.44,Default,,0000,0000,0000,,I should have been able\Nto do that in my head.
Dialogue: 0,0:08:21.44,0:08:25.44,Default,,0000,0000,0000,,Now you have 1.44 e minus 5.
Dialogue: 0,0:08:25.44,0:08:30.12,Default,,0000,0000,0000,,Times 10 to-- let\Nme write that.
Dialogue: 0,0:08:30.12,0:08:32.55,Default,,0000,0000,0000,,So this is 1.44.
Dialogue: 0,0:08:32.55,0:08:36.26,Default,,0000,0000,0000,,And of course all of\Nthis is over 2a.
Dialogue: 0,0:08:36.26,0:08:37.39,Default,,0000,0000,0000,,So let's see.
Dialogue: 0,0:08:37.39,0:08:38.79,Default,,0000,0000,0000,,This is my x value.
Dialogue: 0,0:08:38.79,0:08:41.86,Default,,0000,0000,0000,,My concentration of OH.
Dialogue: 0,0:08:41.86,0:08:42.35,Default,,0000,0000,0000,,So let's see.
Dialogue: 0,0:08:42.35,0:08:51.38,Default,,0000,0000,0000,,I have 3.24 times 10\Nto the minus 10.
Dialogue: 0,0:08:51.38,0:08:53.02,Default,,0000,0000,0000,,That's that.
Dialogue: 0,0:08:53.02,0:09:03.87,Default,,0000,0000,0000,,Plus 1.44 times 10 to the minus\N5 is equal to that.
Dialogue: 0,0:09:03.87,0:09:05.53,Default,,0000,0000,0000,,So that's this whole thing\Nunder the radical.
Dialogue: 0,0:09:05.53,0:09:08.18,Default,,0000,0000,0000,,And I want to take the\Nsquare root of that.
Dialogue: 0,0:09:08.18,0:09:12.52,Default,,0000,0000,0000,,And so that is to\Nthe 0.5 power.
Dialogue: 0,0:09:12.52,0:09:16.68,Default,,0000,0000,0000,,So I get 0.00379.
Dialogue: 0,0:09:16.68,0:09:19.59,Default,,0000,0000,0000,,So I'll switch colors.
Dialogue: 0,0:09:19.59,0:09:25.82,Default,,0000,0000,0000,,So I get x is equal to a minus\N1.8 times 10 to the minus 5
Dialogue: 0,0:09:25.82,0:09:33.32,Default,,0000,0000,0000,,plus 0.003794.
Dialogue: 0,0:09:33.32,0:09:35.66,Default,,0000,0000,0000,,All of that over 2.
Dialogue: 0,0:09:35.66,0:09:36.33,Default,,0000,0000,0000,,Do the math.
Dialogue: 0,0:09:36.33,0:09:42.66,Default,,0000,0000,0000,,So to that I'm going to subtract\Nminus this point
Dialogue: 0,0:09:42.66,0:09:43.06,Default,,0000,0000,0000,,right here.
Dialogue: 0,0:09:43.06,0:09:43.75,Default,,0000,0000,0000,,I have this value.
Dialogue: 0,0:09:43.75,0:09:44.99,Default,,0000,0000,0000,,I'm just subtracting this.
Dialogue: 0,0:09:44.99,0:09:53.77,Default,,0000,0000,0000,,Minus 1.8 e 5 negative\Nis equal to that.
Dialogue: 0,0:09:53.77,0:09:54.75,Default,,0000,0000,0000,,This is the whole numerator.
Dialogue: 0,0:09:54.75,0:09:57.47,Default,,0000,0000,0000,,And now I need to just\Ndivide it by 2.
Dialogue: 0,0:09:57.47,0:10:03.68,Default,,0000,0000,0000,,Divided by 2 is equal\Nto 0.001.
Dialogue: 0,0:10:03.68,0:10:04.43,Default,,0000,0000,0000,,Let me write that.
Dialogue: 0,0:10:04.43,0:10:05.91,Default,,0000,0000,0000,,So x.
Dialogue: 0,0:10:05.91,0:10:08.14,Default,,0000,0000,0000,,So I'll switch colors\Narbitrarily again.
Dialogue: 0,0:10:08.14,0:10:18.48,Default,,0000,0000,0000,,x is equal to 0.001888-- I mean,\Nthen there's a 3 and so
Dialogue: 0,0:10:18.48,0:10:19.69,Default,,0000,0000,0000,,forth and so on.
Dialogue: 0,0:10:19.69,0:10:21.74,Default,,0000,0000,0000,,But if you remember from\Nour original equation.
Dialogue: 0,0:10:21.74,0:10:22.49,Default,,0000,0000,0000,,What was x?
Dialogue: 0,0:10:22.49,0:10:26.76,Default,,0000,0000,0000,,It was what's both the ammonium\Nconcentration and the
Dialogue: 0,0:10:26.76,0:10:27.86,Default,,0000,0000,0000,,hydroxide concentration.
Dialogue: 0,0:10:27.86,0:10:30.53,Default,,0000,0000,0000,,We care about the hydroxide\Nconcentration.
Dialogue: 0,0:10:30.53,0:10:36.06,Default,,0000,0000,0000,,So this is equal to my\Nconcentration of hydroxide.
Dialogue: 0,0:10:36.06,0:10:40.58,Default,,0000,0000,0000,,Now if I want to figure out my\NpOH, I just take the minus log
Dialogue: 0,0:10:40.58,0:10:44.61,Default,,0000,0000,0000,,of this number, which\Nis equal to--
Dialogue: 0,0:10:44.61,0:10:46.83,Default,,0000,0000,0000,,So let's just take\Nthe log of it.
Dialogue: 0,0:10:46.83,0:10:49.18,Default,,0000,0000,0000,,The log is that, and then I\Ntake the minus of that.
Dialogue: 0,0:10:49.18,0:10:55.18,Default,,0000,0000,0000,,So it's 2.72.
Dialogue: 0,0:10:55.18,0:11:00.42,Default,,0000,0000,0000,,And now if we want to figure out\Nthe pH, my concentration
Dialogue: 0,0:11:00.42,0:11:03.37,Default,,0000,0000,0000,,of hydrogen ions-- just\Nremember, when you're in an
Dialogue: 0,0:11:03.37,0:11:10.74,Default,,0000,0000,0000,,aqueous solution at 25 degrees\NCelsius, your pK of water is
Dialogue: 0,0:11:10.74,0:11:15.87,Default,,0000,0000,0000,,equal to your pOH\Nplus your pH.
Dialogue: 0,0:11:15.87,0:11:19.73,Default,,0000,0000,0000,,This at 25 degrees is 14.
Dialogue: 0,0:11:19.73,0:11:22.82,Default,,0000,0000,0000,,Because you have 10 to the minus\N14 molar concentration--
Dialogue: 0,0:11:22.82,0:11:24.58,Default,,0000,0000,0000,,well no, actually, I don't\Nwant to go into that.
Dialogue: 0,0:11:24.58,0:11:26.72,Default,,0000,0000,0000,,You have 10 to the minus\N7 of each of these.
Dialogue: 0,0:11:26.72,0:11:28.51,Default,,0000,0000,0000,,But anyway, this is equilibrium\Nconstant for the
Dialogue: 0,0:11:28.51,0:11:30.31,Default,,0000,0000,0000,,disassociation of water.
Dialogue: 0,0:11:30.31,0:11:37.17,Default,,0000,0000,0000,,This, when water's neutral is 7\Nor a concentration of OH of
Dialogue: 0,0:11:37.17,0:11:38.28,Default,,0000,0000,0000,,10 to the minus 7.
Dialogue: 0,0:11:38.28,0:11:39.76,Default,,0000,0000,0000,,We can take the minus\Nlog, this becomes 7.
Dialogue: 0,0:11:39.76,0:11:44.07,Default,,0000,0000,0000,,But now we know we have a much\Nhigher concentration of OH.
Dialogue: 0,0:11:44.07,0:11:45.08,Default,,0000,0000,0000,,2.72.
Dialogue: 0,0:11:45.08,0:11:47.55,Default,,0000,0000,0000,,Remember, that minus log\Nkind of flips it.
Dialogue: 0,0:11:47.55,0:11:52.67,Default,,0000,0000,0000,,So a lower pOH means a higher\Nconcentration of pOH.
Dialogue: 0,0:11:52.67,0:11:53.23,Default,,0000,0000,0000,,Right?
Dialogue: 0,0:11:53.23,0:11:57.20,Default,,0000,0000,0000,,And a lower pOH, if this\Nis lower, right?
Dialogue: 0,0:11:57.20,0:11:58.48,Default,,0000,0000,0000,,This is a lower pOH.
Dialogue: 0,0:11:58.48,0:12:00.29,Default,,0000,0000,0000,,That means your pH is higher.
Dialogue: 0,0:12:00.29,0:12:03.57,Default,,0000,0000,0000,,
Dialogue: 0,0:12:03.57,0:12:04.87,Default,,0000,0000,0000,,So what is your pH\Ngoing to be?
Dialogue: 0,0:12:04.87,0:12:15.02,Default,,0000,0000,0000,,So your pH is going to be\Nequal to 14 minus 2.72.
Dialogue: 0,0:12:15.02,0:12:21.71,Default,,0000,0000,0000,,So let me do the minus\Nplus 14 is equal to--
Dialogue: 0,0:12:21.71,0:12:23.57,Default,,0000,0000,0000,,let's just say 11.3.
Dialogue: 0,0:12:23.57,0:12:26.65,Default,,0000,0000,0000,,So your pH is equal to 11.3.
Dialogue: 0,0:12:26.65,0:12:30.85,Default,,0000,0000,0000,,Which makes sense, because we\Nsaid this was a weak base.
Dialogue: 0,0:12:30.85,0:12:33.68,Default,,0000,0000,0000,,Ammonia is a weak base.
Dialogue: 0,0:12:33.68,0:12:35.07,Default,,0000,0000,0000,,So it's basic.
Dialogue: 0,0:12:35.07,0:12:39.12,Default,,0000,0000,0000,,So it should increase your\NpH above the neutral 7.
Dialogue: 0,0:12:39.12,0:12:42.80,Default,,0000,0000,0000,,So the pH should be greater than\N7, but as you compare it
Dialogue: 0,0:12:42.80,0:12:45.28,Default,,0000,0000,0000,,to some of the strong bases\Nbefore that took our pH when
Dialogue: 0,0:12:45.28,0:12:49.41,Default,,0000,0000,0000,,you added a molar to 14, this\Ntook our pH-- although we only
Dialogue: 0,0:12:49.41,0:12:54.01,Default,,0000,0000,0000,,did add 0.2 molar\Nof it to 11.3.
Dialogue: 0,0:12:54.01,0:12:56.97,Default,,0000,0000,0000,,Anyway, this is more of a math\Nproblem than chemistry, but
Dialogue: 0,0:12:56.97,0:12:59.80,Default,,0000,0000,0000,,hopefully it clarified\Na few things as well.