Determine the oh ph and poh of a solution with a h of 36106...

Determine the [OH-], pH, and pOH of a solution with a [H+] of 3.6 × 10^-6 M at 25 °C. [OH-] = M, pH = , pOH = Determine the [H+], pH, and pOH of a solution with an [OH-] of 5.0 × 10^-12 M at 25 °C. [H+] = M, pH = , pOH = Determine the [H+], [OH-], and pOH of a solution with a pH of 13.06 at 25 °C. [H+] = M, [OH-] = M, pOH = Determine the [H+], [OH-], and pH of a solution with a pOH of 6.62 at 25 °C. [H+] = M, [OH-] = M, pH =

Transcript

0:00 All right.

00:01 So we're given four separate equations, and in each one we're given either the h plus concentration, the oh concentration, the ph, the ph, and so i have found it easiest to kind of sort this into a table where i included the start values for each associated thing, where this is the h plus, this is the oh minus, this is the ph, and this is the p -o -h here.

00:26 And so that way i can kind of easily solve each one as we go.

00:31 And so the first one we're given an h plus conservation of 3 .6 times the n .a .6 and we need to fill out the rest of the chart here.

00:40 So there's lots of different approaches at acid bases, lots of different equations that can kind of, you can choose one and get to the same point as if you chose another one.

00:50 So i'm going to try to do a broad sampling of all possibilities here, but i may not hit them all.

00:56 But starting at 3 .6 times 10 .6 h plus concentration.

01:00 Well, the kw of water is equal to, or rather, water has a constant of 1 times 10 and negative 14.

01:14 And that's going to equal the h plus concentration times the oh concentration.

01:21 All right.

01:22 So if we have one, we can simplify on the other by dividing 1 times 10 to negative 14 by that number.

01:28 So dividing 1 times 10 to the 814, divided by 3 .6 times 10 and a 9 would give us our o -h concentration.

01:36 And what we find is our age concentration is 2 .78 times 10.

01:41 Oh wait, we're down to 2 -6 -fakes, so 2 .8 times 10 to the negative 9.

01:49 And so now we can solve for the ph and the ph.

01:52 So the ph is this negative log, rh plus concentration.

01:57 Nalog 3 .6 times 10 to the negative 6.

02:00 And that gives us a ph of 5 .4.

02:04 Yeah, 2 sig figs, 5 .44.

02:06 The first before the decimal doesn't count with ph units only after the decimal count.

02:12 So 5 .44 is our 2 -sig ph.

02:17 And then with p -o -h, we can solve for by taking negative log our o -h concentration at 2 .8 times 10 to the and that would give us a value of 8 .56.

02:32 Alright, so we can do a quick double check.

02:35 If you're to add this number to this number, we should get 14, and that is true.

02:42 That's what we get.

02:44 So this is our first example here.

02:50 So now what happens if we're given the oh concentration first? well, we can still use the same equation, except now we'll divide by oh, instead of solving for oh.

03:01 So we'll have 1 times 10 to the negative 14, and that's a constant.

03:06 It holds true for any ph question you may need to solve for.

03:14 And we divide by 5 .0 times 10 and negative 12, and that comes out to 2 .0 times 10 to negative 3.

03:26 So 2 .0 times 10 to negative 3 is our h -plus concentration...

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