Experiment 3: DETERMINATION OF THE DISSOCIATION CONSTANT AND...

Experiment 3: DETERMINATION OF THE DISSOCIATION CONSTANT AND THE CONCENTRATION OF A WEAK ACID REPORT SHEET mL NaOH | pH 0.0 | 2.92 2.0 | 3.51 4.0 | 3.84 6.0 | 4.12 8.0 | 4.32 10.0 | 4.49 11.0 | 4.53 12.0 | 4.61 13.0 | 4.68 14.0 | 4.76 15.0 | 4.81 16.0 | 4.85 17.0 | 4.91 18.0 | 4.98 19.0 | 5.07 mL NaOH | pH 20.0 | 5.15 20.5 | 5.22 21.0 | 5.24 21.5 | 5.30 22.0 | 5.34 22.5 | 5.40 23.0 | 5.44 23.5 | 5.52 24.0 | 5.58 24.5 | 5.68 25.0 | 5.81 25.5 | 5.87 26.0 | 6.14 26.5 | 6.49 27.0 | 10.18 mL NaOH | pH 27.5 | 10.80 28.0 | 11.43 28.5 | 11.32 29.0 | 11.70 29.5 | 11.48 30.0 | 11.86 31.0 | 11.68 32.0 | 12.09 33.0 | 11.84 34.0 | 12.22 35.0 | 11.74 36.0 | 12.32 37.0 | 12.03 38.0 | 12.39 39.0 | 12.09 Volume of NaOH at equivalence point = ________ mL Volume of NaOH at one half equivalence point = ________ mL pKa = ________ , Ka = ________ MHB = (VNaOH x MNaOH) / VHB = ________________________________ M

Transcript

00:01 Okay, so for this problem you are trying to determine some information.

00:12 You're trying to find a dissociation constant, trying to find the molarity of your acid.

00:16 The only thing that we are missing that i can't seem to locate is the initial concentration of your sodium hydroxide, which we would need in order to do that final calculation, but i can show you everything you need for that to get to that point.

00:31 So looking at your data, you want to look where you have a big jump in your ph.

00:36 And so that big jump happens between 26 .5 and 27 milliliters.

00:41 So your volume for sodium hydroxide at the equivalence point is going to be at 27 .0 milliliters, when you had that big ginormous jump.

00:53 The volume for the half equivalence point, so volume for half equivalence point, is going to be half of that, so that's at 13 .5.

01:03 So if you look at that, 13 .5, it looks like your ph was between 4 .68 and 4 .76.

01:15 So let's just find the halfway point of that.

01:18 4 .76 plus 4 .68, if i'm reading that correctly, divided by 2, would be 4 .72.

01:30 So at the half equivalence point, your ph equals 4 .72 at half equivalence point.

01:43 Now one thing to note is that if you have a weak acid and a strong base, in this case we have the strong base sodium hydroxide, and they tell us that we're titrating a weak acid, the ph at the half equivalence point equals the pka.

01:58 So pka will also equal 4 .72.

02:02 So then your ka will be 10 to the negative 4 .72.

02:08 So we just have to unlog that, we have to exponentiate is the real word.

02:13 So second log negative 4 .72, and it's 1 .91 times 10 to the negative fifth.

02:20 1 .91 times 10 to the negative fifth is our ka.

02:27 So here's our pka, here's our ka, here's our volume at the half equivalence point, here's our volume at the equivalent.

02:34 Now for this it says the molarity of your acid equals the volume of your base times the molarity of your base.

02:45 Well we don't know the volume of your base.

02:49 I also don't know the volume of the acid that you are using.

02:53 But in this case to find the molarity of your acid, the volume of your naoh is going to be this 27 .0 milliliters, and then you are going to take that times the molarity of your base, molarity of your sodium hydroxide.

03:11 You should have known the concentration, it was given to you somewhere in your lab experiment.

03:17 Or it should have been labeled on the containers as well.

03:20 And then divided by the volume of that acid that you were titrating to start with.

03:26 So i don't know if you put in 10 milliliters or 15 milliliters or how much acid.

03:32 So your volume of your acid.

03:36 So that's how you will find your molarity of your acid for this one.

03:41 Now you also said make a graph.

03:44 So it says kindly note the following points.

03:47 You didn't tell me which points, but i'm assuming that when you're talking about this, it would be milliliters of your base added.

04:00 Usually it's the equivalence point and half equivalence point starting and ending ph...

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