A determine the ph of the solution that results from the...

A) Determine the pH of the solution that results from the mixing of 60.0 mL of 0.100 M NaOH, 94.0 mL of 0.0500 M KOH, 62.5 mL of 0.075 M HCl, 37.0 mL of 0.065 M HNO3, and 3.00 quarts of distilled water. B) Calculate the pH of a titration of 50.00 mL of 0.100 M Phenylacetic acid, Ka = 4.9 x 10-5, with 0.100 M NaOH at the following points: a. Before any NaOH is added. b. After 18.7 mL of NaOH are added. c. After 25.00 mL of NaOH are added. d. After 50.00 mL of NaOH are added. e. After 53.00 mL of NaOH are added. C) What is the pKa of the acid (show calculation)?

Transcript

00:01 Phenyl acetic acid, which i have abbreviated as hph, with sodium hydroxide, and we are asked to calculate the ph at several different points.

00:09 When you have a sequence of calculations like this that you need to do, some important things to find at the beginning, save yourself time later, we have the ka that's given to us, we need to find the kb, so 1 times 10 to the negative 14th, divided by that ka, 4 .9 times 10 to the negative 5th, will give you your kb, so this would be 2 .04 times 10 to the negative 10th.

00:35 We also need to find our equivalence point and our half equivalence point.

00:39 So to find that, you simply multiply these and set them equal to what these would be multiplied, this would be your volume of your base, right here.

00:48 So 50 times .1 is going to equal 50 times .1, so our equivalence point will be at 50 milliliters.

00:57 That means our half equivalence point is at 25 milliliters.

01:00 Now that's important because we are given those points to find the ph, so you need to do different things at different points.

01:14 So to find our initial ph, it's just the ph of this weak acid, so we will simply use our ka equals our hydrogen ion concentration squared over the concentration of our weak acid.

01:29 So ka is 4 .9 times 10 to the negative 5th, equals our h plus concentration squared, over initial concentration is .100.

01:43 We will solve this for h plus, so 4 .9 times 10 to the negative 5th, times .1, and then the square root of that is going to give you .002214.

02:03 We'll just hang on to four sig figs right now.

02:06 It's probably too many, but we'll just go with it.

02:09 So then to find the ph, ph is simply the negative log of that h plus value, so .002214.

02:19 So log of that is 2 .655, or 2 .66 if you want to round that, or 2 .7 if you want to go that short.

02:36 That's a little bit short.

02:38 Okay, so now once we've added one drop of base up to but not including the equivalence point, we are in the buffer region, the lambda buffer, and you can use one equation for that the whole time.

02:49 That is a variation of the henderson -hasselbalch equation, which is h plus equals your ka times your acid info over your base info.

03:01 Don't ever change that to kb, don't change your b and a around.

03:04 This is what it is.

03:05 Okay, so our ka is 4 .9 times 10 to the negative 5th.

03:12 Okay, so we started out with just acid down in our flask, and we have base up in the burette that we're going to add in.

03:19 So we need to know how many millimoles of acid we have.

03:24 So to find that, we simply take our concentration times our milliliters, and that gives us millimoles.

03:32 So 50 times .1 is going to be 5, so i have 5 millimoles of this.

03:41 I had 0 millimoles of the base in the flask at the start.

03:45 I am adding base, so it gets a plus sign.

03:48 That will use up some of the acid, some of those hydrogen ions.

03:53 So to figure out how much base we're adding, we need millimoles of that.

03:58 So we will take our 18 .7 times .1, and that gives us 1 .87 millimoles.

04:07 So let's go ahead and calculate that.

04:09 5 minus 1 .87 divided by 1 .87, and then we'll take that times our ka, 4 .9 times 10 to the negative 5th, and that gives you an h plus concentration of 8 .20 times 10 to the negative 5th molar.

04:27 It is a concentration.

04:28 Then we take the negative log of that to find our ph, just as we did up above.

04:33 So the negative log of that answer is 4 .09.

04:42 Let's go 86.

04:43 4 .086.

04:44 You can figure out where you want to round this to.

04:47 All right, so we could do the very same thing for this half equivalence point, and we'll just do that really quickly.

04:56 H plus equals 4 .9 times 10 to the negative 5th.

05:02 We still have 5 millimoles.

05:04 Now if we take 25 times .1, we will get 2 .5.

05:10 So we're adding 2 .5 millimoles, subtracting 2 .5 millimoles.

05:14 Hey, look at that.