Name Section Experiment 24 Advance Study Assignment: Molar...

Name Section Experiment 24 Advance Study Assignment: Molar Mass of an Acid 1. 7.0 mL of 6.0 M NaOH are diluted with water to a volume of 400 mL. You are asked to find the molarity of the resulting solution. a. First find out how many moles of NaOH there are in 7.0 mL of 6.0 M NaOH. Use Equation 1. Note that the volume must be in liters. moles b. Since the total number of moles of NaOH is not changed on dilution, the molarity after dilution can also be found by Equation 1, using the final volume of the solution. Calculate that molarity. M 2. In an acid-base titration, 23.91 mL of an NaOH solution are needed to neutralize 24.58 mL of a 0.1002 M HCl solution. To find the molarity of the NaOH solution, we can use the following procedure: a. First note the value of MH+ in the HCl solution. M b. Find MOH- in the NaOH solution. (Use Eq. 3.) M c. Obtain MNaOH from MOH-. M 3. A 0.3174 g sample of an unknown acid requires 26.23 mL of 0.1056 M NaOH for neutralization to a phenolphthalein end point. There are 0.27 mL of 0.1096 M HCl used for back-titration. a. How many moles of OH- are used? How many moles of H+ from HCl? moles OH- moles H+ b. How many moles of H+ are there in the solid acid? (Use Eq. 5.) moles H+ in solid c. What is the molar mass of the unknown acid? (Use Eq. 4.) g/mol

Transcript

00:01 For question one, we have 7 milliliters of a 6 .0 molar solution of sodium hydroxide that's been diluted to 400 milliliters.

00:07 So we have to find the moles of sodium hydroxide in the undiluted solution.

00:12 So the first thing we're going to do is we're going to convert these milliliters to liters by using dimensional analysis, or you can just say dividing by a thousand.

00:21 So this is going to be 7 .0 times 10 to the minus 3.

00:24 We're then going to multiply that by 6 .0, and then this is molarity or moles per one liter.

00:32 If we do this, our liters actually cancel out and you'll be left with just moles.

00:38 So this is a 42 .0 times 10 to the minus 3 moles of sodium hydroxide.

00:48 But if you are uncomfortable with having that 42, what you can do is just say that it's going to be 4 .2 times 10 to the minus 2.

01:00 Either one is fine.

01:02 So let's write that.

01:03 We have 4 .2 times 10 to the minus 2 for 0 .042 moles of naoh.

01:13 Now we have to find the moles in the diluted solution.

01:19 So we're going to use c1v1 is equal to c2v2.

01:24 Now that c, you might recognize it as concentration, but you can also use it for moles.

01:30 So that's what we're going to do.

01:31 So the concentration in our initial undiluted solution was 0 .042, and then we had 7 milliliters of that.

01:40 We're looking for c2, and we have 400 milliliters of that.

01:46 Notice you don't need to convert to liters because the units are going to be the same on both sides, so they will cancel them out.

01:52 So we have c2 is equal to 0 .042 times 7 divided by 400.

02:01 So let's do this math.

02:03 We have 0 .042 times 7 divided by 400.

02:07 This is going to be 7 .35 times 10 to the minus 4 moles of naoh.

02:15 So let's write that in.

02:20 And we know that this is the case because we diluted it, so it has to be smaller than the previous answer.

02:26 So now we are given for question 2, 23 .91 milliliters of a naoh solution needed to neutralize 24 .5 milliliters of a 0 .1002 molar hydrochloric acid.

02:42 So we have to find the molarity of the naoh.

02:46 First, we're going to start with the molarity of hcl.

02:50 Well, the molarity of hcl is 0 .1002 molarity.

02:56 And we can convert this to moles of h by using the molarity and the volume.

03:05 So it's going to be 0 .1002.

03:08 Remember, that's moles per 1 liter.

03:11 We're going to multiply that by 24 .58 times 10 to the minus 3 liters.

03:18 That gets rid of your liters, and you are left with just moles.

03:22 So we'll take 24 .58 times 10 to the minus 3, and then we're going to multiply that by 0 .1002.

03:31 And the moles of hcl are going to be 0 .002462916.

03:40 So if we were to combine naoh plus hcl, we would get a sodium ion, a chlorine ion, and water.

03:53 And if you notice, everything is at 1 to 1.

03:56 So our moles of hcl are going to equal the moles of oh.

04:02 So the moles of oh are 0 .002462916 moles of naoh.

04:14 Or you can just say oh.

04:17 So now the molarity of naoh is just moles divided by a volume.

04:23 And in this case, it has to be liters.

04:25 So we're going to take our previous answer with all of its value, and we're going to divide it by 23 .91 times 10 to the minus 3 liters.

04:42 And when we do this, we get a molarity of 0 .10300778.

04:50 Let's round this to 0 .1030 molarity naoh.

04:56 Now let's scroll down and check out the last one...

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