# Solutions and Molarity – Other Ionic Reactions

by Adam Le Gresley, PhD

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00:01 The majority of ionic chemical reactions that you will encounter take place in water or aqueous solution.

00:08 And it's important to be aware of how you calculate and to choose molarity.

00:14 You'll recall from the previous slide when I talked about acids and bases, we talked about concentrations.

00:19 That is in moles per decimeter cubed, or moles per liter, or just simply given the abbreviation M for molar concentration.

00:31 Molarity is the measurement of that concentration of a chemical, any chemical, in solution.

00:36 And in water, the units of molarity is given the molar M where we have, in this equation, the number of moles of a solute, let's say, sodium chloride, per the number in liters of solution.

00:51 Now, seldom do we operate in liters so it's important to make sure that you break that down.

00:56 If you're dealing, for example, with a hundred mL, then we're talking about 0.1 of a liter, 10 mL, 0.01, and so forth.

01:05 But for molarity, to be accurate, you need to calculate your volume in liters.

01:11 So, let's give you an example here.

01:14 Calculate the molarity of a solution made by dissolving 12.98 grams of calcium hydroxide in enough water to make 1.29 liters of solution.

01:27 Okay, so what is the first thing we need to do? The first thing we need to do is we need to work out how many moles of calcium hydroxide we actually have in 12.98 grams.

01:40 And you should be aware already that the way in which you calculate moles is take the mass of the material you have in front of you and divide it by the formula mass, relative formula mass.

01:56 You can calculate the relative formula mass by adding up all of the atomic masses of the elements in your formula's unit.

02:06 So, in this particular case of calcium hydroxide, we would look at the periodic table, find the atomic mass of calcium, and add it to two equivalents of the atomic mass of oxygen and two equivalents of the atomic mass of hydrogen.

02:22 Note the reason we add those in the case of oxygen and hydrogen, is because calcium hydroxide has two hydroxide per one equivalent of calcium.

02:32 This gives us a formula mass for calcium hydroxide of 74.1 grams.

02:39 So therefore, this allows us to calculate what percentage of that concentration we have.

02:47 In this particular case, we take 12.98 dividing it by 74.1, gives us a number of moles of 0.175, okay? Otherwise known as 175 mmol. We then take our 0.175 moles of calcium hydroxide and as per the previous equation, we divide it by 1.29 liters.

03:15 This gives us a molar concentration of 0.135 molar of calcium hydroxide.

03:23 Let's try another example as well.

03:27 How many grams of ammonium nitrate are in 172.7 milliliters sample of 1.21 molar ammonium nitrate solution? Right, so where do we need to start? First and foremost, what we need to do is we need to work out what, in terms of grams, this correlates to.

03:49 So, if we look at the equation that's shown below, we see that 1.2 molar concentration of ammonium nitrate in 0.1727 liters of solution correlates to 0.209 moles of sodium nitrate.

04:08 So, if you look, what we've effectively done is we've worked out that in one liter, we have 1.21 molars and so therefore, as a percentage of that one liter, we have 0.209 molars in 172 mL sample.

04:24 If it's difficult for you to get your head around, imagine it simply like this: if we had 1.21 molar solution of ammonium nitrate and I gave you 600 mL, you would know that you would have approximately--sorry, 500 mL, you'd know you have approximately 0.6 of a mole in that solution.

04:46 All we've done is we've taken as a proportion of that particular value for volume of that particular concentration.

04:56 This gives us a value of 0.209 moles of ammonium nitrate.

05:01 That's great. Now, what we need to do is we need to multiply that by the relative formula mass.

05:07 Remember what we said about calcium hydroxide.

05:09 In this particular case, we're using ammonium nitrate.

05:13 So to determine the relative formula mass of this case, we need to take one equivalent--two equivalents of nitrogen with atomic mass of 14, four equivalents of hydrogen, atomic mass of one, and three equivalents of oxygen with an atomic mass of 16.

05:32 This gives us an overall formula mass of 80 grams of ammonium nitrate.

05:37 And so therefore, the number of grams we have comes from multiplying the number of moles we have in that small sample by the mass of that as a formula unit.

05:48 This gives us a mass within 172.7 mL sample of our solution of 16.7 grams of ammonium nitrate.

05:58 Let's get on to something that is found often in certainly when it comes to nursing and so forth is dilution. Dilution factors.

06:09 When diluting a solution, you have to bear in mind that the number of moles, just crucial, remains constant.

06:15 And therefore, the equation exists that molarity divided by volume-- sorry, times by volume, equals a number of moles.

06:23 So, M1 times V1, which is prior to dilution, equals a number of moles which is equal to M2 times V2, which are our molarity and our volume after dilution.

06:37 As water is added, increasing the final volume, the molarity therefore must decrease in order for that equation to be true.

### About the Lecture

The lecture Solutions and Molarity – Other Ionic Reactions by Adam Le Gresley, PhD is from the course Ionic Chemistry.

### Included Quiz Questions

1. The number of moles of a solute per liter of a solution
2. The ratio of total numbers of moles of all the components to the number of moles of one component present in a solution
3. The number of moles of a solute per kilogram of the solvent
4. The number of mole equivalents of solute per liter of the solution
5. The number of parts by mass of solute per million parts by mass of the solution
1. 0.25 M HCl
2. 2.5 M HCl
3. 1 M HCl
4. 0.5 M HCl
5. 0.125 M HCl
1. 24.37 g of NaCl
2. 0.417 g of NaCl
3. 14 g of NaCl
4. 140 g of NaCl
5. 2.480 g of NaCl

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