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molarity = moles of solute

MOLALITY

An alternative unit of concentration to molarity is molality. The molalityof a solute is the number of moles of that solute divided by the weightof the solvent in kilograms. For water solutions, 1 kg of water has avolume close to that of 1 liter, so molality and molarity are similar indilute aqueous solutions.To figure out the molarity of a solution, simply work out the number ofmoles of the solute (Probably from the molecular weight) and divide bythe weight of the solvent. It's probably the case that you're given avolume of solvent rather than the weight: use the density to convertbetween the two

molality (M) = moles solute/kg of solutionTo convert a volume and a molality of a solution to moles of solute,simply solve the above equation for moles of solute:

moles solute = molality * kg of solutionExample 1: If you have 10.0 grams of Br2 and dissolve it in 1.00 L ofcyclohexane, what is the molality of the solution? The density ofcyclohexane is 0.779 kg/l at room temperature.Solution 1: First, work out the number of moles of bromine. Br2 has amolecular weight of 159.8 g/mole, so we have

10 g / (159.8 g/mole) = 0.063 moles BR2Next, convert the volume of solvent to the weight of solvent using thedensity

1.00 L * 0.779 kg/l = 0.779 kgNow just divide the two to get the molality

0.063 moles Br2/ 0.779 kg cyclohexane = 0.080molal

MOLARITY

Another way of expressing concentration, the way that we will usemost in this course, is called molarity. Molarity is the number of molesof solute dissolved in one liter of solution. The units, therefore aremolesperliter, specifically it's molesofsolute perliterofsolution.Rather than writing out moles per liter, these units are abbreviated asM orM. We use a capital M with a line under it or a capital Mwritten initalics. So when you see Mor Mit stands for molarity, and itmeans molesperliter(not justmoles).You must be very careful to distinguish between moles and molarity."Moles" measures the amount or quantity of material you have;"molarity" measures the concentration of that material. So when you'regiven a problem or some information that says the concentration of thesolution is 0.1 M that means that it has 0.1 mole for every liter ofsolution; it does not mean that it is 0.1 moles. Please be sure to makethat distinction.

NormalityWhen you need to compare solutions on the basis ofconcentrationofspecificions or the amount of charge that the ions have, a differentmeasure of concentration can be very useful. It is called normality.We will deal with normality more completely in the lesson on acid-basetitrations and give it just a cursory mention in this lesson. For thatreason you can ignoreobjectives 6-9 andexamples 13-17 in yourworkbook for this lesson.The normality of a solution is simply a multipleofthemolarity of thesolution. Generally, the normality of a solution is just one, two or threetimes the molarity. In rare cases it can be four, five, six or even seventimes as much. The symbol for normality is N orN.

Whether the multiplying factor isone, two or three depends on theformula of the chemical and what it

is being compared to. Note thatcalcium chloride has two moles ofchloride ions for every mole ofCaCl2. Because of that, themultiplying factor for calciumchloride is two.

CaCl2 Ca2+ + 2 Cl-

1 M CaCl2 = 2 N CaCl22.4 M CaCl2 = 4.8 N CaCl2

etc.

Similarly, for aluminum chloride themultiplying factor is three.

AlCl3 Al3+ + 3 Cl-

1 M AlCl3 = 3 N AlCl32.4 M AlCl3 = 7.2 N AlCl3

etc.

Hopefully, this gives you an idea of the nature and value of normalityand its relationship to molarity. As mentioned earlier, it will be coveredin more detail in the lesson involving acid-base titrations.

MOLE FRACTIONThe concentration of a solution refers to the amount of solute in agiven amount of solvent. There are many ways to express,quantatively, the concentration of a solution. Some solution propertiesdepend on the relative amounts of all the solution components in termsof moles. The mole fraction of a solution component Xi is the fractionof moles of component iof the total number of moles of all componentsin solution.

The mole fraction is:moles of target substance divided by total moles involvedThe symbol for the mole fraction is the lower-case Greek letter chi, .You will often see it with a subscript: solute is an example.Example #1: 0.100 mole of NaCl is dissolved into 100.0 grams of pureH2O. What is the mole fraction of NaCl?Solution:100.0 g / 18.0 g mol1 = 5.56 mol of H2OAdd that to the 0.100 mol of NaCl = 5.56 + 0.100 = 5.66 mol totalMole fraction of NaCl = 0.100 mol / 5.66 mol = 0.018What is the mole fraction of the H2O?5.56 mol / 5.66 mol = 0.982

pH

Problem #1: A weak acid has a pKa of 3.994 and the solution pH is4.523. What percentage of the acid is dissociated?A comment before discussing the solution: note that the pKa is given,rather than the Ka. The first thing we will need to do is convert the pK ato the Ka. Then, the two values we need to obtain to solve the problemgiven just above are [H

+], which is pretty easy and [HA], which is only a

tiny bit more involved.Solution:1) Convert pKa to Ka:Ka = 10

pKa = 10

3.994 = 1.0139 x 104Often the identity of the weak acid is not specified. That is because,with few exceptions, all weak acids behave in the same way and so thesame techniques can be used no matter what acid is used in theproblem. In cases where no acid is identified, you can use a generic

weak acid, signified by the formula HA. Here is the dissociationequation for HA:H+ + A HAHere is the equilibrium expression for that dissociation:Ka = ([H

+] [A]) / [HA]

2) The pH will give [H+] (and the [A]):

[H+] = 10

pH= 10

4.523= 3.00 x 10

5M

Because of the 1:1 molar ratio in the above equation, we know that[A] = [H+] = 3.00 x 105 M.3) This means that the only value left is [HA], so we will use theequilibrium expression to calculate [HA].1.0139 x 10

4= [(3.00 x 10

5) (3.00 x 10

5)] / x

x = 8.88 x 105

M4) Percent dissociation for an acid is [H

+] / [HA] and then times 100.

3.00 x 105 / 8.88 x 105 = 33.8%

Problem #2: A solution of acetic acid (Ka = 1.77 x 105) has a pH of

2.876. What is the percent dissociation?Solution:1) Calculate the [H

+] from the pH:

[H+] = 10

pH= 10

2.876= 1.33 x 10

3M

2) From the 1:1 stoichiometry of the chemical equation, we know thatthe acetate ion concentration, [Ac] equals the [H

+]. Therefore,

[Ac] = 1.33 x 103 M3) We need to determine [HAc], the acetic acid concentration. We usethe Ka expression to determine this value:1.77 x 10

5= [(1.33 x 10

3) (1.33 x 10

3)] / x

x = 0.09993 M = 0.100 M4) Percent dissociation:(1.33 x 10

3/ 0.100) times 100 = 1.33%

Comment: the first example is somewhat artifical, in that the percentdissocation is quite high. The second example is more in line with whatteachers usually ask. The usual percent dissociation answer is

moles ofcomponent i

Xi = ------------------------

total moles ofsolution

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between 1 and 5 per cent. However, the 33.8% answer, while notcommonly found in introductory chemistry classes, is possible.

Problem #3: A generic weak acid (formula = HA) has a pKa of 4.401. Ifthe solution pH is 3.495, what percentage of the acid is undissociated?Solution:1) Convert pKa to Ka:Ka = 10pKa = 104.401 = 3.97 x 10

52) The pH gives [H+] (and the [A]):[H+] = 10pH = 103.495 = 3.20 x 104 M3) Determine the concentration of the weak acid:Ka = ([H

+] [A]) / [HA]

3.97 x 105

= [(3.20 x 104

) (3.20 x 104

)] / xx = 0.00258 M4) Determine percent dissociation:3.20 x 104 / 0.00258 = 12.4%5) Determine percent undissociated:100 - 12.4 = 87.6%Comment: the calculation technique discussed above determines thepercent dissociation. Notice that the above problem asks for thepercent undissociated.Be aware! The problem above goes one step beyond what is normallytaught. This might show up as a test question.

Problem #4: A weak acid has a pKa of 4.289. If the solution pH is3.202, what percentage of the acid is dissociated?Solution:1) Convert pKa to Ka:Ka = 10pKa = 104.289 = 5.14 x 10

5

2) The pH gives [H+] (and the [A]):[H+] = 10pH = 103.202 = 6.28 x 104 M3) Determine the concentration of the weak acid:Ka = ([H

+] [A]) / [HA]5.14 x 10

5= [(6.28 x 10

4) (6.28 x 10

4)] / x

x = 0.00767284 M (kept a few guard digits on this one4) Determine percent dissociation:6.28 x 104 / 0.00767284 = 8.2%

pOH