Lecture 2 & 3

Solutions and their concentrations

Azad H. Mohammed

Azadalshatteri@Hotmail.com

Chemistry Department

University of Garmian

This lecture

Solutions and their concentrations

Common units for expressing concentration

1. Mass per volume

2. Parts per million

3. Parts per billion

4. Percent concentration

5. molarity

Solution and their Concentration

Concentration is a general measurement unit stating the amount of

solute (solu.) present in a known amount of solution (soln.)

𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 =𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 (𝑠𝑜𝑙𝑢. )

𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 (𝑠𝑜𝑙𝑛. )

Common units for expressing concentration

1. Mass per volume

In this unit we use the gram and its derivatives to express the mass of solute

(wt. of solu.):

1 gram (g) = 103 milligram (mg) = 106 microgram (µg) = 109 nanogram (ng) =

1012 picogram (pg),

and we use the liter and its derivatives to express the volume of solution (vol.

of soln.):

One liter (L) = 103 milliliter (mL) = 106 microliter (µL) = 109 nanoliter (nL)

= 1012 picoliter (pL)

We can use several expressions such as

𝑔/𝐿 =wt. of solu. g

vol. of soln. L𝑜𝑟 𝑚𝑔/𝑚𝐿 =

wt. of solu. (mg)

vol. of soln. (mL)

and so on.

Example2.1

How many grams are needed to prepare 500 mL solution of 5 (g/L) of

substance A?

Solution

𝑔/𝐿 =wt. of solu. g

vol. of soln. L⇒ 5 𝑔/𝐿 =

X g

500 mL 𝑥 (1 𝐿

1000 𝑚𝐿)

⇒ 𝑋 = 2.5 𝑔

Example 2.2

10 g of substance B have been dissolved in water and the volume is completed

with distilled water to 200 mL. Calculate the concentration of B in the following

terms: g/L, mg/mL, µg/mL, ng/mL, mg/dL and pg/µL?

Solution

1. 𝑔/𝐿 =10 𝑔

200 𝑚𝐿 𝑥 10−3𝐿𝑚𝐿

= 50 𝑔/𝐿

2. 𝑚𝑔/𝑚𝐿 =10 𝑔 𝑥 103

𝑚𝑔𝑔

200 𝑚𝐿= 50 𝑚𝑔/𝑚𝐿

3. µ𝑔/𝑚𝐿 =10 𝑔 𝑥 106 µ𝑔

200 𝑚𝐿= 50 𝑥 103 µ𝑔/𝑚𝐿

4. 𝑛𝑔/𝑚𝐿 =10 𝑔 𝑥 109 𝑛𝑔

200 𝑚𝐿= 50 𝑥 106 𝑛𝑔/𝑚𝐿

5. 𝑚𝑔/𝑑𝐿 =10 𝑔 𝑥 103 𝑚𝑔

200 𝑚𝐿 𝑥 10 −2 𝑑𝐿𝑚𝐿

= 5 𝑥 103 𝑚𝑔/𝑑𝐿

6. 𝑝𝑔/µ𝐿 =10 𝑔 𝑥1012 𝑝𝑔

200 𝑚𝐿 𝑥 103µ𝐿𝑚𝐿

= 50 𝑥 106 𝑝𝑔/µ𝐿

Example 2.3

You have a solution of solute B its concentration is 10 µg/mL. Calculate

the mass of B in 5 mL of this solution?

Solution

10µ𝑔

𝑚𝐿=

𝑋 µ𝑔 𝐵

5 𝑚𝐿⇒ 𝑋 = 10 𝑥 5 = 50 µ𝑔 𝐵

2. Parts per million (ppm)

This expression means parts of solute in a million parts of the solution. If we

say 500 ppm of solute B that means 500 parts (g, mg, or µg … etc.) in a

million parts (g, mg, or µg … etc.) of the solution, that means also every

million parts of solution contains 500 parts of solute B. it can be expressed

according to the type of solutions as follows:

a). solid solute in a solid solution: such as an element in an alloy

𝑝𝑝𝑚 =wt. of solu. (g. , mg. , … )

wt. of soln. (g. , mg. , … )𝑥 106

Where the weights of both the solute and the solution are expressed with the

same unit (g, mg or µg …. etc.) and multiply by 106.

Instead of multiplying by 106, we can use a weighing unit for the solution that

is equal to million units used for the solute:

𝑝𝑝𝑚 =𝑊𝑡. 𝑜𝑓 𝑠𝑜𝑙𝑢. (𝑚𝑔)

𝑊𝑡. 𝑜𝑓 𝑠𝑜𝑙𝑛. (𝑘𝑔)=

𝑊𝑡. 𝑜𝑓 𝑠𝑜𝑙𝑢. (µ𝑔)

𝑊𝑡. 𝑜𝑓 𝑠𝑜𝑙𝑛. (𝑔)

Where one kg = 106 mg, and one gram = 106 µg

b). liquid-liquid solution: such as dissolving an alcohol in water. The same way asbefore, but here we use volume units instead of mass units, and either we use thesame units for both the solute and the solution (L, or mL, etc.) and multiply by 106

thus:

𝑝𝑝𝑚 =𝑉𝑜𝑙. 𝑜𝑓 𝑠𝑜𝑙𝑢. (𝑚𝐿, 𝐿)

𝑉𝑜𝑙. 𝑜𝑓 𝑆𝑜𝑙𝑛. (𝑚𝐿, 𝐿)𝑥106

or we use a volume unit for the solution equals million of the unit used for the solute

thus:

𝑝𝑝𝑚 =𝑉𝑜𝑙. 𝑜𝑓 𝑠𝑜𝑙𝑢. (µ𝐿)

𝑉𝑜𝑙. 𝑜𝑓 𝑆𝑜𝑙𝑛. (𝐿)=

𝑉𝑜𝑙. 𝑜𝑓 𝑠𝑜𝑙𝑢. (𝑛𝐿)

𝑉𝑜𝑙. 𝑜𝑓 𝑆𝑜𝑙𝑛. (𝑚𝐿)

Where one liter = 106 µL, and one mL = 106 nL

c). solid-liquid solution: such as dissolving a salt in water. Water is a

common solvent and its density is 1 g/mL at room temperature. Therefore,

one liter of water solution weigh 1000 g which is one kilogram (kg)

assuming its density will not be affected by dissolving trace amount of

solute in it. Consequently, we can deal with the solid-liquid solution in the

same manner as we did with solid-solid or liquid-liquid solutions:

𝑝𝑝𝑚 =𝑊𝑡. 𝑜𝑓 𝑠𝑜𝑙𝑢. (𝑚𝑔)

𝑉𝑜𝑙. 𝑜𝑓 𝑠𝑜𝑙𝑛. (𝐿)=

𝑊𝑡. 𝑜𝑓 𝑠𝑜𝑙𝑢. (µ𝑔)

𝑉𝑜𝑙. 𝑜𝑓 𝑠𝑜𝑙𝑛. (𝑚𝐿)

3. Parts per billion (ppb):

The same as mentioned in the above ppm term, but instead of using 106 we

use 109 or we use a unit for the solution is equal to one billion unit used for

the solute as can be seen in the following formulas:

a). solid-solid solution:

𝑝𝑝𝑏 =𝑊𝑡. 𝑜𝑓 𝑠𝑜𝑙𝑢. (𝑔. , 𝑚𝑔. )

𝑊𝑡. 𝑜𝑓 𝑠𝑜𝑙𝑛. (𝑔. , 𝑚𝑔. )𝑥 109

or

𝑝𝑝𝑏 =𝑊𝑡. 𝑜𝑓 𝑠𝑜𝑙𝑢. µ𝑔

𝑊𝑡. 𝑜𝑓 𝑠𝑜𝑙𝑛. 𝑘𝑔=

𝑊𝑡. 𝑜𝑓 𝑠𝑜𝑙𝑢. 𝑛𝑔

𝑊𝑡. 𝑜𝑓 𝑠𝑜𝑙𝑛. 𝑔

Where one kg = 109 µg and one g = 109 ng.

b). liquid-liquid solution:

𝑝𝑝𝑏 =𝑉𝑜𝑙. 𝑜𝑓 𝑠𝑜𝑙𝑢. (𝑚𝐿. , 𝐿. )

𝑉𝑜𝑙. 𝑜𝑓 𝑠𝑜𝑙𝑛. (𝑚𝐿. , 𝐿. )𝑥 109

or

𝑝𝑝𝑏 =𝑉𝑜𝑙. 𝑜𝑓 𝑠𝑜𝑙𝑢. (𝑛𝐿. )

𝑉𝑜𝑙. 𝑜𝑓 𝑠𝑜𝑙𝑛. (𝐿. )=

𝑉𝑜𝑙. 𝑜𝑓 𝑠𝑜𝑙𝑢. (𝑝𝐿. )

𝑉𝑜𝑙. 𝑜𝑓 𝑠𝑜𝑙𝑛. (𝑚𝐿. )

Where one L = 109 nL, and one mL = 109 pL.

c). solid-liquid solution:

𝑝𝑝𝑏 =𝑊𝑡. 𝑜𝑓 𝑠𝑜𝑙𝑢. (𝑔. , 𝑘𝑔. )

𝑉𝑜𝑙. 𝑜𝑓 𝑠𝑜𝑙𝑛. (𝑚𝐿. , 𝐿. )𝑥 109

or

𝑝𝑝𝑏 =𝑊𝑡. 𝑜𝑓 𝑠𝑜𝑙𝑢. (µ𝑔. )

𝑉𝑜𝑙. 𝑜𝑓 𝑠𝑜𝑙𝑛. (𝐿. )=

𝑊𝑡. 𝑜𝑓 𝑠𝑜𝑙𝑢. (𝑛𝑔. )

𝑉𝑜𝑙. 𝑜𝑓 𝑠𝑜𝑙𝑛. (𝑚𝐿. )

Where one L of aqueous solution = one kg = 109 nL.

4. Percent concentration

It is one of the methods for expressing the concentration of the

following percentages expression:

a) Weight per weight percentage % (w/w): the weight of solute is

divided by the weight of solution (both have the same weighing unit)

and multiplying by 100.

% 𝑤 𝑤 =𝑊𝑡. 𝑜𝑓 𝑠𝑜𝑙𝑢. 𝑔… . .

𝑊𝑡. 𝑜𝑓 𝑠𝑜𝑙𝑛. 𝑔… . .𝑥 100

b) Weight per volume percentage % (w/v): the weight of solute usually in

grams is divided by volume of solution which is usually in milliliters and

multiplying by 100.

% 𝑤 𝑣 =𝑊𝑡. 𝑜𝑓 𝑠𝑜𝑙𝑢. 𝑔… . .

𝑉𝑜𝑙. 𝑜𝑓 𝑠𝑜𝑙𝑛. 𝑚𝐿… . .𝑥 100

c) Volume per volume percentage % (v/v): the volume of solute usually in mL

is divided by volume of solution that is usually in mL then multiplying by 100.

% 𝑣 𝑣 =𝑉𝑜𝑙. 𝑜𝑓 𝑠𝑜𝑙𝑢. 𝑚𝐿… . .

𝑉𝑜𝑙. 𝑜𝑓 𝑠𝑜𝑙𝑛. 𝑚𝐿… . .𝑥 100

Example 2.4

How many milligrams are required to prepare 500 mL solution of substance

B of each of the following concentration units? a) 2000 ppb, b) 500 ppm, c) 5

g/L, and d) 10% (w/v)

Solution

1) 2000 𝑝𝑝𝑏 =Wt. of solu. (µg)

500 mL 𝑥 10−3 (𝐿)

Wt. of solu. µg = 2000 x 500 x 10−3 = 1000 µg = 1 mg

2) 500 𝑝𝑝𝑚 =Wt. of solu. (mg)

500 mL 𝑥 10−3 (𝐿)

Wt. of solu. mg = 500 x 500 x 10−3 = 250 mg

3) 5(𝑔

𝐿) =

Wt. of solu. g

500 mL 𝑥 10−3 (𝐿)

Wt. of solu. g = 5 x 500 x 10-3 = 2.5 g = 2500 mg

4) 10 % =Wt. of solu. (g)

500 (mL)𝑥 100

Wt. of solu. g =10 x 500

100= 50 𝑔 𝑥 1000 = 50000 𝑚𝑔

Example 2.5

5 g of an alloy containing 10 mg of copper. Calculate the concentration of

copper using ppm and ppb units in this alloy?

Solution

This is an example of solid-solid solution:

𝑝𝑝𝑚 =10 (𝑚𝑔)

5 𝑔 𝑥 10−3(𝑘𝑔)=

10 𝑥 103 (µ𝑔)

5 (𝑔)= 2 𝑥 103 𝑚𝑔/𝑘𝑔 𝑜𝑟 µ𝑔/𝑔

𝑝𝑝𝑏 =10 𝑚𝑔 𝑥 103 (µ𝑔)

5 𝑔 𝑥 10−3 (𝑘𝑔)=

10 𝑥 106 (𝑛𝑔)

5 (𝑔)= 2 𝑥 106 µ𝑔/𝑘𝑔 𝑜𝑟 𝑛𝑔/𝑔

Example 2.6

How many grams there is in 50 mL of 500 ppm solution of a substance?

Solution

500 𝑝𝑝𝑚 =𝑊𝑡. 𝑜𝑓 𝑠𝑜𝑙𝑢. (𝑚𝑔)

50 𝑚𝐿 𝑥 10−3 (𝐿)

𝑊𝑡. 𝑜𝑓 𝑠𝑜𝑙𝑢. 𝑚𝑔 = 500 𝑥 50 𝑥 10−3 = 25 𝑚𝑔 𝑥 10−3 = 0.025 𝑔

Example 2.7

2 g of a substance have been dissolved in water and the volume was

completed to 100 mL. Calculate the concentration of this substance in this

solution using the following units: a) g/L, b) mg/mL, c) %(w/v), d) ppm,

and ppb.

Solution

1)𝑔

𝐿=

2 (𝑔)

100 𝑚𝐿 𝑥 10−3(𝐿𝑚𝐿

)= 20 𝑔/𝐿

2)𝑚𝑔

𝑚𝐿=

2 𝑔 𝑥 103 (𝑚𝑔𝑔

)

100 𝑚𝐿= 20 𝑚𝑔/𝑚𝐿

3) % w/v =2 𝑔

100 𝑚𝐿𝑥100 = 2% (𝑤/𝑣)

4) 𝑝𝑝𝑚 =2 𝑔 𝑥 103 (

𝑚𝑔𝑔

)

100 𝑚𝐿 𝑥 10−3(𝐿𝑚𝐿

)= 20 𝑥 103 𝑚𝑔/𝐿

5) 𝑝𝑝𝑏 =2 𝑔 𝑥 106 (

µ𝑔𝑔)

100 𝑚𝐿 𝑥 10−3 (𝐿𝑚𝐿

)= 20 𝑥 106 µ𝑔/𝐿

5. Molarity (M) or molar concentration:

The molar concentration cx of a solution of a solute species X is the number ofmoles of that species which is contained in 1 liter of the solution (not 1 liter ofsolvent).

𝑀 =no.mole solu. mol

Vol. of soln. L=

[𝑊𝑡. 𝑜𝑓 𝑠𝑜𝑙𝑢. 𝑔

𝑀.𝑀. 𝑜𝑓 𝑠𝑜𝑙𝑢.𝑔

𝑚𝑜𝑙

]

Vol. of soln. (L)

The unit of molar concentration is molar, symbolized by M, which has thedimensional of mol/L. molarity is also the number of millimoles of solute permilliliter of solution

𝑀 =no.mole solu. mmol

Vol. of soln. mL=

[𝑊𝑡. 𝑜𝑓 𝑠𝑜𝑙𝑢. 𝑚𝑔

𝑀.𝑀. 𝑜𝑓 𝑠𝑜𝑙𝑢.𝑚𝑔

𝑚𝑚𝑜𝑙

]

Vol. of soln. (mL)

Example2.8

Calculate the molar concentration of ethanol in an aqueous solution that contains

2.3 g of C2H5OH (46.07 g/mol) in 3.5 L of solution.

Solution

To calculate molar concentration, we must find both the mole of ethanol and the

volume of the solution. The volume was given as 3.5 L, so all we need to do is

convert the mass of ethanol to the corresponding amount of ethanol in moles.

𝑚𝑜𝑙𝑒 𝑜𝑓 𝐶2𝐻5𝑂𝐻 = 2.3 𝑔 𝐶2𝐻5𝑂𝐻 𝑥1 𝑚𝑜𝑙 𝐶2𝐻5𝑂𝐻

46.07 𝑔 𝐶2𝐻5𝑂𝐻= 0.04992 𝑚𝑜𝑙 𝐶2𝐻5𝑂𝐻

To obtain the molar concentration, we divide the mole by the volume. Thus,

𝑀 =𝑛𝑜.𝑚𝑜𝑙𝑒 𝑠𝑜𝑙𝑢 (𝑚𝑜𝑙)

𝑉𝑜𝑙. 𝑜𝑓 𝑠𝑜𝑙𝑛 (𝐿)=

0.04992 𝑚𝑜𝑙

3.5 𝐿= 0.0143 𝑚𝑜𝑙 𝐶2𝐻5𝑂𝐻/𝐿

There are two ways of expressing molar concentration: molar analytical

concentration and molar equilibrium concentration.

1. Molar analytical concentration is the total number of moles of a

solute, regardless of its chemical state, in 1 L of solution. The molar

analytical concentration describes how a solution of a given

concentration can be prepared. Note that in the above example, the

molar concentration that we calculated is also the molar analytical

concentration 𝑐𝐶2𝐻5𝑂𝐻 = 0.0143 𝑀 because the solute ethanol

molecules are intact following the solution process. In another

example, a sulfuric acid solution that has an analytical concentration of

𝑐𝐻2𝑆𝑂4= 1𝑀 can be prepared by dissolving 1 mole, or 98 g, of H2SO4

in water and diluting to exactly 1 L.

2. Molar equilibrium concentration is the initial (analytical) concentration

minus the amount reacted. The analytical concentration is given by the

notation Cx, while equilibrium concentration is given by [x]. a solution of 1 M

CaCl2 (analytical molarity) gives at equilibrium, 0 M CaCl2, 1 M Ca2+, and 2 M

𝐶𝑙− (equilibrium molarities). Hence, we say the solution is 1 M 𝐶𝑎2+.

Example 2.9

Calculate the analytical and equilibrium molar concentrations of the solute species in

an aqueous solution that contains 285 mg of trichloroacetic acid, Cl3CCOOH (163.4

g/mol), in 10.0 mL (the acid is 73% ionized in water).

Solution

We calculate the number of moles of Cl3CCOOH, which we designate as HA, and divide

by the volume of the solution, 10.0 mL, or 0.0100 L. Therefore,

𝑚𝑜𝑙𝑒 𝐻𝐴 = 285 𝑚𝑔 𝐻𝐴 𝑥1 𝑔 𝐻𝐴

1000 𝑚𝑔 𝐻𝐴𝑥

1 𝑚𝑜𝑙 𝐻𝐴

163.4 𝑔 𝐻𝐴= 1.744 𝑥 10−3 𝑚𝑜𝑙 𝐻𝐴

The molar analytical concentration, cHA, is then

𝑐𝐻𝐴 =1.744 𝑥 10−3 𝑚𝑜𝑙 𝐻𝐴

10 𝑚𝐿𝑥1000 𝑚𝐿

1 𝐿= 0.174 𝑚𝑜𝑙 𝐻𝐴/𝐿 = 0.174 𝑀

In this solution, 73% of the HA dissociates, giving H+ and A-:

The equilibrium concentration of HA is then 27% of cHA. Thus,

𝐻𝐴 = 𝑐𝐻𝐴 𝑥 (100 − 73)/100 = 0.174 𝑥 0.27 = 0.047 𝑚𝑜𝑙/𝐿 = 0.047 𝑀

The equilibrium concentration of A– is equal to 73% of the analytical

concentration of HA, that is,

𝐴− =73 𝑚𝑜𝑙 𝐴−

100 𝑚𝑜𝑙 𝐻𝐴𝑥 0.174

𝑚𝑜𝑙 𝐻𝐴

𝐿= 0.127 𝑀

Because 1 mole of 𝐻+ is formed for each mole of 𝐴−, we can also write

𝐻+ = 𝐴− = 0.127 𝑀

Example 2.10

Describe the preparation of 2.0 L of 0.108 M BaCl2 from BaCl2.2H2O (244.3

g/mol)

Solution

𝑚𝑜𝑙𝑒 𝑜𝑓 𝐵𝑎𝐶𝑙2. 2𝐻2𝑂 = 2.0 𝐿 𝑥0.108 𝑚𝑜𝑙 𝐵𝑎𝐶𝑙2. 2𝐻2𝑂

𝐿= 0.216 𝑚𝑜𝑙 𝐵𝑎𝐶𝑙2. 2𝐻2𝑂

The mass of BaCl2.2H2O is then

0.216 𝑚𝑜𝑙 𝐵𝑎𝐶𝑙2. 2𝐻2𝑂 𝑥244.3 𝑔 𝐵𝑎𝐶𝑙2. 2𝐻2𝑂

𝑚𝑜𝑙 𝐵𝑎𝐶𝑙2. 2𝐻2𝑂= 52.8 𝑔 𝐵𝑎𝐶𝑙2. 2𝐻2𝑂

Dissolve 52.8 g of BaCl2.2H2O in water and dilute to 2.0 L

Example 2.11

Describe the preparation of 500 mL 0f 0.0740 M 𝐶𝑙− solution from solid

BaCl2.2H2O (244.3 g/mol).

Solution

𝑚𝑎𝑠𝑠 𝐵𝑎𝐶𝑙2. 2𝐻2𝑂 =0.0740 𝑚𝑜𝑙 𝐶𝑙−

𝐿𝑥 0.5 𝐿 𝑥

1 𝑚𝑜𝑙 𝐵𝑎𝐶𝑙2. 2𝐻2𝑂

2 𝑚𝑜𝑙 𝐶𝑙−

𝑥244.3 𝑔 𝐵𝑎𝐶𝑙2. 2𝐻2𝑂

𝑚𝑜𝑙 𝐵𝑎𝐶𝑙2. 2𝐻2𝑂= 4.52 𝑔 𝐵𝑎𝐶𝑙2. 2𝐻2𝑂

Dissolve 4.52 g of BaCl2.2H2O in water and dilute to 500 mL or 0.5 L.

Lecture 3

6. Molality (m) molal concentration:

It is one of the concentration units which is one molal solution contains

one mole per 1000 g of solvent (mol/kg). It does not change with

temperature.

𝑚 =𝑚𝑜𝑙𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 (𝑚𝑜𝑙)

𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 (𝑘𝑔)

Example 3.1

Calculate the number of grams of NaCl (M.M. = 58.5 g/mol) in 50 mL of 0.25 M NaCl?

Solution

𝑀 =no. of moles of solu.

Vol. of Soln. (L)=

Wt. of solu. (g)

𝑀.𝑀. 𝑜𝑓 𝑠𝑜𝑙𝑢. 𝑔

𝑚𝑜𝑙

Vol. of Soln. (L)

0.25 =[𝑊𝑡. (𝑔)

58.5]

50 (mL)𝑥 10−3(𝐿) ⇒ 𝑊𝑡. (𝑔) = 0.73 𝑔

Example 3.2

How many mL should be taken from 0.1 M solution of Na2SO4 (M.M. = 142 g/mol) to obtain 5g of Na2SO4?

Solution

0.1 =(

5 𝑔142 𝑔/𝑚𝑜𝑙

)

Vol. (L)⇒ 𝑉𝑜𝑙. 𝐿 = 0.352 𝐿 = 352 𝑚𝐿

Example 3.3

Determine the molality of a solution prepared by dissolving 75.0g Ba(NO3)2(s) (261.32 g/mol)into 374.00g of water at 25oC.

Solution

𝑚𝑜𝑙𝑒 𝐵𝑎(𝑁𝑂3)2 =75 𝑔

261.32𝑔

𝑚𝑜𝑙

= 0.287 𝑚𝑜𝑙 ⇒ 𝑚𝑜𝑙𝑎𝑙𝑖𝑡𝑦 =0.287 𝑚𝑜𝑙

0.374 𝑘𝑔= 0.767 𝑚

Example 3.4

Calculate the weight of Na2SO4 (M.M. = 106 g/mol) required to prepare 250 mL

solution of 0.1 M Na+ using a Na2SO4 reagent that has a purity of 90% w/w?

Solution

Note that each mmole of Na2SO4 contains 2 mmoles of Na+. Therefore, the

following equation can be used:

𝑚𝑚𝑜𝑙 𝑜𝑓 𝑁𝑎2𝑆𝑂4 =mmol of 𝑁𝑎+

2=

0.1 𝑥 250

2= 12.5 𝑚𝑚𝑜𝑙

𝑤𝑡. 𝑜𝑓𝑝𝑢𝑟𝑒 𝑁𝑎2𝑆𝑂4 = 12.5 𝑚𝑚𝑜𝑙 𝑥 106𝑚𝑔

𝑚𝑚𝑜𝑙= 1325 𝑚𝑔 = 1.325 𝑔

𝑤𝑡. 𝑜𝑓 90% 𝑤/𝑤 𝑜𝑓 𝑁𝑎2𝑆𝑂4 =1.325 x 100

90= 1.47 𝑔

Example 3.5

Find the molarity of 21.4 m HF (20.01 g/mol). This aqueous solution has a densityof 1.101 g/mL.

Solution

21.4 m means 21.4 mol in 1 kg of solvent,

𝑚𝑎𝑠𝑠 𝑜𝑓 𝐻𝐹 = 21.4 𝑚𝑜𝑙 𝑥20.01 𝑔

1 𝑚𝑜𝑙= 428.21 𝑔 𝑜𝑓 𝐻𝐹 (𝑠𝑜𝑙𝑢𝑡𝑒)

𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑛. = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢. +𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑣. = 428.21 + 1000

= 1428.21 𝑔

𝑣𝑜𝑙. 𝑜𝑓 𝑠𝑜𝑙𝑛. 𝐿 = 1428.21 𝑔 𝑥1 𝑚𝐿

1.101 𝑔= 1297.193 𝑚𝐿 = 1.297 𝐿

𝑀 =𝑚𝑜𝑙𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 (𝑚𝑜𝑙)

𝑣𝑜𝑙. 𝑜𝑓 𝑠𝑜𝑙𝑛. (𝐿)=

21.4

1.297= 16.49 𝑚𝑜𝑙/𝐿

Example 3.6

Calculate the molar concentration of 2000 ppm of Pb2+ (A.M. = 207

g/mol)?

Solution

𝒑𝒑𝒎 = 𝑴 𝒙𝑴.𝑴. 𝒙 𝟏𝟎𝟎𝟎

𝑀 =𝑝𝑝𝑚

𝑀.𝑀. 𝑥 1000=

2000

207 𝑥 1000= 0.009 𝑚𝑜𝑙/𝐿

8. Normality or normal concentration (N):

The normal concentration refers either: the number of equivalent (eq.) of

solute per liter of a solution, or can also be obtained by dividing the number

of milliequivalent on milliliter of a solution.

𝑁 =Eq. of solu. (eq)

vol. of soln. (L)=

(𝑊𝑡. 𝑜𝑓 𝑠𝑜𝑙𝑢. 𝑔𝐸𝑞.M. 𝑔/𝑒𝑞

)

𝑉𝑜𝑙. 𝑜𝑓 𝑠𝑜𝑙𝑛. (𝐿)=

(𝑊𝑡. 𝑜𝑓 𝑠𝑜𝑙𝑢. 𝑚𝑔𝐸𝑞.𝑀. 𝑚𝑔/𝑚𝑒𝑞

𝑉𝑜𝑙. 𝑜𝑓 𝑠𝑜𝑙𝑛. (𝑚𝐿)

Relationship between molarity and normality for the same solute in the same

solution:

𝑀 =𝑚𝑜𝑙

𝑉𝑜𝑙. (𝐿)=

(Wt.𝑀.𝑀.

)

Vol. (L), 𝑁 =

𝐸𝑞.

𝑉𝑜𝑙. (𝐿)=

(𝑊𝑡.

𝐸𝑞.M.)

𝑉𝑜𝑙. (𝐿)

So

𝐸𝑞.𝑤𝑡. =M.M.

n⇒ 𝑁 =

(𝑊𝑡.𝑀.𝑀.𝑛

)

Vol. (L)= 𝑛 𝑥

(𝑊𝑡.𝑀.𝑀.

)

𝑉𝑜𝑙. (𝐿)

Thus: 𝑵 = 𝒏𝑴 n ≥ 1 therefore N ≥ M

Equivalent mass (Eq. M.): equivalent mass of a substance can be calculated

by dividing its molar mass per the number of active units in one molecule of

this substance (n) thus:

𝐸𝑞.M.=M.M.

no. of active unit in one molecule of solu. (n)

The active unit in the acid-base reactions are the number of hydrogen

ions liberated by a single molecule of an acid or reacted with a single

molecule of a base.

Acid: one definition of acid is any substance that ionizes in water to form

H+, includes monoprotic and polyprotic. Monoprotic acid like

hydrochloric acid (HCl) which include only one proton (hydrogen ion),

and polyprotic acid like sulfuric acid (H2SO4) that contains two protons

(hydrogen ion).

The active unit in the oxidation-reduction (redox) reactions is the

number of electrons (e) transferred from one reactant to another during

the reaction.

Example 3.7:

Calculate the normality of 0.53 g/100 mL solution of Na2CO3 (M.M. = 106

g/mol) as the following reaction:

Solution

𝑒𝑞. 𝑤𝑡 𝑜𝑓 𝑁𝑎2𝐶𝑂3 =106

2= 53 𝑔/𝑒𝑞

𝑁 =0.53 𝑔

53𝑔𝑒𝑞

𝑥 (100 𝑚𝐿 𝑥1𝐿

1000𝑚𝐿)=

0.53 𝑔 𝑥 1000

53 𝑔/𝑒𝑞 𝑥 100 (𝐿)= 0.1 𝑒𝑞/𝐿

Example 3.8

Calculate the normality of 5.267 g/L solution of K2Cr2O7 (M.Wt. = 294.2

g/mol) when K2Cr2O7 is reduced to Cr3+

Solution

To find the number of active unit in oxidation reduction reaction. Firstly, you

have to find the number of electrons has been transferred. So

K2Cr2O7 → Cr3+, K1+ O2-

+1 x 2 = 2 -2 x 7 = -14

The overall charge of a compound should be zero so the charge of Cr2 is

equal to 12, and the Cr = 12/2 = +6

K1+ Cr6+ O2-

+1 x 2 = +2 +6 x 2 = +12 -2 x 7 = -14

2+12-14 = 0

So the number of active unit (n) in this reaction is equal to 6, because in

each Cr, 3 electrons were transferred (Cr2 = 2 x 3 = 6 e-). It means six

electrons were transferred during the reaction.

𝐸𝑞.𝑤𝑡. 𝑜𝑓 𝐾2𝐶𝑟2𝑂7 =294.2 𝑔/𝑚𝑜𝑙

6 𝑒𝑞/𝑚𝑜𝑙= 49.03 𝑔/𝑒𝑞

𝑁 =5.267 𝑔

49.03 𝑔/𝑒𝑞 𝑥 1 𝐿= 0.1074 𝑒𝑞/𝐿

Example 3.9:

From the above example if the concentration of K2Cr2O7 solution is 0.5

M, calculate its normality?

Solution:

𝑁 = 𝑛 𝑀 = 6 𝑥 0.5 = 3 𝑒𝑞/𝐿

Density calculations

Density is the mass per unit volume at a specific temperature, usually g/mL

or g/cm3 at 20oC (remember 1 mL = 1 cm3)

Preparing Solutions

Preparing a solution of known concentration is perhaps the most common

activity in any analytical lab. The method for measuring out the solute and

solvent depend on the desired concentration unit and how exact the

solution’s concentration needs to be known. Pipets and volumetric flasks are

used when a solution’s concentration must be exact; graduated cylinders,

beakers and reagent bottles sufficient when concentrations need only be

approximate. Two methods for preparing solutions are described in below:

1. Preparing Stock Solutions

A stock solution is prepared by weighing out an appropriate portion of a

pure solid or by measuring out an appropriate volume of a pure liquid

and diluting to a known volume. Exactly how this is done depends on the

required concentration unit. For example, to prepare a solution with a

desired molarity you weigh out an appropriate mass of the reagent,

dissolve it in a portion of solvent, and bring to the desired volume. To

prepare a solution where the solute’s concentration is a volume percent,

you measure out an appropriate volume of solute and add sufficient

solvent to obtain the desired total volume.

Example 3.10

Describe how to prepare the following three solutions: (a) 500 mL of

approximately 0.20 M NaOH using solid NaOH; (b) 1 L of 150.0 ppm Cu using

Cu metal; and (c) 2 L of 4% v/v acetic acid using concentrated glacial acetic

acid (99.8% w/w acetic acid).

Solution

a) Since the mass of NaOH and the volume of solution do not need to be

measured exactly. The desired mass of NaOH is

𝑚𝑎𝑠𝑠 𝑜𝑓 𝑁𝑎𝑂𝐻 =0.2 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻

1 𝐿𝑥 0.5 𝐿 𝑥

40 𝑔 𝑁𝑎𝑂𝐻

1 𝑚𝑜𝑙= 4.0 𝑔 𝑁𝑎𝑂𝐻

To prepare the solution, place 4.0 grams of NaOH in a bottle or beaker and

add approximately 500 mL of distilled water.

b) Since the concentration of Cu has four significant figures, the mass of Cu and

the final solution volume must be measured exactly. The desired mass of Cu metal

is

𝑚𝑎𝑠𝑠 𝑜𝑓 𝐶𝑢 𝑚𝑒𝑡𝑎𝑙 =150 𝑚𝑔 𝐶𝑢

1 𝐿𝑥 1.000 𝐿 = 150.0 𝑚𝑔 𝐶𝑢 = 0.1500 𝑔 𝐶𝑢

To prepare the solution we measure out exactly 0.1500 g of Cu into a small

beaker and dissolve using small portion of concentrated HNO3. The resulting

solution is transferred into a 1-L volumetric flask. Rinse the beaker several times

with small portions of water, adding each rinse to the volumetric flask. This

process, which is called a quantitative transfer, ensures that the complete

transfer of Cu2+ to the volumetric flask. Finally, additional water is added to the

volumetric flask’s calibration mark.

c) The concentration of this solution is only approximate so it is not

necessary to measure the volumes exactly, nor is it necessary to account

for the fact that glacial acetic acid is slightly less than 100% w/w acetic

acid (it is approximately 99.8% w/w). The necessary volume of glacial

acetic acid is

𝑣𝑜𝑙. 𝑜𝑓 𝑠𝑜𝑙𝑢. 𝐶𝐻3𝐶𝑂𝑂𝐻 =4𝑚𝐿 𝐶𝐻3𝐶𝑂𝑂𝐻

100 𝑚𝐿𝑥 2000 𝑚𝐿

= 80 𝑚𝐿 𝐶𝐻3𝐶𝑂𝑂𝐻

To prepare the solution, use a graduated cylinder to transfer 80 mL of glacial acetic acid to a container that holds approximately 2 L and add sufficient water to bring the solution to the desired volume.