1

Find the number of grams of Na2SO4 required to

prepare 500 mL of 0.1 M solution.

 

First, we find mmoles needed from the relation

mmol = M (mmol/mL) x VmL

 

mmol Na2SO4 = 0.1 mmol/mL x 500 mL = 50 mmol

 

mmol = mg substance/FW substance

 

?mg Na2SO4 = 50 mmol x 142 mg/mmol = 7100 mg or

7.1 g

2

One can use dimensional analysis to find the answer in one step as

 

? g Na2SO4 = (0.1 mol Na2SO4/1000 mL) x

500 mL x (142 g Na2SO4/mol) = 7.1 g•

3

When two or more solutions are mixed, one can find the final concentration of each ion. However, you should always remember that the number of moles ( or mmoles ) is additive. For example

4

Find the molarity of K+ after mixing 100 mL of 0.25 M KCl with 200 mL of 0.1 M K2SO4.

 The idea here is to calculate the total mmol of K+ and divide

it by volume in mL.

 mmol K+ = mmol K+ from KCl + mmol K+ from K2SO4

= 0.25 mmol/ml x 100 mL + 2x0.1 mmol/mL x 200 mL = 65 mmol

Molarity = 65 mmol/(200 + 100) mL = 0.22 M

Note that the concentration of K+ in 0.1M K2SO4

is 2x0.1M (i.e. 0.2 M)

5

Normality

We have previously talked about molarity as a method for expressing concentration. The second expression used to describe concentration of a solution is the normality. Normality can be defined as the number of equivalents of solute dissolved in 1 L of solution. Therefore, it is important for us to define what we mean by the number of equivalents, as well as the equivalent weight of a substance as a parallel term to formula weight.

6

An equivalent is defined as the weight of substance giving an Avogadro’s number of reacting units. Reacting units are either protons (in acid base reactions) or electrons (in oxidation reduction reactions). For example, HCl has one reacting unit (H+) when reacting with a base like NaOH but sulfuric acid has two reacting units (two protons) when reacting completely with a base.

7

Therefore, we say that the equivalent weight of HCl is equal to its formula weight and the equivalent weight of sulfuric acid is one half its formula weight. In the reaction where Mn(VII), in KMnO4, is reduced to

Mn(II) five electrons are involved and the equivalent weight of KMnO4 is equal to its

formula weight divided by 5.

8

Number of equivalents = Normality x VL = (eq/L) x L

 Number of milliequivalents = Normality x

VmL = (meq/mL) x mL Also, number of equivalents =

wt(g)/equivalent weight (g/eq) Or, number of milliequivalents =

wt(mg)/equivalent weight (mg/meq)

9

Equivalent weight = FW/n

No. eq = wt/eqw substitute for eqw = FW/n gives: No. eq = wt/(FW/n), but mol = wt/FW, therefore: No. eq = n x number of moles dividing both sides by volume in L, we get:N = n M Where n is the number of reacting units ( protons or

electrons ) and if you are forming factors always remember that a mole contains n equivalents. The factor becomes (1 mol/n eq) or (n eq/1 mol).

• 

10

One last thing to keep in mind is that when dealing with normality problems always 1 eq of A reacts with 1 eq of B regardless of the stoichiometry of the reaction since this stoichiometry was used in the calculation of normalities.

11

Find the equivalent weights of NH3 (FW = 17.03),

H2C2O4 (FW = 90.04) in the reaction with excess

NaOH, and KMnO4 (FW = 158.04) when Mn(VII)

is reduced to Mn(II).

 

Solution

 

Ammonia reacts with one proton only

 

Equivalent weights of NH3 = FW/1 = 17.03 g/eq•  

12

Two protons of oxalic acid react with the base

 

Equivalent weights of H2C2O4 = FW/2 =

90.04/2 = 45.02 g/eq

 

Five electrons are involved in the reduction of Mn(VII) to Mn(II)

 

Equivalent weights of KMnO4 = FW/5 =

158.04/5 = 31.608 g/eq

13

Find the normality of the solution containing 5.300 g/L of Na2CO3 (FW = 105.99),

carbonate reacts with two protons. 

Normality is the number of equivalents per liter, therefore we first find the number of

equivalents 

eq wt = FW/2 = 105.99/2 = 53.00eq = Wt/eq wt = 5.300/53.00 = 0.1000

N = eq/L = 0.1000 eq/1L = 0.1000 N

14

The problem can be worked out simply as below

 

? eq Na2CO3 /L = (5.300 g Na2CO3 /L ) x (1 mol Na2CO3 /105.99 g Na2CO3 ) x (2 eq

Na2CO3 /1 mol Na2CO3) = 0.1 N

A further option is to find the number of moles first followed by multiplying the

result by 2 to obtain the number of equivalents. 

15

The other choice is to find the molarity first and the convert it to normality using the

relation

N = n M 

No of mol = 5.300 g/(105.99 g/mol)

M = mol/L = [5.300 g/(105.99 g/mol)]/ 1L

 

N = n M = 2 x [5.300 g/(105.99 g/mol)]/ 1L = 0.1000

16

Example

Find the normality of the solution containing 5.267 g/L K2Cr2O7 (FW = 294.19) if Cr6+ is reduced to

Cr3+. 

The same as the previous exampleN = eq/L, therefore we should find the number of eq where eq = wt/eq wt, therefore we should find

the equivalent weight; where eq wt = FW/n. Here; each Contributes three electrons and

since the dichromate contains two Cr atoms we have 6 reacting units

 

17

Eq wt = (294.19 g/mol)/(6 eq/mol) 

Eq = 5.267 g/ (294.19 g/mol)/(6 eq/mol) 

N = eq/L = (294.19 g/mol)/(6 eq/mol)/1L = 0.1074 eq/L

 

Using the dimensional analysis we may write 

? eq K2Cr2O7 /L = (5.267 g K2Cr2O7 /L) x (mol

K2Cr2O7 /294.19 g K2Cr2O7 ) x (6 eq K2Cr2O7

/mol K2Cr2O7 ) = 0.1074 eq/L

18

Again one can choose to calculate the molarity then convert it to normality

 

mol = 5.267 g/(294.19 g/mol)

 

M = mol/L = [5.267 g/(294.19 g/mol)]/L

N = n M

N = (6 eq/mol)x [5.267 g/(294.19 g/mol)]/L = 0.1074 eq/L