1. b) Preparation of solutions from known concentration of liquids.
In this case we don’t need for the first step about determination of concentration of bottle, only needed to the dilution step.
Only this relation we needed in these types of examples.
(M1 V1)concentrated = (M2 V2)diluted
known unknown known

2. Ex: Describe a method for the preparation of 0
Ex: Describe a method for the preparation of 0.1M HCl in 250ml volume from the concentrated acid solution (2M).
Ans:
(M1 V1)concentrated = (M2 V2)diluted
When
M1 = 2M
M2 = 0.1M
V2 = 250ml
V1 = ? x 250
(2 x V1)conc. = (0.1 x 250)dil V1 = = 12.5ml 2
That is mean for the preparation of 0.1M HCl in 250ml we dilute 12.5ml concentrated acid to 250ml in the proper volumetric flask.

3. Ex:
Prepare 0.5mol/L in 0.5L of KOH from the concentrated solution of 5mol/L.
Ans: V = 50ml
That is mean for the preparation of 0.5 mol/L (M) KOH in 0.5L we dilute 50ml concentrated KOH to 500ml in the proper volumetric flask.

4. Ex:
Describe the method for preparation of 500 ml of 0.03M K+ from the 0.7M K4Fe(CN)6
Ans: V = 5.35ml
Taking 5.35ml from the 0.7M K4Fe(CN)6 then diluted to 500ml with water to prepare 0.03M K+.

5. Ex:
From 1M FeCl3 prepare 0.06M Cl- ion in 100ml
Ans: V = 2ml
Taking 2ml from the 1 M FeCl3 then diluted to 100 ml with water to prepare 0.06M Cl- .

6. Q1) A concentrated solution of aqueous ammonia is 28% NH3 and has a specific gravity of and (M.Wt. = 17 g/mol). What is the molar concentration of NH3 in this solution, and how can you prepare 0.5M solution in 250ml volume of NH3?
Q2) Prepare 2mol/L in 100ml of H2SO4 from the concentrated solution (bottle). When percentage=98%, Sp.Gr.=1.83, M.Wt = 98g/mol.
Q3) Describe the method for preparation of 500 ml of 0.1M OH- from the 10M Ca(OH)2 .
Ca(OH) Ca OH-

7. Normality (N):
The number of equivalents of solute per liter of solution, or the number of milliequivalents in ml of solution.
no.of equivalent of solute no. of eq Wt(g)
N = = no. of eq.=
Volume of solvent by liter V(L) Eq.Wt
Wt(g) / Eq.Wt M.Wt
n: Is based on a reaction unit, which is that part of a chemical species involved in a reaction.
N = Eq.Wt =
V(L) n Or
no.of milliequivalent of solute no. of meq.
N = =
Volume of solvent by milliliter V(ml)

8. Equivalent weight for different reaction types:
1- Acid-base reaction (neutralization reaction)
Reactive species (n) in acid base reactions were H+ and OH-, respectively.
M.Wt of acid M.Wt of base
Eq.Wt of acid = Eq.Wt of base =
no.of H no.of OH-
Ex:
For HCl no.of H+ =1, n= For NaOH no.of OH- =1, n=1
For H2SO4 no.of H+ =2, n= For Ca(OH)2 no.of OH- =2, n=2
For H3PO4 no.of H+ =3, n= For Al(OH)3 no.of OH- =3, n=3

9. 2- Precipitation reaction
Reactive species (n) in this case was cation or anion charge
M.Wt of salt
Eq.Wt of salt =
no.of cation or anion x charge
Ex:
For NaCl n= 1 x For Na2CO3 n= 2 x1
For FeSO4 n= 1 x For Fe2(SO4)3 n= 2 x 3

10. 3- Oxidation-reduction (Redox) reaction
Reactive species (n) was the number of electrons donated or accepted in the reaction.
M.Wt
Eq.Wt =
no.of donated or accepted electrons
Ex:
Fe Fe n= Mn Mn6+ n= 1
Cr Cr n= Mn n= 3
Mn n= 5

11. Q1) Calculate the number of milliequivalent for Oxalic acid (H2C2O4
Q1) Calculate the number of milliequivalent for Oxalic acid (H2C2O4.2H2O , M.Wt. = 126.1) in 0.5 g pure material.
Q2) Calculate the normality of sodium sulphate salt (Na2SO4) in a 1L solution that contains 165 mg (Na2SO4) [M.Wt = 142 g/mol].
Q3) Prepare 500 milliliter of 0.1N Al2(SO4)3 [M.Wt = 342 g/mol] from the solid salt material.
Q4) Prepare 500ml (6N) from the concentrated H3PO4 .
Sp.gr. = , percentage = %80 , M.Wt = 98 g/mol.

12. Density and Specific gravity of solution
Density (d): Is the weight (g) of a substance per unit volume (mL).
Specific gravity (sp.gr): Is the ratio of the density of substance to the density of an equal volume of water.

13. The density of water is approximately 1
The density of water is approximately 1.00 g/mL at 4◦C, since we employ density and specific gravity interchangeably.
Sp.gr ≈ density of substance at 4◦C
For water d = 1.00 = wt (g) / V (mL) wt(g) = V(mL)

14. Ex:
Calculate weight of 100ml HCl. If you know that (dwater = 1g/ml and Sp.gr.(HCl) = 1.18)
Ans: = 118g

15. Ex:
The weight of 1L from H3PO4 is equal to 1710g, what is specific gravity of H3PO4 ?
Ans: = 1.71

16. Ex:
Specific gravity of 0.5 kg liquid HNO3 is equal to 1.4, calculate volume of this solution in (ml).
Ans: = ml

17. Percentage concentrations:
a) Weight-weight percent (% w/w):
Grams (g) of solute per 100 g of solution.
Weight of solute (g)
Weight% = % (w/w) = x 100
Weight of solution (g)
For example:
65% HNO3 (w/w): Means that 100 g of solution contains 65g of HNO3 .

18. Ex:
How can you prepare 100ml of a solution of 5%(w/w) sodium hydroxide (NaOH)?
Ans:
It can be prepared by addition of 5g of the sodium hydroxide to a little amount of distilled water then dilute the solution to 100ml (final volume).

19. Ex:
How can you prepare 200ml of a solution of 15%(w/w) potassium iodide (KI)?
Ans:
It can be prepared by addition of 30g of the (KI) to a little amount of distilled water then dilute the solution to 200ml (final volume).

20. Ex:
Express the weight percent %( w/w) for a 200g solution which contain 25g of KCl?
Ans: %

21. Ex:
Calculate the weight percent for a solution prepared by dissolving 5g AgNO3 in 100mL of distilled water (density of water = 1 g/mL)?
Ans = 5 %

22. Ex:
Calculate the weight percent for a solution prepared by dissolving 35g Fe(NO3)3 in 250mL of HNO3 (density of HNO3 = 1.4 g/mL)?
Ans = 10 %

23. Ex:
Calculate weight of solute (NaCl) in a 20% (w/w) 500g of solution.
Ans = 100 g