Clinical Chemistry MLS

1. Preparation of Solutions
By
Amir Mohammed Albushra
MBBS, MSc in Clinical Chemistry.

2. Solution
A solution is a homogeneous mixture composed of two or more
substances. In such a mixture, a solute is dissolved in another
substance, known as a solvent

3. Solute:
The substance which dissolves in a solution
Solvent
The substance which dissolves another to form a solution
Saturation
Saturation is the point at which a solution of a substance can dissolve
no more of that substance and additional amounts of it will appear as a
precipitate.

4. Types of solutions
• Percentage solution
• Molar solution
• Normal solution

5. A)Weight/volume solution % (w/v)
Weight-volume percentage, (sometimes referred to as mass-volume
percentage and often abbreviated as % m/v or % w/v) describes the
mass of the solute in g per 100 ml of the resulting solution.
Example:
Preparation of 10% (W/V) NaCl solution
Calculations:
Wt of solute (g )=
C (required Conc.) X V of solution (ml)
100
So 10% (W/V) NaCl solution has 10 grams of sodium chloride dissolved
in 100 ml of solution.

6. Procedure:
1) Weigh 10g of sodium chloride.
2) Pour it into a graduated cylinder containing about 80ml of water.
3) Once the sodium chloride has dissolved completely add water to
bring the volume up to the final 100 ml.
Note
Do not simply measure 100ml of water and add 10g of sodium
chloride. This will introduce error because adding the solid will change
the final volume of the solution and throw off the final percentage.

7. B) Volume /volume solution % (v/v)
Volume-volume percentage (abbreviated as % v/v) describes the
volume of the solute in ml per 100 ml of the resulting solution.
Example:
• Preparation of 30% (V/V) sulfuric acid
Calculations:
V1 of solute (ml ) =
C (required Conc.) X V Solution(ml)
100
Volume of Solution (mL)= volume of Solute (mL) + volume of Solvent (mL)
So 30% (V/V) sulfuric acid has 30 ml of sulfuric acid dissolved in 70 ml
of water.

8. Procedure:
• Calculate the required volume of solute
• Subtract the volume of solute from the total solution volume
• Dissolve 30 ml sulfuric acid in a 70 ml of water to bring final volume
of solution up to 100ml.

9. Percentage solution
C) Weight/ Weight solution % (w/w)
This type of solution is rarely if ever prepared in the laboratory since it
is easier to measure volumes of liquids rather than weigh the liquid on
an analytical balance.
This type of percent solution is usually expressed as (w/w), where "w"
denotes weight (usually grams) in both cases.
Example:
An example of a correct designation for this type of solution is as
follows: 10
g/100 g (w/w), which indicates that there are 10 grams of solute for
every 100 grams total

10. Molar solution
Molarity
Number of moles of a given substance dissolved per liter of solution.
Molar solution
It is a solution that contains 1 mole of solute in each liter of solution.
Units of molarity
mol/L.

11. 1 mol/l = 1000 mmol/l = 1000000 μmol/l = 1000000000 nmol/l

12. Wt of solute (g) = M(mole/L) X M.wt (g/mole) X V (L)
▪M: the required molarity
▪ M.wt: molecular weight of the solute
▪ V: total volume in liters

13. Example:
• Preparation of 1M NaCl solution in 1000 ml
Calculations:
Dissolve 58.5 g of NaCl in a 1000 ml (1 liter) of water to prepare 1M
NaCl

14. Normal solution
Equivalent weight
An equivalent weight is equal to the molecular weight divided by the
valence.
Normal &Normality
A normal is one gram equivalent of a solute per liter of solution.
Normal solutions
Is a solution that contains 1 gram equivalent weight (gEW) per liter
solution.

15. Units of Normality
The units for normal concentration are Eq/L.

16. Eq.wt for Acids and Bases
Eq.wt =
M.wt of acid or base
No of H+ in acids or OH− in bases
Examples:
• HCL the MW= 36.5 the EW = 36.5
• H₂SO4 the MW = 98 the EW = 49
• H3PO4 the MW = 98 the EW = 32.7
• NaOH the MW = 40 the EW = 40
• Ca (OH)₂ the MW = 74 the EW = 37

17. Eq.wt for salts
Eq.wt =
𝑀𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑎𝑙𝑡
𝑁𝑜 𝑜𝑓 𝑐𝑎𝑡𝑖𝑜𝑛𝑠 ×𝑣𝑎𝑙𝑒𝑛𝑐𝑦
or
𝑀𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑎𝑙𝑡
𝑁𝑜 𝑜𝑓 𝑎𝑛𝑖𝑜𝑛𝑠 ×𝑣𝑎𝑙𝑒𝑛𝑐𝑦
Examples
• KCL the MW= 74.55 the EW = (74.55/1 X 1) = 74.55
• CaCl₂ the MW= 110.98 the EW (110.98/1 X 2) or (110.98/2 X 1) = 55.49
• Na₂CO3 the MW= 105.98 the EW = (105.98/1 X 2) or (105.98/2 X 1) =
52.99

18. Preparation of Normal solution
▪N: the required normality
• Eq.wt: equivalent weight of the solute
▪V: total volume in liters.
Wt of solute (g )= N (Eq/l) X Eq.wt XV (L)

19. Examples:
• Preparation of 1N NaOH solution
• Preparation of 1N Ca(OH)₂ solution
• Preparation of 1N KCL solution
Procedure
• Dissolve 40 g of NaOH in a 1000 ml (1 liter) of water to prepare 1N NaOH
solution.
• Dissolve 37 g of Ca (OH)2 in a 1000 ml (1 liter) of water to prepare 1N Ca(OH)₂
solution.
• Dissolve 74.55 g of KCI in a 1000 ml (1 liter) of water to prepare 1N KCI solution.

20. Dilution
• It is used when we need to make a specific volume of known
concentration from stock solutions.
• To do this we use the following formula:
V°=
RV
𝑂
• V°: the volume of stock we start with.
• O: Original concentration of stock solution.
• V: Required volume needed.
• R: the required concentration( new concentration.
( V= V°+ volume of diluent).

21. Example
Suppose we have a stock solution with concentration of 100 mg/dL and
we want to make 200 ml of solution having 25 mg/dL.
V°=
RV
𝑂
V°=
25mg/dL 𝑋 200𝑚𝐿
100𝑚𝑔/𝑑𝐿
= 50ml from stock
V= V°+ volume of diluent 200ml = 50ml + volume of diluent
volume of diluent= 200 ml ‒50 ml = 150ml
So, we would take 50 ml stock solution and dilute it with 150 ml of
solvent to get the 200 ml of 25 mg/ ml solution needed