Solutions:- Part 1 - Solutions Preparation used in Clinical Laboratory, and Dilution Formulas (2024)

Definition of the solution:

• A solution is a hom*ogenous mixture of one or more substances called solutes dispersed molecularly in a sufficient quantity of dissolving medium called solvent.
  • When a solution holds as much dissolved solute as it can at a specific temperature, it is called a saturated solution.
  • The solution may be gaseous, liquid, or sold.

Summary of the different solutions:

• Mole = Mass in grams (g)/gram molecular weight (g).
• The molarity of a solution = Numbers of moles of solute/Number of liters of solution.
• The molality of a solution = Number of moles of solutes Number of kilograms of solvent.
• Normality of a solution = Number of gram equivalents of solute/Number of liters of solution.
• Gram equivalent weight = Weight of the formula of substance/ Difference in oxidation state

Saturation of the solution depends upon the following:

1. Temperature.
2. Atmospheric pressure.
3. Nature of the solute.
4. Nature of the solvent.

Solutions Preparation precautions foraccuracy:

1. Us the balance of good sensitivity.
2. Use chemicals of analytical grade. Always use pure reagents.
3. In the case of hygroscopic chemicals, weigh those rapidly.
4. Use calibrated, clean glassware.
5. Use carefully automatic pipettes.
6. Clean the weighing containers, and if possible, clean them with the solvent.
7. Ensure the bottom of the fluid meniscus is on the graduation mark when viewed at eye level.

Solutions Preparation: Precaution to read the solution accurately.

The following are the type of solutions commonly used in the clinical laboratory:

Percent solution

1. This is defined as parts per 100, representing the percent (%).
   1. This is independent of the molecular weight of the substance.
   2. This is expressed as a solute concentration as a percent % (per hundred parts of the total solution).
2. If you want to make a 5% glucose solution, you will add 5 grams of glucose in 100 ml of distal water.

Percent solution types are:

Weight/weight (weight per unit weight)

1. The % of a solute or grams of solute per 100 grams of the final solution.
2. Both solutes and the solvent are weighed, and the total equals 100 g.
   1. Example:
   2. To make 5% of an aqueous solution of NaCl containing 50 grams of NaCl and 950 g of diluent.
3. Calculation =
   1. 5% = 5/100 = 0.05
   2. Therefore 0.050 x 100 = 5 grams of NaCl.

solutions preparation: Solution weight/weight

1. Another example is:
   1. 5 grams of Na₂SO₄ dissolvedin 95 grams of water (roughly 95 mL). Totalweight of 100 grams (solute + solvent).

Weight/volume (weight per unit volume):

1. This is an expression of weight (mass) per unit volume. This W/V is often used when the solid chemical is diluted in liquid. Weight per unit volume is expressed as g%.
2. It is grams per dL (g/dL), milligrams per dL (mg/dL), or µg/dL. In this case, SI units are weight per µL, or liter (L).
3. Examples:
4. This is usually expressed as gram/100 mL of diluent.
   1. To make a 10 % solution, add 10 grams of the substance to a final volume of 100 mL of liquid.
5. If you want to prepare 100 mL of 100 g/L of NaCl.
   1. Weigh 10 g of NaCl and dilute to volume in a 100 mL flask.

Volume/volume (volume per unit volume V/V):

1. This is convenient for the compositionof twoliquids.
2. Example:
3. If you want to make 50 mL of 2% HCl.
   1. Calculation:
      1. 0.20 x 50 = 1 mL
      2. Therefore add 49 mL of water to 1 mL of HCL.

solutions preparation: Solution volume by volume

4. 5% of the glacial acetic acid solutions:
   1. 5 mL of glacial acetic acid diluted with distal water to a volume of 100 mL.

Molar solution

1. This is defined as units of moles per liter (mol/L).
2. The symbol M indicates molarity is replaced by mol/L
   1. or millimoles/millimeter (mmol/mL).
3. 1 Mol of a substance = gram molecular weight of that substance.
   1. Example:
      1. Make up 250 mL of a 4.8 molar solution of HCl.
      2. HCL molar weight = 36.5 g.
      3. 36.5 HCL/mol x 4.8 mol HCL/L x 250/1000 mL = 43.8 g HCL
      4. 250 mL H2O + 43.8 = 4.8 Molar solution.
   2. The One molar solution of H2SO4:
      1. It contains 98.08 g/L of the solution.

Normal solution

1. Normality (normal solution) is a gram equivalent weight per liter (eq.wt/L).
   1. OR milliequivalent weight/milliliter (meq /mL).
2. Equivalent weight = gram weight/valency.
3. Example:
   1. NaCl gram weight = 58 gram and valency = 1
   2. 58/1 = 58 grams equivalent weight per liter.

Simple Dilution

1. This is defined as the total volume desired and the amount of stock needed.
2. There is most commonly used equation is:
   1. V1 x C1 = V2 x C2
   2. Where V1 is volume, C1 is the concentration of solution 1, and V2 and C2 are the concentration and volume of the diluted solution.
   3. The basic equation is V1/V2 = C1/C2
   4. Using the above equation, C1 x V1 = C2 x V2
   5. Example: Prepare 250 mL of 0.1 M HCL from stock 1 M HCL.
      1. V1 = unknown
      2. V2 = 250 mL
      3. C1 = 1.0 mol/L
      4. C2 = 0.1 mol/L
         Use the formula C1 x V1 = C2 x V2 = Where V1 is unknown.
   6. V1 = 0.1 x 250 /1.0 = 25 mL
   7. Measure 25 ml of 1 M HCL; dilute to 250 mL with distle water.
   8. This diluted solution has a concentration of 0.1 M HCL
3. Another Example:
4. 1: 10 dilution = ratio of 1 : 9 = 1 : 10 = one part of serum + 9 parts of diluent.
   1. 100 μL + 900 μL of saline.
   2. 1 mL serum + 9 mL of saline.
   3. 2 mL serum + 18 mL of saline.

Simple dilution calculation method

Serial Dilution

1. This is defined as multiple progressive dilutions ranging from a more concentrated solution to a less concentrated solution.
2. Serial dilutions are useful in a small quantity of serum to find the titer of antibodies.
3. The first dilution is made just like the simple dilution.
4. Now subsequent dilutions are made from each preceding dilution.
5. If you want to make a serial dilution of 1: 2, 1: 4, 1: 8, 1: 16, and so on.
   1. The total volume is fixed; suppose it is 1 mL.
   2. (Initial dilution factor) (next dilution factor) = final dilution factor
   3. 1 : 2 x 1 : 2 = 1 : 4.
6. Procedure
   1. First tube = (1 mL serum + 1 mL diluent) = 1 : 2
   2. Second Tube = 1 mL from from ist tube + 1 mL diluent = 1 : 4
   3. Third tube = 1 mL from tube 2 + 1 mL diluent = 1 : 8
   4. Fourth tube = 1 mL from tube 3 + 1 mL diluent = 1 : 16

Serial dilution calculation method

Question 1: What is the normal solution?

Question 2: What is the molar solution?