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Solutions
A solution is formed when two substances are mixed together and
here are some examples:-
In ( Solid + Liquid ) = Solution
Hospital ( Liquid + Liquid ) = Solution
( Solid + Solid ) = Solution
Solid = Solute )
+ = Solution
Liquid = Solvent)
Example - Saline = Salty water
SYMBOL 175 \f "Symbol" SYMBOL 173 \f "Symbol"
Salt + Water = Solution
Solute Solvent
Chlorhexidine Solution
SYMBOL 175 \f "Symbol"
Chlorhexidine + Water = Chlorhexidine Solution
Solute + Solvent
Alcohol is sometimes used as solvent
When alcohol is used, the Solution is usually called a tincture -
Nurses -
Prepare solution for disinfecting purposes
Penicillin - adding water to drugs for injection
Solutions to apply to the skin :
SYMBOL 92 \f "Symbol" Needs accurate mixing
Two abbreviations: -
V/V Solution
W/V Solution
1. V/V Solution
Volume to volume solutions are those in which the amount of solute and the amount of solvent can both be measured in volume units.
e.g. ml or Litres
2. W/V Solutions are those in which the solute is a solid and must be measured in weights units.
The solvent is measured in volume units.
For calculation - The strength of the solvent is converted to weight units.
Ratio - Strengths
Strength of a solution can be expressed in 3 ways -
1. Ratio strength
2. Percentage strength
3. Dose strength.
1 - Ratio-Strength
To find the Ratio strength of a solution -
The Total volume of Solution is divided by volume of Solute
and the answer is expressed in 1 in whatever the figure may be.
Dilution of Lotions
2 Rules -
1. Ensure that the strength required and the strength available are expressed in the same units before you work out the amount
required.
2. Divide the strength required by the strength available and multiply by the volume you are preparing. This gives the amount of concentrate you will need to make up the Final Solution.
Strength Required x Final Volume
Strength available
There are times when a nurse may be called upon to prepare a weak lotion from a more concentrated solution which she has in the ward cupboard.
2 Rules which must be used to complete the calculation.
Ensure that the strength required is in the same unit before you work out the amount required. That is the strengths must either be all expressed as proportions or all as percentages. If necessary, convert one to the other before applying the formula.
2. Divide the strength required by the strength available and multiply by the volume you are preparing. This gives the amount of concentrate you will need to make up the final solution.
Strength required x final volume
Strength available
= amount of concentrate required to make up final solution
An aside
In summary, the concept of concentration is about the number of particles present in a given volume.
There are two ways of determining this:-
Weight approach
Example:
Imagine a container which contains a number of marbles and it weighs 1000 g and the weight of each marble is 10 g
You can find the number of marbles in the container by dividing the total weight of the container by the weight of each marble.
i.e. EMBED Equation.DSMT4 = 100 marbles
By counting each marble at a time
i.e. Take one marble out at a time (Boring?)
Similarly the number of particles present in a solution can be calculated.
However, with the introduction of the SI units, this has been made easier by stipulating the number of particles, thus eliminating the need to count them.
Avogadros Number
This is a large convenient number depicting the number of particles and is given as 6 x 1023
In Chemistry, large number of molecules exists in a small sample of substance and if 6x 1023 molecules is present in a container, then one can say that there is a mole of molecules present. Therefore, if there is 6 x 1023 particles present, the term one mole is used.
Relationship between Moles and Concentration
The concentration of a solution is determined by the number of moles of substance present in one litre of that solution.
This concentration is referred to as a molar solution
Remember:
One Molar solution contains one mole of substance in one Litre of that solution.
Molecular Weight
This is the weight of a molecule. It is the total weight of each atom which makes up the molecule.
For Example:-
A glucose molecule is composed of 6 atoms of Carbon, 12 atoms of Hydrogen and 6 atoms of Oxygen and is given the formula C6H12O6.
In order to find the molecular weight of Glucose, we need to find the atomic mass of each element present in glucose.
The atomic mass of each element, i.e. Carbon, Hydrogen, Oxygen, etc is listed in the periodic table and can usually be found in any anatomy & physiology books which include the chemistry of the human body.
Once the atomic mass is obtained, the molecular weight of Glucose can be worked out
ElementNumber of atomsAtomic MassTotalCarbon6126x12= 72Hydrogen12112x1= 12Oxygen6166x16= 96Sum= 180
The Molecular weight of glucose is therefore 180 and one mole of glucose is present in 180 g.
Therefore a one molar solution of glucose contains 180 g of glucose in one litre of solution.
Human Body
The concentration of substances within the body and used in intravenous fluid is normally less than a mole
Therefore other units such as the millimole per litre, micromole per litre and nanomole per litre are used.
Further reading
Hinwood B 1993 A textbook of science for the health professions. London. Chapman & Hall.
Example 1 - V/V Solution
Using a measuring jug pour out 20 ml of red ink. Empty this into a litre jug. Add water to the ink to make up a total of one litre.
SYMBOL 92 \f "Symbol" You have now made V/V solution of red ink.
Strength is 20 ml of Red in 1 litre: This is 1 in 50
Example 2
W V
Add 4 g of salt to 300 ml of water.
1 ml of water weighs 1 g SYMBOL 92 \f "Symbol" 300 ml of water weigh 300 g
Weight of solution = 300 g + 4g = 304 g
Water + Salt
SYMBOL 92 \f "Symbol"4 g of salt in 304 g which is 1 in 76g
Example :-
Find the Ratio strength of a solution in which 7 ml of pure substance have been added to 203 ml of Water
Volume of solute = 7 ml
Total Volume of solution = 203 + 7 = 210 ml
SYMBOL 92 \f "Symbol" Strength =Total Volume of Solution = 210 = 30
Volume of Solute 7
SYMBOL 92 \f "Symbol" Ratio Strength = 1: 30
SYMBOL 92 \f "Symbol" 30 part altogether but 1 part of that is pure drug
Find the Ratio Strength when 300 ml of eusol is made up to 3 litres with water.
Volume of solute = 300 ml
Total Volume of solution = 3L = 3000 ml
Ratio = 3000 = 10
300
SYMBOL 92 \f "Symbol" Strength = 1: 10 SYMBOL 92 \f "Symbol" 10 parts altogether 1 part of that is pure
Examples
Prepare 100 ml of 2% solution using a concentrate of 1 in 5
Strength required = 2%
Strength available = 1 in 5 = 20%
SYMBOL 92 \f "Symbol" using formula
2 x 100 ml = 10 ml
20
Take 10 ml of concentrate and dilute to 100 ml with water
Converting the other way -
Strength required = 2% = 1 in 50
Strength available = 1 in 5
SYMBOL 92 \f "Symbol" 1 1 x 100 = 10 ml
50 5
Self Assessment Questions
1. Dilute a 1 in 500 solution to 1 in 4000 solution and prepare 800 ml
Prepare 600 ml of a 1 in 3000 solution using a 0.5% concentrate
Self assessment Question
How many molecules will be present if there were 4 moles of a substance in a container?
Self assessment Question
How many moles of substance would there be in a 2 Molar solution?
Self assessment Question
Find the molecular weight of Sodium Chloride with the formula NaCL, i.e. one atom of sodium and one atom of Chlorine
Self assessment Questions
What fractions of a mole are the following:-
A millimole = of a mole----- Problem-refer to the section on indices
A micromole= of a mole
A nanomole = of a mole
How many grams of glucose would there be in a one millimolar solution of glucose?
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