# Making solutions

**Introduction to preparation of solutions.**

Many experiments involving chemicals call for their use in solution form. That is, two or more substances are mixed together in known quantities. This may involve weighing a precise amount of dry material or measuring a precise amount of liquid. Preparing solutions accurately will improve an experiment's safety and chances for success.

**Solution 1:** Using percentage by weight (w/v)

The percentage concentration refers to the **mass or volume of solute** in a final volume of solution.

**Example**
So, to make up a 1% ‘w/v’ solution of a solid, (1 g solid in this case - glucose) will be need to be dissolved in, say, 95 ml of solvent (water in this case I assume) and the resulting solution made up to 100 ml with water.

**Another Example**
The formula for Weight percent (w/v) is: Mass of solute (g) / Volume of solution (cm^{3}) x 100

Example:

A 10% NaCl solution has 10g of sodium chloride dissolved in 100cm^{3}of solution.

**Procedure**

- Weigh out 10g of sodium chloride.
- Pour it into a volumetric flask containing about 80cm
^{3}of water. - Once the sodium chloride has dissolved completely
*(use a electrical stirrer if necessary)*, add more water to bring the volume up to the final 100cm^{3}mark.

Caution: 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.

**Solution 2:** Using percentage by volume (v/v)

When the solute is a liquid, it is sometimes convenient to express the solution concentration as a volume percent.

**What you should remember!**:
The formula for volume percent (v/v) is: volume of solute (ml) / Volume of solution (ml) x 100

Example:

Make 1000ml (1 litre) of a 5% by volume solution of ethylene glycol (ethane-1,2-diol) in water.

**Procedure**

- First, express the percent of solute as a decimal: 5% = 0.05
- Multiply this decimal by the total volume: 0.05 x 1000ml = 50ml (of solution needed).
- Subtract the volume of solute (ethylene glycol) from the total solution volume:
- 1000ml (total solution volume) - 50ml (ethylene glycol volume) = 950ml (water needed)
- Dissolve 50ml ethylene glycol in a little less than 950ml of water.
- Now bring final volume of solution up to 1000ml with the addition of more water. (This eliminates any error because the final volume of the solution may not equal the calculated sum of the individual components).
- So, 50ml ethylene glycol / 1000ml solution x100 = 5% (v/v) ethylene glycol solution.

**Solution 3:** Molar Solutions

Molar solutions are the most useful in chemical reaction calculations because they directly relate the moles of solute to the volume of solution.

**What you should remember!:**
The formula for molarity (M) is: Mole of solute / 1 litre of solution (or gram-molecular mass of solute) / 1 litre of solution.

Example 1;

The molecular weight of a sodium chloride molecule (NaCl) is 58.44, so one gram-molecular mass (= 1 mole) is 58.44g. We know this by looking at the periodic table. The atomic mass (or weight) of Na is 22.99, the atomic mass of Cl is 35.45, so 22.99 + 35.45 = 58.44.

If you dissolve 58.44 g of NaCl in a final volume of 1 litre, you have made a 1M NaCl solution, a 1 molar solution.

**Procedure**

To make molar NaCl solutions of other concentrations dilute the mass of salt to 1000ml of solution as follows:

- 0.1M NaCl solution requires
**0.1**x 58.44 g of NaCl = 5.844g - 0.5M NaCl solution requires
**0.5**x 58.44 g of NaCl = 29.22glite - 2M NaCl solution requires
**2.0**x 58.44 g of NaCl = 116.88g

Example 2;

How would you dilute concentrated propanone to 1 molar for the kinetics experiment with iodine and propanone?

1 molar propanone would be 58 g in 1 litre of solution. So you could weigh out 58g, add it to some distilled water and add more distilled water to bring the volume up to exactly 1 litre. Alternatively, as the density of propanone is 0.789 g/ cm^{3}, you could measure out 73.5 cm^{3}.

Example 3;

You are asked to make up 0.1M solution of propanoic acid for a freezing point experiment.

A one molar solution is made up by **weighing out 74.06g** of propanoic acid into a stopped weighing bottles ideally in the fume cupboard with the fan off whilst weighing. Add washing of the stopped weighing bottles to the volumetric flask, and finish of as normal. This would be called your Standard or stock solution. From this you can dilute using distilled water to any other molar strength you wish. Alternatively you could simply weigh out 74g of propanoic acid and dilute to 1 litre with distilled water.

**Solution 4:** Working with "Specific Gravity"

Specific gravity or relative density is an archaic method of referencing strengths of solutions. However, because of the ease of use, it is still common in certain industries. Calculations can be treated in a similar way to concentration.

This method **does not work where volumes are not conserved**, For example: adding water to ethanol.

Here 50.00cm^{3} of pure ethanol + 50.00 cm^{3} of water will not give 100.00 cm^{3}.

Consider the case of diluting concentrated sulphuric acid (specific gravity 1.84 *(no units as this is relative to water)*) to give a solution of specific gravity 1.25.

- Defining the situation

Sg_{1} - specific gravity of denser material (conc. acid) with a volume v_{1}

Sg_{2} Specific gravity of dilutant (water =1.00 ) having a volume v_{2}

diluting to a final volume v_{3} and Sg3 final specific gravity required.

we must assume the sum of the two initial volumes:

v_{1}+v_{2} = (v_{1}+v_{2}) =v_{3}, the final volume,

v_{1} x Sg_{1} + v_{2} x Sg_{2} = (v _{1} + v_{2}) x Sg_{3}

Expanding LHS to give v_{1} x Sg_{3} + v_{2} x Sg_{3},

then collecting terms of volume together v_{1}(Sg_{1} - Sg_{3}) = v_{2}(Sg_{3} - Sg_{2})

rearranging gives:

v_{2} = v_{1}(Sg_{1}-Sg_{3})/(Sg_{3}-Sg_{2})

or in words:

Volume of water = volume of acid x (difference between acid and final sg)/(difference between final sg and water).

In the example required assume that 100ml of acid is to be diluted.

v_{2} = v_{1}(Sg_{1}-Sg_{3})/(Sg_{3}-Sg_{2})

=100ml x (1.84-1.25)/(1.25-1.00)

= 100ml x (0.59/0.25) =100ml x2.36

=236ml of water into which the acid should be poured, preferably in an ice bath.

verification of calculation:

100cm^{3} x 1.84 g/cm^{3}= 184g

236cm^{3} x 1.00 g/cm^{3} = 236g

vol =100+236 = 336cm^{3}

mass= 184+236 = 420g

density =420g/336cm^{3} = 1.25g/cm^{3}

or a specific gravity of 1.25

The final solution should be tested with a hydrometer before use.

Note: making any solutions containing volatile or corrosive nature should be done with stopped weighing bottles and using the fume cupboard for safety reasons.

**Special Safety Considerations**

Some solutions can cause particular safety considerations. The most common one found in school labs is the dilution of concentrated (conc) acids. When these are diluted they give off heat due to **heat of dilution** effects.

In the case of diluting conc sulphuric acid (H_{2}SO_{4}) this can even cause the water to boil. The hazard from diluting acids can be **minimised** by **ALWAYS adding the acid to the water**, slowly and with lots of stirring.

If large quantities of conc. acid are being diluted, then you should consider standing the beaker being used in an ice bath.

The wearing of suitable Personal Protective Equipment (PPE) must always be used.

Another dilution that can cause severe heat of dilution problems is dissolving sodium (or potassium) hydroxide. Once again, using the same methods of precaution outlined above, the hazards will be minimised.

It must also be born in mind that in both cases cited above that besides the hazards of heat, the solution will be corrosive. If you are unfamiliar with any solution preparation (or dilution) it is always wise to check the relevant CLEAPSS for any special safety considerations. This forms part of your own personal Health and Safety risk assessment

## Diluting a solution given as a percentage solution

This section is to illustrate how to calculate the amount of water required to make a solution to a given percentage from a stronger solution. It will also extend into calculating how to make a specific volume.

Unfortunately the easiest way to express how to do this is with algebra, but a worked example should make it clear.

Suppose that you have a stock solution of x% and you wish to have a solution of y%. Each unit of solution simply needs to be made up to x/y units.

So, as an easy example, you have a stock solution of 50% and you require a solution of 10%. Take one volume of the stock solution and dilute it so that the total volume is 50%/10% (=5) units. Remember that diluting acids requires that the acid is added to the water.

Now lets do a more likely scenario using the 50% solution.

You have a stock solution of 50% and you want a litre of 1.5%.

The dilution is one volume to be made up to 50%/1.5% (= 33.33) volumes.

To calculate the required volume needed to make 1 litre (1000ml) the stock solution needs to be 1000/33.33 = 33ml

See also;

- CLEAPSS CDROM (2007) "Making up Solutions"
- Volumetric glassware
- Molecular Weight
- Risk Assessments
- Health and Safety at Work

**Return to: Chemistry Theory**

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