Module 2: Units of concentration Flashcards
What are the terms used when discussing units of concentration?
The dissolved component is called the solute; the liquid in which it is dissolved is the solvent, and the overall combination is called the solution.
The amount of solute in a given quantity of the solution (not solvent) is called the concentration of the solute. It may be a solid that is dissolved, as is the case with acids in wine, or sodium hydroxide in solutions in a winery laboratory. Alternatively, the dissolved component may be a liquid, such as the alcohol of wine.
How do solids have their concentration expressed?
Solid substances usually have their concentration expressed as the mass of the solid dissolved in a given volume of solution; the units are therefore of mass per volume:
e.g. grams per litre (g L-1)
milligrams per millilitre (mg mL-1)
Concentration = mass of solute / volume of solution
How do liquids have their concentration expressed?
Liquid substances usually have their concentration expressed as the volume of liquid dissolved in a given volume of solution; the units are therefore of volume per volume:
e.g. millilitres per litre (mL L-1)
litres per hectolitre (L hL-1)
- Concentration = volume of solute / volume of solution
What is percentage (by volume), %(v/v)
This measures concentration as the number of parts by volume of a solute in one hundred of the same parts by volume of solution.
For example:
the number of millilitres of solute in 100 millilitres of solution(i.e., mL/100mL); or
the number of litres of solute in 100 litres of solution (i.e., L/100L).
How is percentage (by volume), % (v/v) calculated?
% (v/v) value = mass of the solute alone / volume, in equivalent units of the solution x 100
If 250.0 mL solution needs to be prepared whereby the concentration of tartaric acid is 10.0 %(m/v) in water, what mass of tartaric acid is required to prepare the solution?
10.0 %(m/v) = 10.0 g tartaric acid in 100 mL solution
We want 250 mL of solution, and as 250 mL / 100 mL = 2.5, this means we need 2.5 times the tartaric acid in 250 mL as in 100 mL to achieve a 10.0 %(m/v) solution.
Therefore: 2.5 x 10.0 g = 25.0 g tartaric acid is required in the 250 mL solution to achieve the 10.0 %(m/v) solution.
OR
% (v/v) value = mass of the solute alone / volume, in equivalent units of the solution x 100
Where, %(m/v) = 10, volume of solution = 250 mL, and the mass of the solute is the unknown, but the units must be in ‘g’ if the volume is in units of mL.
Therefore: 10 %(m/v) = (mass of tartaric acid / 250 mL ) x 100
(10 %(m/v) / 100 ) x 250 mL = mass of tartaric acid
25.0 g = mass of tartaric acid
What is percentage (by mass), %(m/m)?
This measures concentration as the number of parts by mass of a solute in one hundred of the same parts by mass of solution. For example:
the number of grams of solute in 100 grams of solution (i.e., g/100g),
or
the number of kilograms of solute in 100 kilograms of solution (i.e., kg/100kg).
How is percentage (by mass), %(m/m) calculated?
% (m/m) value = mass of the solute alone / mass, in the same units, of solution x 100
The sulfur dioxide content of potassium metabisulfite is 57.6% (m/m). How much potassium metabisulfite is needed to provide 134 g of sulfur dioxide?
mass of sulfur dioxide component / mass of potassium metabisulfite x 100 = 57.6
Therefore, mass of potassium metabisulfite =
mass of sulfur dioxide x 100 / 57.6
For 134 g of sulfur dioxide,
mass of potassium metabisulfite
= 134g x 100 / 57.6
= 233 g
Define parts per million
The number of milligrams of solute per litre of solution
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The number of parts by mass of a solute in one million equivalent parts by volume of the solution, for example:
- the number of milligrams of solute in 1 litre of solution (i.e., mg/L), or
- the number of grams of solute in 1000 litres of solution(i.e., g/1000L).
How is parts per million calculated?
It can be calculated in the following way:
ppm (m/v) value = mass of solute alone / volume, in equivalent units, of solution x 10 ^ 6
Example
What mass of potassium metabisulfite must be added to 850 litres of wine if it is desired to increase its sulfur dioxide content by 10.0 ppm (m/v)?
- Since the ppm (m/v) value indicates the number of milligrams of solute (sulfur dioxide) in 1 litre of solution (wine), the increase in sulfur dioxide content of 10.0 ppm (m/v) corresponds to 10.0 milligrams of sulfur dioxide being added to 1.00 litre of wine.
For 850 L of wine, sulfur dioxide to be added
= 10.0 mg x 850
= 8.50 g
The sulfur dioxide content of potassium metabisulfite is 57.6 (m/m)
= sulfur dioxide content / mass of potassium metbisulfite x 100 = 57.6
= sulfur dioxide required x 100 / 57.6
= 8.50 x 100 / 57.6
mass of PMS = 14.8 g
What is the generic single/equation approach for dilution calculations?
C1 x V1 = C2 x V2
C1 = the concentration of a solute in the concentrated solution.
V1 = the volume of the concentrated solution to be used.
C2 = the concentration of the solute in the final diluted solution (i.e., wine)
V2 = the volume of the final diluted solution (i.e., wine)
In the expression, the units of concentration and volume are not restricted to any particular unit. However, the C1 and C2 concentration units must be identical to each other, and the V1 and V2 concentration units must also be identical to each other.
The expression can be used then to solve any one of C1, V1, C2 or V2, provided the other three values are known.
Work out the volume of 10.0 %(m/v) tartaric acid solution to add to 125 mL of wine to achieve a 0.50 g/L increase in the tartaric acid concentration of the wine.
1) We list the information that we have:
C1 = 10.0 %(m/v), V1 = unknown, C2 = 0.50 g/L, and V2 = 125 mL.
2) We check units. The two concentration values we have are in different units (i.e., %(m/v) vs g/L).
3) Convert to common units. We will convert 10.0 %(m/v) to g/L. Whereby 10.0 %(m/v) = 10.0 g/100mL = 100 g/1000mL = 100 g/L. C1 = 100 g/L.
4) Place the appropriate values into C1 x V1 = C2 x V2 and solve for the unknown.
100 g/L x V1 = 0.50 g/L x 125 mL
V1 = 0.50 g/L x 125 mL / 100 g/L
(note the ‘g/L’ units will cancel out)
V1 = 0.625 mL
This tells us that adding 0.625 mL of the 10.0 %(m/v) tartaric acid solution to 125 mL of wine will increase the tartaric acid concentration by 0.50 g/L. Note, we have ignored the impact of the 0.625 mL addition on the final volume of the solution.
What is Avogradro’s number?
6.02 x 1023
Avogradro’s number is a unit that corresponds to a specific number of molecules, or other chemical entities.
Chosen as the standard is the number of atoms of carbon in exactly 12.000 g of pure carbon.
What is a mole?
This number of any chemical entities is called a mole of the substance.
What is Atomic Mass?
Different atoms have different masses. Their relative masses are indicated by their atomic mass, 1.00 for hydrogen, 12.00 for carbon and 16.00 for oxygen. The size of the mole has been chosen such that for carbon, hydrogen, oxygen or any other atom, the mass of a mole of atoms is equal to the atomic mass when expressed in grams.
What is Molecular Mass?
The relative mass of different molecules is indicated by their molecular mass, which is equal to the sum of the atomic masses of all the atoms in the molecule.
The size of the mole is such that the mass of a mole of a molecular substance is equal to the molecular mass in grams.
How is the number of moles calculated for any chemical substance?
no. of moles = mass in grams / atomic mass or molecular mass
Which wine has the greater molar concentration of acid:
a. one with 4.00 g L-1 of tartaric acid and 2.00 g L-1 of malic acid, or
b. one with 2.00 g L-1 of tartaric acid and 4.00 g L-1 of malic acid?
The molecular masses of tartaric and malic acids are 150.1 g mol-1 and 134.1 g mol-1 respectively.
Calculating the no. of moles of each acid in one litre of wine;
For wine (a) 4.00/150.1 of tartaric aplus 2.00 / 134.1 of malic, ie/ 0/0266 + 0.0149 = 0.0415 mol
For wine (b) 2.00 / 150.1 of tartaric plus 4.00 / 134.1 of malic i.e. 0.0133 + 0.0298 = 0.0431 mol
Thus the wine with the higher malic acid content contains more molecules of acid per litre. Since 1 mole = 6.02 x 10^23 molecules, it can be seen that wine
a. contains 2.50 x 10^22 acid molecules per litre whereas wine;
b. contains 2.59 x 10^22 acid molecules per litre.
What is Molarity?
molarity value = no. of moles of the substance in one litre of solution
molarity value = no. of moles of the substance / no. of litres of its solution
Determination of sulfur dioxide levels in wine requires the use of a 0.0100M solution of sodium hydroxide.
How much sodium hydroxide needs to be dissolved in water to give 5.00 litres of a solution of this strength?
The molecular mass of sodium hydroxide is 40.0 g mol-1.
From the relationship:
sodium hydroxide solution molarity = no of mole of sodium hydroxid / no. of litres of its solution
0.0100 = no of mole of sodium hydroxide / 5.00
= 0.0500
From the relationship: no of moles = mass in grams / molecular mass
For the sodium hydroxide, 0.0500 = mass in grams / 40.0
Therefore, mass of sodium hydroxide = 0.0500 x 40.0 g = 2.00 g
