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chem unit 2 > ch 12- measuring solubility and concentrations > Flashcards

ch 12- measuring solubility and concentrations Flashcards

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1
Q

Define concentration

A

Amount of chemical substance in a unit of volume

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2
Q

Define concentrated solution

A

Solution with a relatively high amount of solute in comparison to the volume of solution

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3
Q

Define dilute solution

A

Solution with a relatively low amount of solute in comparison to the volume of solution

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4
Q

Define dilution

A

Process of decreasing concentration of a solution by adding more solvent

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5
Q

Define molarity (M)

A

Also called molar concentration, is a measure of the concentration of a chemical species in terms of amount of substance per unit volume of solution, measured in moles per litre (mol L-1)

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6
Q

Define parts per billion (ppb)

A

One in one billion parts of a solution

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7
Q

Define parts per million (ppm)

A

One in one million parts of a solution

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8
Q

Define solute

A

Substance that dissolves into the solvent

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9
Q

Define solution

A

Homogenous mixture in which a substance is dissolved in another

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10
Q

Define % (m/m)

A

Mass percent concentration of solute in solution

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11
Q

Define % (w/w)

A

Weight percent concentration of solute in solution

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12
Q

Define % (m/v)

A

Mass percent concentration of solute per unit of volume of solution

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13
Q

Define % (v/v)

A

Volume percent concentration of solute in solution 6

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14
Q

Contrast solute and solvent

A

A solute is a substance that can be dissolved into a solution by a solvent.A solvent is a substance in which a solute is dissolved

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15
Q

What does concentration or dilution depend on in a solution

A

Whether a solution is concentrated or diluted doesn’t depend on the number of particles but rather the ratio of solute particles to the total volume of the solution

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16
Q

How to calculate the concentration of a solution in mol L-1

A

c=n/V
where c= concentration mol L-1
where n= amount mol
where v= volume L

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17
Q

Calculate the concentration, in mol L−1, of a 100 mL solution containing 0.25 mol of NaCl.

A

V=100mL/1000
V= 0.1L 
c=n/V
c= 0.25 mol/0.1 L
c= 2.5 mol L-1

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18
Q

How to calculate the concentration of a solution in a unit of mass per a unit of volume?

A

c=m/V
where c= concentration mol L-1
where m= mass in mg/g
where v= volume L/mL

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19
Q

Calculate the concentration in mg L−1 of a liquid medication that contains 40.0 mg of paracetamol
in 100 mL.

A

V= 100ml/1000
V=0.1 L
c=m/V
c= 40mg/0.1L
c=400mg L-1

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20
Q

Different ways to express ppm

A

1kg=1,000,000 mg 
mass of solute (mg)/mass of solute (kg)=ppm= mass of solute (mg)/volume of solution (L) 
1L= 1,000,000mg

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21
Q

Different ways to express ppb

A

1kg=1,000,000,000µg
mass of solute (µg/mass of solute (kg)=ppb= mass of solute (µg)/volume of solution (L) 
1L= 1,000,000,000µg

22
Q

Express the concentration of a 2.0 × 10−9 mol L−1 MgCl2 solution in ppm.

A

m = n × M
m(MgCl2) = 2.0 × 10−9 mol × (24.3 + (2 × 35.5)) g mol−1
m(MgCl2) = 1.906 × 10−7 g
m(MgCl2) = 1.906 × 10−7 g × 1000 mg
m(MgCl2) = 1.91 × 10−4 mg
Therefore, the concentration of the solution is
1.9 × 10−4 ppm.

23
Q

What are some common percentage concentrations

A

%(m/m)
%(w/w)
%(m/v)
%(v/v)

24
Q

What is %(mass/mass)

A

Both solute and solution are expressed as a unit of mass e.g. mg, g, kg
c=mass of solute/mass of solution x 100% 
masses must be in the same units

25
Q

What is %(weight/weight)

A

Both solute and solution are expressed as a unit of mass e.g. mg, g, kg 
c=weight of solute/weight of solution x 100%
weights must be in the same units

26
Q

What is % (mass/volume)

A

The solute is expressed in grams whereas the solution is expressed in millilitres (g mL-1)
c=mass of solute g/volume of solution mL x 100%

27
Q

What is % (volume/volume)

A

Both solute and solution are expressed as a unit of volume (e.g. mL, L)
c=volume of solute/volume of solution x 100%
volumes must be in the same units

28
Q

What is the process of decreasing the concentration of a solution?

A

Dilution

29
Q

What is the dilution formula

A

(before dilution) c1 x V1= c2 x V2 (after dilution)

30
Q

50.0 mL of water was added to 20.0 mL of 2.0 mol L−1 KCl solution. Calculate the concentration
of the diluted solution.

A

V2 = 50.0 mL + 20.0 mL
V2 = 70.0 mL
V2 =70.0/ 1000 L
V2 = 0.0700 L
c1 = 2.0 mol L−1
V1 = 20.0 mL
V1 =20.0/1000 L
V1 = 0.0200 L
c2 = ?
V2 = 0.0700 L
c1 × V1 = c2 × V2
2.0 mol L−1 × 0.0200 L = c2 × 0.0700 L
c2 =2.0 mol L−1 × 0.0200 L/0.0700 L
c2 = 0.57 mol L−1

31
Q

Define alum

A

Inorganic chemical compound composed of a hydrated aluminium salt

32
Q

Define crystallisation

A

Process by which solute molecules leave solution to form a highly organised structure known as a crystal

33
Q

Define eutrophication

A

Excessive algal growth due to increased nutrient concentration in water

34
Q

Define precipitation reaction

A

Chemical reaction resulting in the formation of a solid from an aqueous solution

35
Q

Define saturated solution

A

Solution containing the maximum amount of dissolved solute

36
Q

Define solubility

A

Ability to be dissolved in a given solvent

37
Q

Define solubility curve

A

Graph plotting the solubility of a substance (y axis) against temperature (x axis)

38
Q

Define solubility table

A

Table listing the solubilities of compounds formed from different ions when dissolved in water

39
Q

Define supersaturated solution

A

Solution containing more than the maximum amount of solute that can be dissolved

40
Q

Define unsaturated solution

A

Solution containing less than the maximum amount of dissolved solute

41
Q

What are some factors that affect the solubility of a substance?

A

Temperature 
solubility tables

42
Q

Trends of solubility regarding temperature

A

While many substances do become more soluble as temperature increases, substances like cerium (III) sulfate and a number of gases instead become less soluble with increasing temperature

43
Q

What do solubility curves show?

A

the relative solubility of a substance,
the general trend in solubility of a substance as temperature changes,
how much solute can be dissolved to create a saturated solution, how much dissolved solute creates a supersaturated solution, and, how much solute would be dissolved in an unsaturated solution.

44
Q

What occurs in unsaturated, saturated and supersaturated solutions?

A

Unsaturated solution- more solute dissolves
Saturated solution- no more solute dissolves
Supersaturated solution- becomes unstable, crystals form

45
Q

What can solubility curves be used to?

A

Enables us to compare the solubility of substances at a given temperature
Enables us to visualise the general trend in the solubility of a compound as temperature changes
predict the amount of solute that can be dissolved at a given temperature
predict the amount of solute that won’t dissolve when added to a solvent
and, predict the mass of crystals formed in a supersaturated solution.
•

46
Q

What is a domestic or industrial application of solubility curves?

A

In the pharmaceutical industry, solubility curves are often used to determine the temperature at which a reaction should take place
Also used to determine the point at which a solution becomes supersaturated, as there are some pharmaceuticals that require crystallisation

47
Q

What is one of the biggest threats to water purity?

A

Eutrophication which is wherein a body of water becomes oversaturated with nutrients, leading to an increased grow of algae.
The algae growth deteriorates the water quality and depletes the amount of dissolved oxygen in bodies of water.

48
Q

How does phosphate impact water purity

A

Phosphate is known to stimulate algal growth and therefore increased phosphate concentration is associated with decreased water purity. Phosphate can enter the water via excretion, waterways, sewage and fertiliser run off

49
Q

What is alum

A

A hydrated salt of aluminium which is used in treatment plants to reduce phosphate concentration.

50
Q

How can alum be used to precipitate excess phosphate in water purification

A

Alum reacts with water to form aluminium hydroxide. As such phosphate is released from lake beds reacts with aluminium hydroxide to form an insoluble aluminium phosphate precipitate. This precipitate then settles at the bottom of lakes and no longer contributes to eutrophication, improving water quality.

chem unit 2 (7 decks)

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