Download Reactions in Aqueous Solution (Brown 13th-Fossum

Chapter 4 Aqueous Reactions and Solution Stoichiometry
4.1 General Properties of Aqueous Solutions
• Solutions are homogeneous mixtures.
• The solvent is present in greatest abundance.
• All other substances are solutes (they are dissolved in the solvent).
Dissociation
• When an ionic substance dissolves in water, the solvent pulls the individual ions from the crystal and solvates them
(dissociation).
KCl(aq) = K+ (aq) + Cl- (aq)
CuSO4(aq) = Cu+2(aq) + SO42-(aq)
K2SO4(aq) = 2 K+ (aq) + SO42-(aq)
Electrolytes
• An electrolyte is a substance that dissociates into ions when dissolved in water; therefore, it will conduct electricity.
Nonelectrolytes
• A nonelectrolyte may dissolve in water, but it does not dissociate into ions when it does so.
• Molecular compounds tend to be nonelectrolytes, except for acids and bases.
Electrolyte’s Strength
• A strong electrolyte dissociates completely when dissolved in water.
• A weak electrolyte only dissociates partially when dissolved in water.
Ionic
Molecular
Strong Electrolyte
Weak Electrolyte
Nonelectrolyte
All
Strong acids (See Table 4.2)
None
Weak acids
Weak bases
None
All other compounds
4.2 Precipitation Reaction
•
•
AX(aq) + BZ(aq) → AZ + BX
If either AZ or BX is an insoluble compound, there is a chemical reaction.
If no precipitate (solid) is formed, then NR.
Solubility of Ionic Compounds
The solubility of ionic compounds varies:
We say a compound is “soluble” if a noticeable amount dissolves.
A compound is “insoluble/slightly soluble” if less than 0.01 mol/L dissolves (often times is much less). We don’t see a change.
Solubility Rules (Handout)
Problem 1
Soluble or Insoluble?
K2CO3
BaSO4
NaOH
CuSO4
FePO4
Ca(C2H3O2) 2
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Precipitation Reaction
KI(aq) + AgNO3(aq) → KNO3(aq) + AgI(s)
A solid (precipitate) forms out of two aqueous solutions.
Process for Predicting the Products of a Precipitation Reaction
• For Reactants: Determine what ions each aqueous reactant has
Ex. Mix: AgNO3 (aq) + KCl (aq)
Ions present: Ag+ (aq), NO3- (aq), K+ (aq), Cl- (aq)
• To get the Products: Exchange ions
(+) ion from one reactant with (-) ion from other
AgCl and KNO3
(Write correct formulas: charges!! Neutral overall.)
•
Determine solubility of each product in water and write their phases.
– Solubility rules
– If product is insoluble or slightly soluble, it will precipitate.
AgCl (s), KNO3(aq)
•
Balance the equation
– Count atoms
AgNO3(aq) + KCl(aq)  AgCl(s) + KNO3(aq)
Example 1
Mix: K2CrO4 (aq) + Ba (NO3)2 (aq)
K2CrO4 (aq) + Ba(NO3)2 (aq) → BaCrO4 (s) + 2 KNO3 (aq)
Example 2
Mix: KNO3 (aq) + BaCl2 (aq)
KNO3 (aq) + BaCl2 (aq) → Ba(NO3)2 (aq) + 2 KCl (aq)
All the dissolved ions remain in solution. NR!
(No chemical change takes place.)
Ionic Equations
Three types of equations are used to describe Reactions in Aqueous Solutions:
1. Molecular Equation – Shows all formulas.
2. Complete Ionic Equation – Shows all compounds as they actually exist.
3. Net Ionic Equation – Shows only those ions that undergo a change (spectator ions are cancelled).
Example 1:
• Molecular:
Na2CO3(aq) + BaCl2(aq)  2NaCl(aq) + BaCO3(s)
• Ionic:
2Na+(aq) + CO32-(aq) + Ba2+(aq) + 2Cl-(aq)  2Na+(aq) + 2Cl-(aq) + BaCO3(s)
• Net ionic:
CO32-(aq) + Ba2+(aq)  BaCO3(s)
(cancel: Na+ and Cl- - unchanged)
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•
•
•
Ionic equation: An equation in which ions are explicitly shown.
Spectator ion: An ion that appears unchanged on both sides of a reaction arrow.
Net ionic equation: An equation that does not include spectator ions.
Example 2:
• Molecular:
2AgNO3(aq) + Na2CrO4(aq) → Ag2CrO4(s) + 2NaNO3(aq)
• Ionic:
2Ag+(aq) + 2NO3-(aq) + 2Na+(aq) + CrO42-(aq) →
Ag2CrO4(s) + 2Na+(aq) + 2NO3-(aq)
• Net ionic:
2Ag+(aq) + CrO42-(aq) → Ag2CrO4(s)
(cancel: Na+ and NO3- - unchanged)
Problem 2
Aqueous nickel (II) nitrate is added to aqueous potassium carbonate, is there a reaction? If so, write the molecular, total ionic
and net ionic equations for it.
Problem 3
Write the net ionic equation for:
a. FeCl3 (aq) + NaOH (aq)
b. Ba(NO3)2 (aq) + (NH4)3PO4 (aq)
c. NaCl (aq) + Cu(NO3)2 (aq)
4.3 Acids, Bases, and Neutralization Reactions
Acids – Taste sour and have a low pH. (Turn litmus paper red.)
• Arrhenius: substances that increase the concentration of H+ when dissolved in water.
• Brønsted and Lowry: proton donors.
Bases – Taste bitter and have a high pH. (Turn litmus paper blue.)
• Arrhenius: Increase the concentration of OH− when dissolved in water.
• Brønsted and Lowry: proton acceptors.
Strong and Weak Acids and Bases
Strong acids and bases ionize 100% in solution.
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Neutralization Reactions
Generally, when solutions of an acid and a base are combined, the products are a salt and water.
CH3COOH (aq) + NaOH (aq) CH3COONa (aq) + H2O (l)
When a strong acid reacts with a strong base, the net ionic equation is…
HCl (aq) + NaOH (aq)  NaCl (aq) + H2O (l)
H+ (aq) + OH- (aq)  H2O (l)
Polyprotic acids (H2SO3,H2CO3, H3PO4) lose one H+ at a time; the first H+ is the easiest to lose. The other protons can be forced
out by an excess of base.
Mix equal # of moles:
1 mol NaOH + 1 mol H2SO3 → H2O + NaHSO3
Using excess base:
2 NaOH + H2SO3 → H2O (l) + Na2SO3 (aq)
Problem 4
Write the molecular, total ionic, and net ionic equations for the reactions of H3PO4 with excess NaOH. (Hint: This is a weak
acid.)
Gas-Forming Reactions
As it turns out, this happens:
H2CO3 (aq)  H2O (l) + CO2 (g)
H2SO3 (aq)  H2O (l) + SO2 (g)
NH4OH (aq)  H2O (l) + NH3 (g)
•
Carbonates and bicarbonates with acids produce H2CO3.
CaCO3 (s) + HCl (aq)  CaCl2 (aq) + CO2 (g) + H2O (l)
NaHCO3 (aq) + HBr (aq) NaBr (aq) + CO2 (g) + H2O (l)
H2CO3 (aq)
Baking soda is used to:
- Neutralize acid spills.
- Make baked goods rise.
•
Here, the expected product H2SO3 (aq), but it decomposes.
SrSO3 (s) + 2 HI (aq) SrI2 (aq) + SO2 (g) + H2O (l)
•
Hydrogen sulfide (H2S) has a low solubility in water; therefore, reactions that produce it will form a gas (smells like
rotten egg).
Na2S (aq) + H2SO4 (aq)  Na2SO4 (aq) + H2S (g)
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• Ammonium salt + strong base.
Here, NH4OH (aq), decomposes:
NH4Cl (aq) + NaOH (aq)  H2O (l) + NH3 (g) + NaCl (aq)
Problem 5
Predict the products. Write the molecular, total and net ionic equations.
a. CuCO3(s) + HBr(aq)
b. (NH4)2S(aq) + HClO4(aq)
4.4 Oxidation-Reduction Reactions
• An oxidation occurs when an atom or ion loses electrons.
• A reduction occurs when an atom or ion gains electrons.
• One cannot occur without the other.
Combustion reactions and corrosion (rusting of metals) are two examples.
To determine if an oxidation-reduction reaction has occurred, we assign an oxidation number to each element in a neutral
compound or charged entity.
Oxidation Numbers
• Elements in natural elemental state have oxidation number = 0.
• The oxidation number of a monatomic ion is the same as its charge.
• Nonmetals tend to have negative oxidation numbers, although some are positive in certain compounds or ions.
 Oxygen has an oxidation number of −2, except in the peroxide ion in which it has an oxidation number of −1.
 Hydrogen is −1 when bonded to a metal, +1 when bonded to a nonmetal.
 Fluorine always has an oxidation number of −1.
 The other halogens have an oxidation number of −1 when they are negative; they can have positive oxidation
numbers, however, most notably in oxyanions.
•
•
The sum of the oxidation numbers in a neutral compound is 0.
The sum of the oxidation numbers in a polyatomic ion is the charge on the ion.
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Problem 6
Determine the oxidation numbers:
Li2O
H3PO4
MnO4–
Cr2O72–
C7H8
Problem 7
Identify what’s oxidized and what’s reduced:
Cu(s) + 2 NO3– + 4H+ → Cu2+ + 2 NO2 + 2 H2O
Single Displacement Reactions
• In General:
A + BZ → AZ + B
Two outcomes:
• If the reaction happens, we say that A is more active than B (and displaces it).
• If the reaction doesn’t happen, then B is more active than A (i.e. A cannot displace it).
Example 1
Cu (s) + HgCl2 (aq)  CuCl2 (aq) + Hg (l)
Orange
colorless
pale blue silvery or black
BUT
Hg (l) + CuCl2 (aq)  NO Reaction!!
Which one is more active?
Example 2
Cu (s) + 2 Ag+ (aq)  Cu2+ (aq) + 2 Ag (s)
(net ionic eq.)
Displacement Reactions
1. Metal and aqueous solution
Fe(s) + CuSO4(aq) → FeSO4(aq) + Cu(s)
Fe(s) + Cu2+(aq) → Fe2+(aq) + Cu(s)
(Net Ionic)
2. Metal and aqueous acid solution
Fe(s) + 2 HCl(aq) → FeCl2(aq) + H2(g)
Fe(s) + 2 H+(aq) → Fe2+(aq) + H2(g)
(Net Ionic)
• Metals less active than (H) show no reaction:
Au(s) + H2SO4(aq) → NR
3. Active metal and water
Ca(s) + 2 H2O(l) → Ca(OH)2(aq) + H2(g)
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Use the Activity Series to Make Predictions
Problem 8
Predict whether a single displacement reaction will occur. Write the equation for the reaction. (Use the activity series)
a. Cu(s) + Pb(NO3)2(aq)
b. Pb(s) + HCl(aq)
4.5 Concentrations of Solutions
Molarity is one way to measure the concentration of a solution.
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 (𝑀) =
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑖𝑛 𝑙𝑖𝑡𝑒𝑟𝑠
The volume of the solution includes the volume of the solute and the solvent.
Problem 9
If you dissolve 32.0 g NaCl in 100. mL of solution, what is the molarity?
Moles of solute
Using Molarity
Molarity (M) =
• Since:
Liters of solution
Knowing two of the three allows us to find the unknown easily. (Simply solve for it.)
Always use consistent units.
Problem 10
What volume of 2.00 M HCl solution contains 15.0 g HCl?
Problem 11
How many grams of K2Cr2O7 are there in 250.0 mL of 0.100 M K2Cr2O7?
Molarity & Dissociation
• CaCl2(aq) = Ca+2(aq) + 2 Cl-1(aq)
• A 1.0 M CaCl2(aq) solution contains 1.0 moles of CaCl2 units in each liter of solution
• Since each CaCl2 furnishes one Ca+2, it follows that 1.0 M CaCl2 = 1.0 M Ca+2
• Since each CaCl2 furnishes 2 Cl-1, it follows that 1.0 M CaCl2 = 2.0 M Cl-1
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Problem 12
If the concentration of a solution of Cr2(SO4)3 is 0.30 M, what are the concentrations of all ions in solution?
Steps involved in the preparation of a standard aqueous solution.
• We have to know the mass (and, therefore, number of moles) of the solute.
• The solute is added to a volumetric flask.
Don’t ever go beyond the calibration mark… Or you’ll have to start all over again...
Dilution
• Dilution: Lowering concentration by adding additional solvent.
• Dilution factor: The ratio of the initial and final solution volumes (V1/V2).
• In the dilution process, the amount of solute remains constant, only the volume is increased.
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 = 𝑀𝑐 ×𝑉𝑐 = 𝑀𝑑 ×𝑉𝑑 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
• We will always know three out of the four so we can always solve for the unknown.
Problem 13
What volume of 6.0 M NaOH needs to be diluted to prepare 5.00 L if 0.10 M NaOH?
How to prepare a diluted solution
Again: Don’t go beyond the calibration mark…
4.6 Solution Stoichiometry and Chemical Analysis
• We have a new way to calculate the number of moles:
aA + bB→ cC + dD
Problem 14
What mass of silver bromide is produced from the reaction of 37.5 mL of 0.100 M aluminum bromide with excess silver
nitrate solution?
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Problem 15
In the precipitation reaction between 35.0 mL of 0.155 M lead(II) nitrate and 45.0 mL of 0.287 M potassium chloride, how
much product is produced?
Titration
A technique used to calculate the concentration of a solute in a solution.
You should get a slightly pink color.
Problem 16
In a titration, 25.00 mL of 0.1000 M H2SO4 requires 32.59 mL of NaOH to reach the endpoint. What is the concentration of the
NaOH?
Problem 17
How many mL of 0.206 M HI solution are needed to reduce 22.5 mL of a 0.374 M KMnO4 solution?
16 HI + 2 KMnO4 → 5 I2+ 2 MnI2 + 2 KI + 8 H2O
Problem 18
An impure sample containing oxalic acid (H2C2O4) weighs 1.278 g. If 32.18 mL of 0.3526 M NaOH is needed to reach the
endpoint, what was the mass-% of oxalic acid in the mixture?
Problem 19
An unknown monoprotic acid is titrated with NaOH. It requires 44.31 mL of 0.1003 M NaOH to titrate 0.8359 g of the acid to
the endpoint. What is the molar mass of the acid?
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Problem 20
50.0 mL of 1.50 M NaOH and 35.0 mL of 1.00 M FeCl3 are mixed. What mass of ppt forms? What are the ion concentrations
after the reaction?
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13.4 Expressing Solution Concentration
Mass Percentage
𝑚𝑎𝑠𝑠 % 𝐴 =
𝑚𝑎𝑠𝑠 𝑜𝑓 𝐴 𝑖𝑛 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
×100
𝑡𝑜𝑡𝑎𝑙 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
;
5.0 % 𝑜𝑓 𝑁𝑎𝐶𝑙 =
5.0 𝑔 𝑜𝑓 𝑁𝑎𝐶𝑙
100 𝑔 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
Parts per Million (ppm)
𝑚𝑎𝑠𝑠 𝑜𝑓 𝐴 𝑖𝑛 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
×106
𝑡𝑜𝑡𝑎𝑙 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
𝑝𝑝𝑚 =
2 𝑝𝑝𝑚 𝑃𝑏 2+ =
2 𝑔 𝑃𝑏 2+
106 𝑔 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
𝑜𝑟
2 𝑚𝑔 𝑃𝑏 2+
𝑘𝑔 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
Since a dilute solution is mostly water (d = 1.00 g/mL).
2 𝑚𝑔 𝑃𝑏 2+
𝐿 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
Parts per Billion (ppb)
𝑝𝑝𝑏 =
2 𝑝𝑝𝑏 𝑃𝑏 2+ =
𝑚𝑎𝑠𝑠 𝑜𝑓 𝐴 𝑖𝑛 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
×109
𝑡𝑜𝑡𝑎𝑙 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
2 𝑔 𝑃𝑏 2+
109 𝑔 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
𝑜𝑟
2 𝜇𝑔 𝑃𝑏 2+
𝑘𝑔 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
2 𝜇𝑔 𝑃𝑏 2+
𝐿 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
ppm and ppb are used for very dilute solutions (toxins).
Mole Fraction (χ):
𝜒𝐴 =
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐴
𝑡𝑜𝑡𝑎𝑙 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
Moles in solution = moles of solute + moles of solvent.
Mole fractions of all components add up to 1.
Molarity (M)
𝑀=
𝑚𝑜𝑙 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝐿 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
;
3.0 𝑀 𝐻𝐶𝑙 =
3.0 𝑚𝑜𝑙 𝐻𝐶𝑙
1 𝐿 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
Since volume is temperature-dependent, molarity can change with temperature; furthermore, volumes often change upon
mixing (they are not additive).
Molality (m)
𝑚=
𝑚𝑜𝑙 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝑘𝑔 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡
;
1.5 𝑀 𝐻𝐶𝑙 =
1.5 𝑚𝑜𝑙 𝐻𝐶𝑙
1 𝑘𝑔 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡
Since both moles and mass do not change with temperature, molality (unlike molarity) is not temperature-dependent.
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Changing Molarity to Molality
If we know the density of the solution, we can calculate the molality from the molarity and vice versa.
Molality
(mol/kg solvent)
Mass of solvent
+
Mass of solute
Molar
mass
Moles of solute
Mass of solution
Density
Volume of solution
Molarity
(mol/L solution)
Problem 21
You dissolve 10.0 g of sugar (C12H22O11) in 250 g of water. Calculate the mole fraction, molality and mass percent of the
solution. (Why can’t we calculate M?)
Problem 22
The maximum allowable concentration of As in drinking water in the US is 0.010 ppm.
a. If a 2.5 L water sample is found to contain 0.030 mg As, is this within the allowable limit?
b. What is the allowable mass of As in 1 large glass of water (500 mL)?
Problem 23
A sucrose solution is 25.0% sucrose (C12H22O11) by mass. Calculate the molality of the solution.
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Fossum-Reyes
Problem 24
An aqueous solution of urea ((NH2)2CO) is 3.42 m. Calculate M for this solution. (dsol = 1.045 g/mL)
Problem 25
An aqueous solution of urea is 2.00 M and has a density of 1.029 g/mL. What is its molality?
M and ppm
1 𝑝𝑝𝑚 =
1𝑔𝑋
1×10−6 𝑔 𝑋
1 𝜇𝑔 𝑋
=
=
6
1×10 𝑔 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 1 𝑔 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 1 𝑔 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
𝐹𝑜𝑟 𝑤𝑎𝑡𝑒𝑟: 𝑑 =
𝟏 𝒑𝒑𝒎 =
1𝑔
1 𝜇𝑔 𝑋
; 1 𝑝𝑝𝑚 𝑋 =
1 𝑚𝐿
1 𝑚𝐿 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
1 𝜇𝑔 𝑋
1𝐿
1 𝑚𝐿 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 (1000 𝑚𝐿)
=
; 𝑀[=]
𝑚𝑜𝑙
𝐿
1,000 𝜇𝑔 𝑋
𝟏 𝒎𝒈 𝑿
=
1 𝐿 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝟏 𝑳 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏
ppm ⇔ M
𝑺𝒊𝒏𝒄𝒆:
𝟏 𝒑𝒑𝒎 =
𝟏 𝒎𝒈 𝑿
𝟏 𝑳 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏
To change ppm’s to M, convert the mg’s to moles.
To change M to ppm’s, convert the moles to mass in grams, then to milligrams.
Odor threshold value (OTV) is defined as the most minimal concentration of a substance that can be detected by a human
nose; it can be expressed as a concentration of water or concentration in air.
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Fossum-Reyes
ppm ⇔ M
The major aromatic constituent of bell pepper (2-isobutyl-3-methoxypyrazine) can be detected at a concentration of 0.01 nM
CH3
(OTV). What is this concentration in ppm?
CH2
N
𝑔
𝑀𝑀 = 9𝐶 + 14 𝐻 + 2 𝑁 + 1 𝑂 = 166.22 ⁄𝑚𝑜𝑙
HC
166.22𝑔 1000 𝑚𝑔
0.01×10−9 𝑚𝑜𝑙 (
)( 1 𝑔 )
𝑚𝑜𝑙
0.01 𝑛𝑀 =
= 2×10−6 𝑝𝑝𝑚
𝐿
CH
C
HC
CH3
C
N
OCH3
How low is this?
An Olympic swimming pool has a volume of 2,500 m3 or 2,500,000 L.
1,000 𝑚𝐿 20 𝑑𝑟𝑜𝑝𝑠
2,500,000 𝐿 (
)(
) = 5.0×1010 𝑑𝑟𝑜𝑝𝑠
1𝐿
1 𝑚𝐿
2×10−6 𝑝𝑝𝑚 =
2×10−6
= 2×10−12
1×106
5.0×1010 𝑑𝑟𝑜𝑝𝑠(2×10−12 ) = 0.1 𝑑𝑟𝑜𝑝𝑠
Like detecting a fraction of a drop in an Olympic swimming pool!!
14
Fossum-Reyes