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Page 1: Chem. 250 – 11/18 Lecture. Announcements I A.Exam 2 Results Average = 73 B.New Homework Set (Text Ch. 4: 25; Ch. 7: 3, 5, 6, 8, 10, 24, 25, 26, 35, 44

Chem. 250 – 11/18 Lecture

Page 2: Chem. 250 – 11/18 Lecture. Announcements I A.Exam 2 Results Average = 73 B.New Homework Set (Text Ch. 4: 25; Ch. 7: 3, 5, 6, 8, 10, 24, 25, 26, 35, 44

Announcements I

A. Exam 2 Results Average = 73

B. New Homework Set (Text Ch. 4: 25; Ch. 7: 3, 5, 6, 8, 10, 24, 25, 26, 35, 44 + Furlough Questions)

Score Range

N

90-92 2

80s 4

70s 5

60s 2

<60 3

Page 3: Chem. 250 – 11/18 Lecture. Announcements I A.Exam 2 Results Average = 73 B.New Homework Set (Text Ch. 4: 25; Ch. 7: 3, 5, 6, 8, 10, 24, 25, 26, 35, 44

Announcements - II

C. Topic Covering1. Cloud Chemistry2. Precipitation Chemistry3. Hydrologic Cycle4. Water Properties5. Water Composition6. Some of the above topics may be

covered in the next lecture

Page 4: Chem. 250 – 11/18 Lecture. Announcements I A.Exam 2 Results Average = 73 B.New Homework Set (Text Ch. 4: 25; Ch. 7: 3, 5, 6, 8, 10, 24, 25, 26, 35, 44

Announcements - III

D. Rough Drafts due today (one copy to me + keep track of who receives other copies)

E. Next Wednesday – Furlough Day- I will give a couple of additional

homework problems for you to do as practice toward understanding of concepts (not collected or graded)

Page 5: Chem. 250 – 11/18 Lecture. Announcements I A.Exam 2 Results Average = 73 B.New Homework Set (Text Ch. 4: 25; Ch. 7: 3, 5, 6, 8, 10, 24, 25, 26, 35, 44

Furlough Problems:

1. A cloud is nucleated on an aerosol containing 8.0 μg m-3 NH4HSO4 with 75% efficiency and reaches a LWC of 0.5 g m-3. The SO2 mixing ratio is present at 2.0 ppb. If P = 0.8 atm and T = 15ºC, calculate the pH of the cloud water. Calculate the pH independently for aerosol only acidity and SO2 only acidity and use only the source which adds the most acidity. Then determine the following: [H2SO3 (aq)], [HSO3

-], [SO4

2-]2. Derive the equation on slide 34.

Page 6: Chem. 250 – 11/18 Lecture. Announcements I A.Exam 2 Results Average = 73 B.New Homework Set (Text Ch. 4: 25; Ch. 7: 3, 5, 6, 8, 10, 24, 25, 26, 35, 44

Cloud Chemistry

• Rationale for Studying- Cloud reactions can be important (e.g. formation of H2SO4)

- Precipitation composition depends on cloud composition- Provide introduction to aqueous chemistry

Page 7: Chem. 250 – 11/18 Lecture. Announcements I A.Exam 2 Results Average = 73 B.New Homework Set (Text Ch. 4: 25; Ch. 7: 3, 5, 6, 8, 10, 24, 25, 26, 35, 44

Cloud Chemistry- Incorporation of Pollutants

Page 8: Chem. 250 – 11/18 Lecture. Announcements I A.Exam 2 Results Average = 73 B.New Homework Set (Text Ch. 4: 25; Ch. 7: 3, 5, 6, 8, 10, 24, 25, 26, 35, 44

Cloud Chemistry- Incorporation of Pollutants

• Main mechanisms- Nucleation of cloud droplets on aerosol particles- Scavenging of gases- Reactions within the droplet

Page 9: Chem. 250 – 11/18 Lecture. Announcements I A.Exam 2 Results Average = 73 B.New Homework Set (Text Ch. 4: 25; Ch. 7: 3, 5, 6, 8, 10, 24, 25, 26, 35, 44

Cloud ChemistryNucleation of Cloud Droplets (some

review?)• Cloud droplets can not form in the absence

of aerosol particles unless RH ~ 300%.• Cloud droplets nucleate on aerosol particles

at RH of ~100.1 to ~101%.• Cloud droplets should nucleate when RH =

100% except that the vapor pressure over a curved surface is less than that over a flat surface (due to water surface tension)

• Smaller particles (d < 50 nm) have more curved surfaces and are harder to nucleate

Page 10: Chem. 250 – 11/18 Lecture. Announcements I A.Exam 2 Results Average = 73 B.New Homework Set (Text Ch. 4: 25; Ch. 7: 3, 5, 6, 8, 10, 24, 25, 26, 35, 44

Cloud Chemistry- Nucleation of Cloud Droplets

• Nucleation more readily occurs with:- Larger particles- Particles with more water soluble compounds (due to growth according to Raoult’s law)- Compounds that reduce surface tension- Smaller aerosol number concentrations (less competition for water so higher RH values)

Page 11: Chem. 250 – 11/18 Lecture. Announcements I A.Exam 2 Results Average = 73 B.New Homework Set (Text Ch. 4: 25; Ch. 7: 3, 5, 6, 8, 10, 24, 25, 26, 35, 44

Cloud Chemistry- Nucleation of Cloud Droplets

• The concentration of constituents incorporated from

nucleation depends on the efficiency of nucleation

and on the liquid water content (or LWC).

• LWC = g liquid H2O/m3 of air

• The higher the LWC, the lower the concentration

(dilution effect)

• Cloud nucleation leads to heterogeneous cloud

droplet composition – Ignored here for calculations

Page 12: Chem. 250 – 11/18 Lecture. Announcements I A.Exam 2 Results Average = 73 B.New Homework Set (Text Ch. 4: 25; Ch. 7: 3, 5, 6, 8, 10, 24, 25, 26, 35, 44

Cloud ChemistryNucleation Example Problems

• Why is a RH over 100% required for cloud droplet nucleation?

• Why is nucleation efficiency higher in less polluted regions?

• An ammonium bisulfate aerosol that has a concentration of 5.0 μg m-3 is nucleated with 50% efficiency (by mass) in a cloud that has a LWC of 0.40 g m-3. What is the molar concentration? What is the cloud pH?

Page 13: Chem. 250 – 11/18 Lecture. Announcements I A.Exam 2 Results Average = 73 B.New Homework Set (Text Ch. 4: 25; Ch. 7: 3, 5, 6, 8, 10, 24, 25, 26, 35, 44

Cloud Chemistry- Scavenging of Gases

• Also Important for covering water chemistry (e.g. uptake of CO2 by oceans)

• For “unreactive” gases, the transfer of gases to cloud droplets depends on: the Henry’s law constant (always)

• In special cases, transfer can depend on LWC (if high), or can be limited by diffusion

• Henry’s Law: X

H PX

K

where KH = constant (at given T) and X = molecule of interest

Page 14: Chem. 250 – 11/18 Lecture. Announcements I A.Exam 2 Results Average = 73 B.New Homework Set (Text Ch. 4: 25; Ch. 7: 3, 5, 6, 8, 10, 24, 25, 26, 35, 44

Cloud Chemistry- Scavenging of Gases: “unreactive” gases

• When LWC and KH are relatively low, we can assume that PX is constantThen [X] = KH∙PX

• When KH is high (>1000 M/atm), conservation of mass must be considered (PX decreases as molecules are transferred from gas to liquid)

• We will only consider 2 cases (low KH case and 100% gas to water case)

• Example Problem (low KH case): What is the concentration of CH3OH in cloud water if the gas phase mixing ratio is 10 ppbv and a LWC of 0.2 g/m3? The Henry’s law constant is 290 M/atm (at given temp.). Assume an atmospheric pressure of 0.9 atm and 20°C.

Page 15: Chem. 250 – 11/18 Lecture. Announcements I A.Exam 2 Results Average = 73 B.New Homework Set (Text Ch. 4: 25; Ch. 7: 3, 5, 6, 8, 10, 24, 25, 26, 35, 44

Cloud Chemistry- Scavenging of Gases “unreactive” gases

• For compounds with high Henry’s law constants, a significant fraction of compound will dissolve in solution

• fA = 10-6KHRT(LWC) where fA = aqueous fraction (not used in assigned problems)

• When fA ~ 1, can use same method as for cloud nucleation From Seinfeld and Pandis (1998)

Page 16: Chem. 250 – 11/18 Lecture. Announcements I A.Exam 2 Results Average = 73 B.New Homework Set (Text Ch. 4: 25; Ch. 7: 3, 5, 6, 8, 10, 24, 25, 26, 35, 44

Cloud Chemistry- Scavenging of Gases: “reactive” gases

• Many of the gases considered are acidic and react further

• Example: Dissolution of SO2 gasReaction: Equation:SO2(g) + H2O(l) ↔ H2SO3(aq) KH = [H2SO3]/PSO2

H2SO3(aq) ↔ H+ + HSO3- Ka1 = [H+][HSO3

-]/[H2SO3(aq)]

HSO3- ↔ H+ + SO3

2- Ka2 = [H+][SO32-]/[HSO3

-]

Note: concentration of dissolved SO2 = [S(IV)]

= [H2SO3] + [HSO3-] + [SO3

2-] = [H2SO3](1 + Ka1/[H+] + Ka1Ka2/[H+]2)“Effective” Henry’s law constant = KH

* = KH(1 + Ka1/[H+] + Ka1Ka2/[H+]2) = function of pH

Page 17: Chem. 250 – 11/18 Lecture. Announcements I A.Exam 2 Results Average = 73 B.New Homework Set (Text Ch. 4: 25; Ch. 7: 3, 5, 6, 8, 10, 24, 25, 26, 35, 44

Cloud Chemistry- Scavenging of Gases: “reactive” gases

• For SO2 problems in homework, assume:– Little SO2 is depleted from gas phase (usually valid)

(This means PSO2 and [H2SO3] are constant)

– pH is just determined from SO2 (usually not valid)– The third reaction can be ignored (dissociation of HSO3

- doesn’t affect pH)

• Dissolution of HNO3

– Because both KH and Ka are large, we can not assume little HNO3 is depleted from gas phase

– Better assumption is 100% transfer to aqueous phase

Page 18: Chem. 250 – 11/18 Lecture. Announcements I A.Exam 2 Results Average = 73 B.New Homework Set (Text Ch. 4: 25; Ch. 7: 3, 5, 6, 8, 10, 24, 25, 26, 35, 44

Cloud Chemistry- Scavenging of Gases: “reactive” gases

• Example problem:Determine the pH and aqueous NO3

- concentration (in M) if air containing 1 ppbv enters a cloud with a pressure of 0.90 atm, a T = 293K, and a LWC of 0.50 g/m3. Assume 100% scavenging.

Page 19: Chem. 250 – 11/18 Lecture. Announcements I A.Exam 2 Results Average = 73 B.New Homework Set (Text Ch. 4: 25; Ch. 7: 3, 5, 6, 8, 10, 24, 25, 26, 35, 44

Cloud Chemistry- Combining two scavenging methods

example including ammonium bisulfate, sulfur dioxide and carbon dioxide

Equilibrium pH where sum of anion charge = sum of cation charge

Calculation method is fairly complex (uses systematic method)

Page 20: Chem. 250 – 11/18 Lecture. Announcements I A.Exam 2 Results Average = 73 B.New Homework Set (Text Ch. 4: 25; Ch. 7: 3, 5, 6, 8, 10, 24, 25, 26, 35, 44

Cloud Chemistry- Reactions in Clouds

• Cloud reactions are important for water soluble species because of higher concentrations in clouds

• Only sulfur chemistry covered here

Page 21: Chem. 250 – 11/18 Lecture. Announcements I A.Exam 2 Results Average = 73 B.New Homework Set (Text Ch. 4: 25; Ch. 7: 3, 5, 6, 8, 10, 24, 25, 26, 35, 44

Cloud Chemistry- Reactions in Clouds

• Reaction of S(IV) and H2O2

- HSO3- + H2O2 → HSO4

- + H2O (acid catalyzed)- Rate = k[HSO3

-][H+][H2O2]

- Rate = k’[H2O2]PSO2

- Effectively pH independent (despite what text says)

Page 22: Chem. 250 – 11/18 Lecture. Announcements I A.Exam 2 Results Average = 73 B.New Homework Set (Text Ch. 4: 25; Ch. 7: 3, 5, 6, 8, 10, 24, 25, 26, 35, 44

Cloud Chemistry- Reactions in Clouds

• Reaction of S(IV) and Ozone- Two main reactions:HSO3

- + O3 → HSO4- + O2 moderately

fast SO3

2- + O3 → SO42- + O2 fast

reaction is faster at high pH because more S(IV) is present in reactive forms

Page 23: Chem. 250 – 11/18 Lecture. Announcements I A.Exam 2 Results Average = 73 B.New Homework Set (Text Ch. 4: 25; Ch. 7: 3, 5, 6, 8, 10, 24, 25, 26, 35, 44

Cloud Chemistry- Reactions in Clouds

1E-14

1E-12

1E-10

1E-08

1E-06

1E-04

1E-02

S(I

V)

loss

rate

(M

/s)

2 3 4 5 6 7 pH

O3 HCHO H2O2

Page 24: Chem. 250 – 11/18 Lecture. Announcements I A.Exam 2 Results Average = 73 B.New Homework Set (Text Ch. 4: 25; Ch. 7: 3, 5, 6, 8, 10, 24, 25, 26, 35, 44

Precipitation Chemistry

• Precipitation Formation– Cloud droplets are collected by collisions

with rain droplets or snow crystals and transfer their contents

– Snow crystals also can form mainly through diffusion from water vapor and are very clean

• Below Cloud Scavenging– Incorporation of gases or particles

Page 25: Chem. 250 – 11/18 Lecture. Announcements I A.Exam 2 Results Average = 73 B.New Homework Set (Text Ch. 4: 25; Ch. 7: 3, 5, 6, 8, 10, 24, 25, 26, 35, 44

Cloud/Precipitation ChemistrySome Questions

1. Which reactant for sulfur dioxide oxidation is likely to be most important if a cloud is nucleated on a soil dust aerosol? on an acidic sulfate aerosol?

2. Two snow events occur down-wind of a pollution source. In one case, the snow is mostly crystals formed from diffusional growth. In the other the snow grew by removing cloud droplets. How will the snow composition be different?

Page 26: Chem. 250 – 11/18 Lecture. Announcements I A.Exam 2 Results Average = 73 B.New Homework Set (Text Ch. 4: 25; Ch. 7: 3, 5, 6, 8, 10, 24, 25, 26, 35, 44

Water ChemistryHydrological Systems

• Most of water on Earth is in the ocean• Much of the freshwater is inaccessible for use• Groundwater is becoming an increasingly

important resource

from Girard

Page 27: Chem. 250 – 11/18 Lecture. Announcements I A.Exam 2 Results Average = 73 B.New Homework Set (Text Ch. 4: 25; Ch. 7: 3, 5, 6, 8, 10, 24, 25, 26, 35, 44

Water ChemistryHydrological Systems

• The hydrologic cycle is the cycle by which water is distributed around the Earth

• Evaporation removes most of the non-volatile constituents of water

• For this reason, atmospheric source of many compounds are not large (although they are important atmospheric sinks)

• As with clouds, regions of heavier precipitation tend to have greater “dilution” of pollutants

• Water flowing through sediments can add or remove constituents from Girard

Page 28: Chem. 250 – 11/18 Lecture. Announcements I A.Exam 2 Results Average = 73 B.New Homework Set (Text Ch. 4: 25; Ch. 7: 3, 5, 6, 8, 10, 24, 25, 26, 35, 44

Water ChemistryHydrological Systems

Sacramento Valley

Mg - Long. Plot

0102030405060708090

120.8121.0121.2121.4121.6121.8122.0122.2

Longitude (deg. W)

Mg

(ppm

)

Tap Water West East Transect

sediments

granite

Groundwater??

River Water??

Data From Chem. 31N

Page 29: Chem. 250 – 11/18 Lecture. Announcements I A.Exam 2 Results Average = 73 B.New Homework Set (Text Ch. 4: 25; Ch. 7: 3, 5, 6, 8, 10, 24, 25, 26, 35, 44

Water ChemistryProperties of Water

• See Text for boiling point/melting point and heat capacity properties

• Temperature – Density Relationship: density maximum occurs at 4°C and ice density is much lower than water density

• Note: if density increases with depth, water is stable.

from GirardKotz et al., “Chemistry and Chemical Reactivity” (6th Ed.)

Page 30: Chem. 250 – 11/18 Lecture. Announcements I A.Exam 2 Results Average = 73 B.New Homework Set (Text Ch. 4: 25; Ch. 7: 3, 5, 6, 8, 10, 24, 25, 26, 35, 44

Water ChemistryProperties of Water

• In the oceans, production of dense water that can sink occurs when warm water evaporates producing cool water with high salinity

• This only occurs in two areas (near Iceland and near Antarctica)

• The volume of deep water formed equals the volume of upwelling water

Page 31: Chem. 250 – 11/18 Lecture. Announcements I A.Exam 2 Results Average = 73 B.New Homework Set (Text Ch. 4: 25; Ch. 7: 3, 5, 6, 8, 10, 24, 25, 26, 35, 44

Water ChemistryWater Composition

• Salt Water- main ions are sodium (1.06%) and chloride (1.9%) with lower amounts of magnesium and sulfate- main compound affecting pH is HCO3

- ion (a weak base)

• Fresh Water- main ions are HCO3

-, Mg2+, Ca2+, Na+, and Cl-

- main source of major ions is dissolution of carbonates

e.g. CaCO3(s) + CO2(g) + H2O(l) ↔ Ca2+ + 2HCO3-

Page 32: Chem. 250 – 11/18 Lecture. Announcements I A.Exam 2 Results Average = 73 B.New Homework Set (Text Ch. 4: 25; Ch. 7: 3, 5, 6, 8, 10, 24, 25, 26, 35, 44

Water ChemistryWater Composition

• Dissolved solids– Mass of material left after evaporating

water– Expressed in ppm– Surrogate measure is electrical

conductivity

Page 33: Chem. 250 – 11/18 Lecture. Announcements I A.Exam 2 Results Average = 73 B.New Homework Set (Text Ch. 4: 25; Ch. 7: 3, 5, 6, 8, 10, 24, 25, 26, 35, 44

Water ChemistryChemical Reactions

• Acid-Base EquilibriaDissociation of water (always important)H2O ↔ H+ + OH-

Carbon dioxide reactions:1) Acid-Base Reactions

CO2 (g) ↔ CO2 (aq) KH = 0.0338 M/atmCO2 (aq) + H2O ↔ H+ + HCO3

- Ka1 = 4.45 x 10-7

HCO3- ↔ H+ + CO3

2- Ka2 = 4.7 x 10-11

Page 34: Chem. 250 – 11/18 Lecture. Announcements I A.Exam 2 Results Average = 73 B.New Homework Set (Text Ch. 4: 25; Ch. 7: 3, 5, 6, 8, 10, 24, 25, 26, 35, 44

Water Chemistry Chemical Reactions

• Acid-Base Properties – continuedNote: If water is in contact with atmosphere, PCO2 = fixed value, so [CO2] = independent of pH

Other equations useful for solving water chemistry equations:

Mass balance: T = [CO2] + [HCO3-] + [CO3

2-]where T = total carbonate concentration

Charge balance equation:

Σ(zi*[cation]i) = Σ(zj*[anion]j) zi = charge of ion i

Page 35: Chem. 250 – 11/18 Lecture. Announcements I A.Exam 2 Results Average = 73 B.New Homework Set (Text Ch. 4: 25; Ch. 7: 3, 5, 6, 8, 10, 24, 25, 26, 35, 44

Water Chemistry Chemical Reactions

• Form of carbonate as a function of pHThe fraction of carbonate species α present in a single form (e.g.

HCO3-) can be calculated as follows:

211

213

3

][)(

aaa

a

KKHKH

HKT

HCOHCO

The right part to the equation

can be derived from equilibrium equations

When pH < pKa1, CO2 is the dominant species, when pKa1 < pH < pKa2, HCO3

- is the dominant species, and when pH > pKa2, CO32-

is the dominant species

Page 36: Chem. 250 – 11/18 Lecture. Announcements I A.Exam 2 Results Average = 73 B.New Homework Set (Text Ch. 4: 25; Ch. 7: 3, 5, 6, 8, 10, 24, 25, 26, 35, 44

Water Chemistry Chemical Reactions

Composition of Carbonates

0.0

0.2

0.4

0.6

0.8

1.0

1.2

0 2 4 6 8 10 12 14

pH

Alp

ha

Val

ues

CO2

HCO3-

CO32-

80% of US surface water

Page 37: Chem. 250 – 11/18 Lecture. Announcements I A.Exam 2 Results Average = 73 B.New Homework Set (Text Ch. 4: 25; Ch. 7: 3, 5, 6, 8, 10, 24, 25, 26, 35, 44

Water Chemistry Chemical Reactions

• Second source of carbonate: dissolution or weathering of carbonate rock/soilCaCO3(s) ↔ Ca2+ + CO3

2- Ksp = 4.6 x 10-

9

This reactions normally must be considered with other reactions (because in most waters, the pH is such that [HCO3

-] >> [CO32-])

Problems normally can be solved using 1) the systematic method or 2) simplifying assumptions

Page 38: Chem. 250 – 11/18 Lecture. Announcements I A.Exam 2 Results Average = 73 B.New Homework Set (Text Ch. 4: 25; Ch. 7: 3, 5, 6, 8, 10, 24, 25, 26, 35, 44

Water Chemistry Chemical Reactions

Example of simplifying assumption:Solubility of CaCO3 in pure water (no CO2 present)Water with carbonate soils is usually in regime where

α(HCO3-) > α(CO2) > α(CO3

2-), so a more representative reaction would result in HCO3

-

By combining CaCO3(s) ↔ Ca2+ + CO32-

with H+ + CO32- ↔ HCO3

- and H2O ↔ H+ + OH-

The following is obtained: CaCO3(s) + H2O ↔ Ca2+ + HCO3- +

OH- where: K = KspKw/Ka2 = 9.7 x 10-13 = [Ca2+ ][HCO3-][OH-]

Assumption that [Ca2+] = [HCO3-] = [OH-] leads to

[Ca2+] = solubility = 9.9 x 10-5 M pH = 10.00Vs. solubility = 6.8 x 10-5 M and pH = 7.00 considering

solubility reaction only

Page 39: Chem. 250 – 11/18 Lecture. Announcements I A.Exam 2 Results Average = 73 B.New Homework Set (Text Ch. 4: 25; Ch. 7: 3, 5, 6, 8, 10, 24, 25, 26, 35, 44

Water Chemistry Chemical Reactions

Simplifying assumption when both CaCO3 and CO2 are present

Combine simplified equation for CaCO3 solubility with first 2 CO2 reactions:

CaCO3(s) + H2O ↔ Ca2+ + HCO3- + OH-

and CO2 (g) + H2O ↔ H+ + HCO3- (and H+ + OH- ↔ H2O)

Net reaction: CaCO3(s) + CO2 (g) + H2O ↔ Ca2+ + 2HCO3-

Notes: 1) increased CO2 leads to increased solubility

2) 2[Ca2+] = [HCO3-] expected

Page 40: Chem. 250 – 11/18 Lecture. Announcements I A.Exam 2 Results Average = 73 B.New Homework Set (Text Ch. 4: 25; Ch. 7: 3, 5, 6, 8, 10, 24, 25, 26, 35, 44

Water Chemistry Chemical Reactions

From Harris, Quantitative Chemical Analysis, 6th Ed., 2003

Low Carbonate Soils

Additional CO2 sources

Page 41: Chem. 250 – 11/18 Lecture. Announcements I A.Exam 2 Results Average = 73 B.New Homework Set (Text Ch. 4: 25; Ch. 7: 3, 5, 6, 8, 10, 24, 25, 26, 35, 44

Water Chemistry Chemical Reactions

Buffering capacity and alkalinityThrough their reactions carbonate soils buffer water

from the addition of acids.Alkalinity is a measure of the buffering capacity of

water.Alkalinity = mmol of acid that can be added to a 1 L

water sample before the pH → 4.5.Alkalinity = [OH-] + 2[CO3

2-] + [HCO3-]

(approximate)Alkalinity = [OH-] + 2[CO3

2-] + [HCO3-] – [H+] (better,

but still approximate equation)

Page 42: Chem. 250 – 11/18 Lecture. Announcements I A.Exam 2 Results Average = 73 B.New Homework Set (Text Ch. 4: 25; Ch. 7: 3, 5, 6, 8, 10, 24, 25, 26, 35, 44

Water ChemistrySome Problems

1. What are first and second largest reservoirs of water on Earth?

2. What two ions are the most prevalent in sea-water?

3. What three ions are the most prevalent in fresh water?

4. How do the three major ions in fresh water generally get into fresh water?

5. How do the concentrations of major ions in rain water compare with fresh water?

Page 43: Chem. 250 – 11/18 Lecture. Announcements I A.Exam 2 Results Average = 73 B.New Homework Set (Text Ch. 4: 25; Ch. 7: 3, 5, 6, 8, 10, 24, 25, 26, 35, 44

Water ChemistrySome Problems - II

1. At 5°C, the water hydrolysis equilibrium constant is 2.0 x 10-15. What is the pH of pure water (no CO2, no other sources of trace species)?

2. Determine the solubility of CaCO3 in water in equilibrium with 380 ppm CO2. What is the pH of the water? What is its alkalinity?

3. Coral is largely CaCO3. As PCO2 goes up, what will happen to the solubility of coral in the ocean? What should happen to the pH of the ocean? What will happen to [Ca2+]?

Page 44: Chem. 250 – 11/18 Lecture. Announcements I A.Exam 2 Results Average = 73 B.New Homework Set (Text Ch. 4: 25; Ch. 7: 3, 5, 6, 8, 10, 24, 25, 26, 35, 44

Water ChemistryOne last problem

1. A water sample has a measured alkalinity of 0.4 mM and a pH of 6.7. Determine the concentration of [OH-], [HCO3

-], [CO32-], and [CO2].