Dissociation of H2O: H2O ↔ H+ + OH-

Kw = aH+ • aOH-

aH2O

Under dilute conditions:ai = [i]

And aH2O = 1Hence:Kw = [H+] • [OH-]

At 25oC Kw = 10^-14

At 25oC Kw = 10^-14

Neutral water [H+] = [OH-] At 25oC pH of neutral water = 7

Kw is temperature dependent

At 0o C Kw = 10^-14.727 and neutral water has a pH of 7.363

At 60o C Kw = 10^-13 and neutral water has a pH of 6.5

National Atmospheric Deposition ProgramData from Denali National Park, Alaska

Measurements of the pH of precipitation at Denali National Park over the last 23 years average approximately 5.3

pH of natural precipitaion

pH at other remote localities:Siberia – 6.0Australian Desert – 5.8Amazon River Basin near coast – 4.7Amazon River Basin 2230 kilometers inland – 5.3Iceland – 5.5

pH of natural precipitation between 5 and 6: this is due largely to the dissolution of atmospheric gases (especially CO2 ) into the precipitation,

River (+ Lake) Water:

Normal range between 6.5 and 8.5 with maximum frequency around 7.5

Why if natural rain water is acidic is river water typically neutral to slightly basic?

Most weathering reactions consume H+

MgSiO3 + 2 H+ + H2O ↔

Mg++ + H4SiO4

pH of ocean water between 7.8 and 8.4: controlled by equilibrium involving CO2

Photosynthesis uses up CO2 in photic zone (near surface) creating higher pH’s. Respiration and decay release CO2 in deeper waters leading to a decrease in pH.

Formation of carbonic acid begins with dissolution of CO2 gas in water:

CO2 (gas) + H2O ↔ H2CO3 (aq)

This dissolution follows Henry’s law:[H2CO3](aq) = KCO2 • PCO2

Where PCO2 is the partial pressure of CO2 in the gas andKCO2 is the Henry’s law constant.

kCO2 is a function of temperature (Table 2.1). At 25oC it is 3.92 • 10^-2 = 10^-1.47

What will the pH of precipitation be in 2100 if PCO2 doubles?

Today pCO2 = 10^-3.5

Equations

1. [H2CO3] = 10^-1.47 • 2 • 10^-3.5

2. [H+] • [HCO3-] = Ka1 = 10 ^ -6.35

[H2CO3]

Assuming that the solution is dilute so that ai ~ [i]

Additional equation comes from charge balance:Number of negative charges = number of positive charges

3. [H+] = [OH-] + [HCO3-] + 2 • [CO3

-2]Where concentrations are in moles/liter

Note that we must multiply the concentration of the ion by the absolute value of its charge

Equations

1. [H2CO3] = 10^-1.47 • 2 • 10^-3.5

2. [H+] • [HCO3-] = Ka1 = 10 ^ -6.35

[H2CO3]3. [H+] = [OH-] + [HCO3

-] + 2 • [CO-2]

Simplifying assumption in an acidic solution [OH-] and [CO3

-2] are very small so charge balance (Equation 3) becomes3’. [H+] ~ [HCO3

-]

Plugging equations 1 and 3’ into 2 we get:

[H+]^2 = 10^-6.35

10^-1.47 • 2 • 10^-3.5 [H+] = 10^-5.51 pH = 5.51

In contrast pH of precipitation in equilibrium CO2 in atmosphere today = 5.66

Next we will consider the effect of carbonate minerals (minerals that contain CO3

-2) on pH of water

2nd most abundant carbonate mineral is:

Dolomite CaMg(CO3)2

Dolomite dissociates through the reaction:CaMg (CO3)2 (solid) ↔ Ca+2(aq) + Mg+2 (aq) + 2 CO3

-2 (aq)Ksp

= 10-17.02

What is the pH of an aqueous solution in equilibrium with dolomite and the atmosphere.

Our equations (note T = 25o):1. aCa++ • aMg++ • aCO3--^2

= Ksp = 10^-17.02

aCaMg(CO3)2

2. [H2CO3](aq) = KCO2 • PCO2 = 10^-1.47PCO2

3. aH+ • aHCO3- = Ka1 = 10 ^ -6.35

aH2CO3

4. aH+ • aCO3-- = Ka2 = 10 ^ -10.33

aHCO3-

5. 2 • [Ca+2] + 2 • [Mg+2] + [H+] = [OH-] + [HCO3-] + 2 • [CO3

-2]

Our simplifying assumptions

i. CaMg(CO3)2 is a pure solid so that its activity = 1

ii. Ai ~ [i] (at least for first pass)

iii. All Ca+2 and Mg+2 comes from the dissociation of dolomite so that [Ca+2] = [Mg+2]

iv. Ca+2 ,Mg+2 and HCO3- are the dominant ions so that

[HCO3-1] ~ 2 • [Ca+2] + 2 • [Mg+2] = 4[Ca+2]

It is now all over but for the algebra

A. Combining equation 1 with simplifying assumptions i, ii and iii we get:

1’. [Ca+2]^2 • [CO3-2]^2 = Ksp or [Ca+2] • [CO3

-2] = √Ksp

B. Use simplifying assumption iv to solve for [Ca+2], and simplifying assumption ii and equation 4 to solve for [CO-2]. Substitute the results into equation 1’ to get: 4’. 0.25 • [HCO3

-]^2 Ka2 = √Ksp

[H+]

C. Substitute the expression for [H2CO3] in equation 2 into equation 3. Then use simplifying assumption ii and the new equation 3 to solve for [HCO3

-]. Substitute the results into 4’ to get: 4”. 0.25 • (Ka1 • kCO2 • PCO2)^2 •Ka2 = √Ksp

[H+]^3

Or 0.25 • (Ka1 • kCO2 • PCO2)^2 •Ka2

√Ksp( )

1/3[H+] =

Plugging the numbers in we get a pH of 8.35

Waters will initially be acidic but once they come into contact with dolomite they will become basic.

As your book shows calcite has a very similar effects.

Interaction with carbonate minerals on of the best ways of countering the effects of acid precipitation.

If water is in contact with the atmosphere pH calculation known as open system calculations if not they are closed system calculations.

Open systems

Closed systems

For closed systems we often do not know PCO2 must have additional information.One assumption often used is the constant total carbon (CT) assumption see book page 47.