Acid Rain: Review of pH & Acid/Base Chemistry
Recap of Lecture 5
Overview of Acid Rain Phenomenon
The Basics: Acidity and pH
Need a refresher on
acid/base chemistry? Try this tutorial!
What is an acid?
A substance that, when dissolved in water, increases
the concentration of hydrogen ion (H+). Similarly a base
is a substance that, when dissolved in water, increases the concentration
of the hydroxide ion (OH-).
In an acidic solution, [H+] > 1.00 x 10-7 M (M º moles L-1)
In a neutral solution, [H+] = 1.00 x 10-7 M
In a basic solution, [H+] < 1.00 x
10-7 M
The pH scale
For convenience, the pH (potential hydrogen) scale was developed. This relates the [H+] to an easier number to manipulate, via:
pH = -log10[H+]
So, if the [H+] of a solution is 3 x 10-4 M, the pH would be
pH = -log10(3 x 10-4) = 3.52
Self-ionization of water
To some small degree, water will break down into ions:
H2O « H+ + OH- Kw = 1 x 10-14
Where Kw = [H+][OH-],
the ion product or equilibrium constant for this reaction.
One can show from this relationship that
pH + pOH = 14.00
Problem: The pH of blood is 7.40. What is the hydrogen ion concentration? What is the hydroxide ion concentration?
pH = -log[H+], so [H+] = 10-pH
[H+] = 10-7.4 = 3.98 x 10-8 M
To determine the [OH-], one can take either of two paths:
(a.) pH + pOH = 14.0, so pOH = 14.0 - pH = 6.6
Then [OH-] = 10-pOH
= 10-6.6 = 2.51 x 10-7 M
(b.) [H+][OH-] = 1 x 10-14, so [OH-]
= 1 x 10-14/[H+]
Then [OH-] = 1 x 10-14/3.98
x 10-8 = 2.51 x 10-7 M
Acids (and Bases) in Water
HA(aq) + H2O(l)« H3O+(aq)+ A- (aq)
K = [H3O+][A-]/[H2O][HA]
Since the water concentration does not change, it can be removed from the equation:
Ka = K[H2O] = [H3O+][A-]/[HA]
The chemical equation is typically shortened to read:
HA(aq)« H+(aq)+ A- (aq)
Ka = [H+][A-]/[HA]
Problem:
A 0.01 M solution of hydrofluoric acid (HF) is used in an experiment.
If Ka = 6.8 x 10-4, what is the pH of this solution?
HF « H+(aq) + F- (aq)
initial: 0.01 0 0
change: -x +x +x
final: 0.01-x x x
Then,
Ka = [H+][F-]/[HF]
6.8 x 10-4 = x2/0.01-x
If x is small compared to 0.01, we can disregard it in the denominator.
x2 = 6.8 x 10-6
x = 2.61 x 10-3 M
pH = -log [2.61 x 10-3] = 2.58
Atmospheric Aqueous Chemistry
Problem: Nitrogen oxides can interact with water to produce nitrous acid:
NO(g) + H2O(l)« HNO2(aq) K1 = 6.5 x 10-2 M atm-1
HNO2(aq)« H+ (aq) + NO2- (aq) K2 = 4.5 x 10-4 M
Suppose there were a raindrop in contact with these
gases; what would its pH be? [NO = 1 x 10-9 atm]
First, set up equilibrium expressions for the two processes:
K1 = [HNO2]/PNO K2 = [H+][NO2-]/[HNO2]
Both expressions contain [HNO2], so we can solve one and substitute into the other:
[HNO2] = K1 PNO ; then K2 = [H+][NO2-]/K1 PNO
However, this still leaves us with an equation with two unknowns. From the second chemical reaction, we can see that for every H+, we must have an NO2-. In other words, [H+] = [NO2-]. This allows us to simplify the equation for K2 to:
K2 = [H+][H+]/K1 PNO
Solving for [H+]:
[H+]2 = K2 K1 PNO
[H+] = (K2 K1 PNO )1/2
[H+] = [(6.5 x 10-2 M atm-1)(4.5 x 10-4M)(1 x 10-9 atm)]1/2 = 1.7 x 10-7 M
pH = -log[H+] = -log[1.7 x 10-7] = 6.77
Problem:
Vitamin
C (ascorbic acid) is a diprotic acid, H2C6H6O6.
What is the pH of a 0.10 M solution? How much ascorbate (C6H6O62-)
is in solution? K1 = 7.9 x 10-5 and K2
= 1.6 x 10-12
Try this one at home... first part of the problem is similar to the
HF problem above. The second part is more challenging. Answers given in
next lecture.
What are the acids in the atmosphere?
Inorganic acids
HSO3 + O2 ® HO2 + SO3
SO3 + H2O ® H2SO4
Organic acids - typically biological in origin
Short answer it depends on the location!!
North America:
Some regional dependence more chloride near coasts, more nitrate in heavily polluted areas (such as LA)
Cations (to balance charge): Na+, Ca2+, NH4+
Similar to the US
China:
Otherwise, afflicted with the same problems as industrialized areas.