Neutralization reactions -
An acid lowers the pH of a solution. A base raises the pH of a solution. What happens if you mix them together?
The reaction of a strong acid with a strong base will result in the formation of water and a salt of neutral pH.
If just enough acid is added to a strong base, a neutral solution will result.
If just enough base is added to a strong acid, a neutral solution will result.
Stoichiometry of neutralization reactions - What is just enough?
HCl + NaOH à
H2O + NaCl (one mole of acid required to neutralize one mole of base)
HCl + NaOH à H2O + NaCl (one mole of acid required to neutralize one mole of base)
The stoichiometry is not always one-to-one
The stoichiometry is not always one-to-one
H2SO4 + 2NaOH à
2H2O + Na2SO4
H2SO4 + 2NaOH à 2H2O + Na2SO4
Notice is takes two moles of the base here to neutralize one mole of the sulfuric acid because it can contribute two protons.
Normality (N)– more convenient for reactions between acids and bases
based on the concept of an equivalent
one equivalent of acid can generate 1 mol of H+
one equivalent of base can generate 1 mol of OH-
Normality = equivalents / L (Volume*normality = VN = moles of equivalents)
1M HCl = 1N
1M H2SO4 = 2N
The normality of a base is the moles of OH- it
can contribute per liter of solution
The normality of a base is the moles of OH- it can contribute per liter of solution
1M NaOH = 1N
1M NaOH = 1N
1M Ca(OH)2 = 2N
To neutralize an acidic or basic solution, an equal number of equivalents of OH- and H+ must react.
Problem: What volume of 1M KOH must be added to neutralize 100mL of 0.25M H2SO4. What will the pH of the resulting solution be?
In neutralization reactions, it is easiest to deal with equivalents:
For neutralization to occur, we need to have the number of equivalents of base = number of equivalents of acid
NbaseVbase = NacidVacid
The 1M KOH solution is 1N
The 0.25 M H2SO4 solution is 0.5N
(1N)(Vbase ) = (0.5N)(0.100L)
Vbase = 0.05 L = 50 mL
Neutralization reactions results in the formation of salts with little or no acid base character and water.
2 KOH + H2SO4 à
K2SO4 + H2O
2 KOH + H2SO4 à K2SO4 + H2O
Consequently, the pH of the solutions will be near neutral
Consequently, the pH of the solutions will be near neutral = 7
Dilution calculations - when working with acids the concentration of acid is often manipulated by dilution:
Ex: 50mL of 1M HCl is diluted with water to a final volume of 100ml. What is the molarity of the final solution.
molarity = mol / L
The number of moles remains unchanged by the dilution; consequently:
M1V1 = moles = M2V2
(1M)(0.050L) = M2(0.100L)
M2 = 0.5M
Controlling the pH of solutions - Buffer solutions
Buffers resist changes in pH when acids or bases are added to them.
Buffers contain a weak acid and its salt or a weak base and its salt:
Buffers contain both a weak acid and a weak base. As both are weak, the buffer solutions are not strongly acidic or basic. Futhermore, the presence of both an acid and base allows for the pH of the solution to be “buffered” (maintains constant pH) against the addition of strong acids or bases.
For instance, CH3COOH / NaCH3COO buffer
CH3COOH / CH3COO-
The acid CH3COOH reacts with added base to remove OH-
CH3COOH + NaOH à NaCH3COO + H2O
instead of generating OH- the NaOH is neutralized
The base CH3COO- reacts with added acid to remove H+
NaCH3COO + HCl à CH3COOH + NaCl
instead of generating H+ the HCl is neutralized.
1. Acids relevant to acid rain
Many compounds of nonmetals with oxygen dissolve in water to form acids:
CO2(aq) + H2O(l) --> H2CO3(aq) (carbonic acid)
SO2(aq) + H2O(l) --> H2SO3 (sulfurous acid)
With oxygen present
2SO2(aq) + 2H2O(l) + O2(g) --> 2H2SO4 (sulfuric acid)
Similarly, NO2 and NO form HNO3 and HNO2
Water in the environment is naturally acidic due to the dissolution of CO2, pH = 5.6
2. 1872 - acid rain - Robert Smith used the term to describe acidic precipitation at the start of the industrial revolution. Acid rain has a pH < 5.6.
3. Primary source of acid rain - burning of fossil fuels, primarily coal, oil (also called petroleum) and natural gas (statistics from the department of energy)
More than 85 percent of the energy used by the United States comes from fossil fuels. Oil supplies about 40 percent of our energy; natural gas provides about 25 percent, while coal provides about 20 percent.
These fuels were formed millions of years ago from plants and animals that died and decomposed beneath tons of soil and rock.
Why fossil fuels?
SO2 primarily from electrical utilities (65%)
Coal is used primarily as a fuel in electric power plants. In fact, more than half of the electricity generated in the United States comes from plants that burn coal. Coal contains sulfur impurities which are converted to SO2 when burned. Coal from eastern states 3-10%(w/w) sulfur. Western states less than 1%(w/w) sulfur.
S(s) + O2(g) --> SO2(g)
At high temperatures (such as when burning coal in an power generation plant or gas in your engine), N2 combines with O2 at an appreciable rate to yield NOxs
N2 + O2 --> NO + NO2 (not balanced)
4. Some info:
a. average annual pH in much of northeastern US and northeastern Europe = 4.0-4.5
b. fog in West Virginia, pH = 1.5, LA pH = 3
c. more than 1000 bodies of water in eastern US are acidified
Ontario - 100 lakes devoid of fish
48,000 threatened in Ohio river valley and Great Lakes region
5. Effects of acid waters
a. damage tree leaves
b. damage tree bark -
black forest in Germany - by 1990 nearly half of the trees in the Black Forest were damaged by acid rain
c. Acid soils harm roots
d. Can leach out necessary minerals
e. Can add unwanted minerals and heavy metals
Ex. Al(OH)3(s) + 3 H+(aq) --> Al3+(aq) + 3 H2O(l) , Al3+ harmful to roots and inhibits absorption of other nutrients
f. Dissolves limestone statues and buildings
CaCO3(s) + 2 H3O+ --> Ca2+ + 3H2O + CO2(g)
6. Tough choices and fixes
a) acid water can be treated with limestone (base) - $$$ and short term
Ironically, acid lakes are beautifully clear
b) stop burning coal, stop using internal combustion engines (as now found in cars, lawn mowers, etc...)
c) Some measures to stop/minimize damage
i. plastic protecting material on statures - short lived
ii. remove pollutants from coal burning and cars
Clean Coal Technology Program - mid-1980s, the U.S. Government has invested more than $2 billion in cleaning up coal burning process - state-of-the-art processes can filter out 99% of particles and more than 95% of acid rain components (private companies and state governments have added another $4 billion)
"flue gas desulfurization units," or "scrubbers