Strong Acids and Bases
Note
• It is important that you don't confuse the
words strong and weak with the terms
concentrated and dilute.
• As you will see, the strength of an acid is
related to the proportion of it that has reacted
with water to produce ions.
• The concentration tells you about how much
of the original acid is dissolved in the
solution.
• It is possible to have a concentrated solution
of a weak acid, or a dilute solution of a
strong acid.
Acids
• When an acid dissolves in water, a proton
(hydrogen ion) is transferred to a water
molecule to produce a hydroxonium ion
and a negative ion depending on what
acid you are starting from.
• In the general case . . .
HA + H2O
H3O+ + A• The strength of an acid is defined by the
equilibrium position of the dissociation
reaction shown above
Strong Acids
• The reaction is reversible, but with a
strong acid the equilibrium lies far to the
right. This means the original acid is
virtually 100% dissociated(ionized)
• For example, when hydrogen chloride
dissolves in water to make hydrochloric
acid, so little of the reverse reaction
happens that we can write:
HCl + H2O
H3O+ +Cl-
Strong Acids
• At any one time, virtually 100% of the
hydrogen chloride will have reacted to
produce hydroxonium ions and chloride
ions. Hydrogen chloride is described as a
strong acid.
• A strong acid is one which is virtually
100% ionized in solution.
• Other common strong acids include: HCl hydrochloric acid, HNO3- nitric acid
H2SO4- sulfuric acid, HBr- hydrobromic
acid, HI- hydroiodic acid, and HClO4perchloric acid
Strong Acid
There are four ways to describe a strong
acid:
1) Ka is large
2) Position of the dissociation equilibrium
lies far to the right
3) The equilibrium concentration of H+
approximately equal to the original
concentration of HA ([H+]≈[HA])
4) The Strength of conjugate base of a
strong acid is a much weaker base than
H2O
Working out the pH of a
strong acid
• With strong acids this is easy.
• Hydrochloric acid is a strong acid virtually 100% ionized. Each mole of HCl
reacts with the water to give 1 mole of
hydrogen ions and 1 mole of chloride ions
• That means that if the concentration of
the acid is 0.001M then the concentration
of hydrogen ions is also 0.001M
• So finding the pH is easy, just use your
equation: pH=-log[H+]
Example-Strong Acids
Calculate the pH of an aqueous
solution containing 2.5x10-5M of HCl.
Solution
So first write the dissociation expression with HCl
and water
HCl + H2O
H3O+ + ClSecond, since we know that all the HCl will be
converted to products, because HCl is a strong
acid, we know that the concentrations of H3O+
and Cl- will be the same as the original acid.
All we need to do now is plug the value into the pH
equation
Recall from yesterday pH=-log[H3O+]
Doing so we get
pH=-log(2.5x10-5)=4.6
Strong Bases
• Much the same as strong acids, strong bases
dissociate 100% into the cation and OH(hydroxide ion). The hydroxides of the Group IA
and Group IIA metals usually are considered to
be strong bases, they include:
Group IALiOH, NaOH,KOH, RbOH and CsOH(only NaOH
and KOH are common because the others are
expensive to work with)
Group IIA- Ca(OH)2, Ba(OH)2, Sr(OH)2
For these soultions 2 moles of hydroxide ion is
produced for every 1 mole of metal hydroxide
dissolved in aqueous solution.
Strong Bases
Much the same as acids, there are
four ways to describe a strong base:
1) Kb is very large
2) Position of the dissociation equilibrium
lies far to the right
3) Equilibrium concentration of OHcompared t original strong base is
approximately equal
4) Strength of conjugate acid is much
weaker acid than H2O
Working out the pH of a
strong base
• With strong bases this is easy as well.
• Sodium hydroxide is a strong base - virtually
100% ionized. Each mole of NaOH reacts with
the water to give 1 mole of hydroxide ions and 1
mole of sodium ions
• That means that if the concentration of the base
is 0.001M then the concentration of hydroxide
ions is also 0.001M
• So finding the pH is easy, use the equation
pH+pOH=14, find the pOH and subtract from 14
to get the pH.(You could also use the Kw
expression to find the [H+] then find the pH,
either way is acceptable)
Example
Calculate the pH of a 5.0x10-2M KOH
Solution
So first write the dissociation expression with KOH
and water
KOH H2O K+ + OHSecond, since we know that all the KOH will be
converted to products, because KOH is a strong
base, we know that the concentrations of K+ and
OH- will be the same as the original base
All we need to do now is find pOH and use the
realtionship between pOH and pH
Recall from yesterday pOH=-log[OH-]
Doing so we get
pOH=-log(5.0x10-2)=1.30
Next subtract from 14 to get pH
pH=14-pOH=12.7
Quizes
• Go to Mr. Richards home page on
Horton website
• Go to AP page and choose from
units tab, acids and bases
• Go to pH tutorial
• Read over sections 4 and 6 and do
quizzes