Reaction Type
 Acid/base reaction
Types of Acids/Bases
 Arrhenius: Acids donate hydrogen ions to solution; bases donate hydroxide ions to
solution.
 Bronstëd-Lowry: Acid donates protons; bases accept protons. This means water can act
as both an acid and a base.
 Lewis: Acid accepts electron pairs; bases donate electron pairs.
Strong vs. Weak Acids/Bases
 Strong acids ionize to completion in aqueous solutions; weak ones ionize only partially.
 Strong acids are:
o HCl, HBr, HI, HClO3, HClO4, HNO3, and H2SO4.
 Metal hydroxides are all strong bases. Most metal hydroxides are only slightly soluble.
Only the hydroxides of group 1A metals, Strontium and Barium have appreciable
solubility; calcium hydroxide is moderately soluble.
 Strong bases are:
o Group 1A and Group 2A Metal Hydroxides (append necessary OHs to 1st and 2nd
columns of periodic table, i.e. LiOH and Ca(OH)2)
 Most weak acids are organic compounds; common weak mineral acids:
o HF, H2CO3, H3PO4, H3AsO4, HClO3, HClO2, and HClO
 All bases related to ammonia, NH3 are weak bases. Organic compounds related to
ammonia are called amines.
 Electrolytes: A substance which disassociates which ionizes completely into two ions in
solution. In essence, this is every strong acid and base.
 To identify electrolytic substances, note a handful of things:
o Bond types: is it ionic or molecular? Ionics are all strong electrolytes.
o Molecular: Is there an H at the beginning or COOH at the end? How large is the
molecule?
Net Ionic Reactions
 Net ionic reactions are the display of acid-base neutralizations with just the net reactants
and products. Two examples:
1. The molecular reaction NaOH + HCl  H2O + NaCl produces the net ionic reaction H+
+ OH-  H2O. if you put just NaOH or HCl in solution, they completely disassociate, so
when they’re put together, they only produce a salt which disassociates and water. The
salt isn’t included in the net ionic reaction because it would be two separate ions whether
or not the two part solution had been created.
2. The molecular reaction Mg(OH)2 + 2HCl  MgCl2 + 2H2O produces what net ionic
reaction? Mg(OH)2, because it’s not a soluble base, would persist in solution; HCl,
however, completely disassociates. So, in the net ionic reaction, we get Mg(OH)2 + 2H+
 Mg2+ + 2H2O. The H2O is the newly created substance, and Mg2+ is the ion newly
released into solution.
Autoionization of Water
 Water naturally lends protons to/receives protons from other water molecules, acting as
Bronstëd acids and bases. This produces a concentration constant of 10-14 for water, being
Kw. This is the ion product constant.

Kw = [OH-] [H3O+], or the multiple of the hydronium and hydroxide concentrations.
Therefore, the concentration of each can be calculated if the other is known.
 When the two concentrations are equal, the solution is neutral, and as one increases the
other decreases so that they always equal Kw.
 pKw = pH + pOH = 14
pH = -log[H+]
Calculations involving Ka and Kb
 Ka = acid dissociation constant. The higher the Ka the stronger the acid.
o HA  H+ + A-
Ka =
[H+ ][A− ]
[HA]
Understanding pH
Salts


When acids and bases react, they produce salts. Generally, they produce neutral salts.
However, when strong acids react with weak bases, they produce acidic salts; likewise,
strong bases reacting with weak acids produces basic salts.
Explanations of Titration Curves




Titration curves display the pH of a titration at any point of the titration when base is
added to an acid solution or vice versa
Titration curves generally take on an S shape or inverted S shape with when titrating
strong acids and strong bases.
The starting point pH is when 0 ml of titrant has been added. The equilibrance point is
when the amount of titrant added is equivalent to amount of analyte in moles of each
substance.
o In monoprotic acids, there is only one equivalence point
Polyprotic Acids
o With polyprotic acids, there are multiple equivalence points
o At each equivalence point except for the last equivalence point, one needs to
determine whether the substance will continue to have a more acidic or basic
tendency by solving for its Ka and Kb.
 If the Ka > Kb then the substance will act as an acid, and vice versa.
Remember: Kw = Ka * Kb

o At the last equivalence point for polyprotic acids, the substance must act as a base
since there are no more H+ ions to donate
o If base is added beyond the final equivalence point, to determine pH, find the
moles of the base added and divide by total volume. Then with the Molarity, one
can take the -log of the value for the pOH. Subtract it from 14 for pH.
Example
Given Ka = 6.6e10-4
R eaction
I nitial Conditions
C hange
E quilibrium
↔
HF
0.200
-x
0.200-x
H+
0.000
+x
+x
+
F0.000
+x
+x
Plugging the equilibrium values into the equation provides
(𝑋)(𝑋)
6.6E10-4=.200−𝑥
Because .200 is much larger than x in the grand scheme of things, the eq’n is simplified to:
(𝑋)(𝑋)
6.6E10-4= .200
And some simple algebra provides x = .0115

Kb = base dissociation constant. The higher the Kb the stronger the base
o B  BH+ + OH-
Example
Given Kb = 4.2e10-4
R eaction
CH3NH2
I nitial Conditions
0.500
C hange
-x
E quilibrium
0.500 - x
Kb =
+
[BH+ ][OH− ]
[B]
H2O
55.5
-x
55.5 - x
↔
CH3NH2+ +
0.000
+x
+x
OH0.000
+x
+x
The greyed column is not included in calculations.
Plugging the equilibrium values into the equation provides
(𝑋)(𝑋)
4.2E10-4= .500
And some simple algebra yields x = .0145


pKa = -logKa; the same is true for pKb
[A− ]
pH = pKa + log[HA]
Examples of Titration Curves
Using a polyprotic acid, H2SeO3, we’ll demonstrate all the different possibilities with titration
curves. There is 20.00 mL of .250 M acid and it’s being titration by .250 M NaOH.
Ka1 = 2.7E10-3 and Ka2 = 2.5E10-7.
First, a RICE (yay Dana!) chart when no base has been added.
+
R eaction
H2SeO3
H2O
↔
H3O+
I nitial Conditions
0.250
0
C hange
-x
+x
E quilibrium
0.250 - x
+x
+
HSeO30
+x
+x
Basic algebra and the usage of the equation gets us x = .02598 M; thus, the pH = 1.59.
Next, we’ll try adding 10 mL of the base.
10.00 mL x .25 M base= 2.5 mmol of base; 20.00 mL x .25 M acid = 5 mmol of base
First, we subtract the base from the acid, because all of the base will go to titrating the acid. This
means, after we’ve added 10 mL NaOH we have 2.5 mmol left. We just plug that in and it’s
business as usual.
In the equation, we’d normally have to divide the new acid amount and the derived acid
amount over 30 mL (the original 20 mL acid plus the 10 mL of base), but because we’d divide
them both by the same thing, it’s unnecessary.
 P.S. From here on out, for simplicity’s sake, we’ll show only the simplifying assumptions in
the RICE (yay Dana!) charts.
+
R eaction
H2SeO3
H2O
↔
H3O+
+
HSeO3I nitial Conditions
2.5
0
2.5
C hange
-x
+x
+x
E quilibrium
0.250
+x
2.5
-3
Again, plug it in to the equation and solve for x, which equals 2.7 x 10 , and the pH = 2.57.
Now, an equivalence point.
We just saw that 10 mLs cuts the acid amount in half, so 20 mLs should completely convert to
the monoprotic derivative. This is an equivalence point. At an equivalence point, we have one
extra step: deciding if this is a base or an acid. So, at the first equivalence point, we look at the
next possible Ka, Ka2, and the Kb which would move the reaction in reverse, which is Kb1.
The given acid constant is 2.5 x 10-7, and Kb1 =
𝐾𝑤
.𝐾𝑎1
=
10−14
2.7 𝑥 10−3
= 3.7 𝑥 10−12 .
The acid constant is greater, thus it’s an acid. From there, it’s the same thing we’ve been doing, but
now starting with the monoprotic acid instead of the diprotic one. 5.00mmol/40 mL = .125 M
+
R eaction
HSeO3H2O
↔
I nitial Conditions
.125
C hange
-x
E quilibrium
.125
Simple algebra again yields x = 1.76E-4, and pH = 3.75.
H3O+
0
+x
+x
+
SeO322.5
+x
2.5
Next is an equivalence point which uses Kb instead of Ka.
Double the original equivalence point is 40 mL, which should be the next one. Because there are
no more hydrogen atoms to pulled away, we know we’re using Kb this time; what’s more, you
can’t go on with acid equivalencies because you just don’t have any more Kas.
So, using Ka2 and the equation from the last example, we get that Kb=4 x 10-8. Now that we’ve
completely titrated away both hydrogens, we start from the end and move backwards with
5mmol of SeO32-. 5mmol/60 mL =.08333M.
+
R eaction
SeO32H2O
↔
OH+
I nitial Conditions
.08333
0
C hange
-x
+x
E quilibrium
.08333
+x
Once more, simple algebra. Really. X=5.77x10-5 M, the pOH=4.24, and pH=9.76.
HSeO30
+x
x
The last thing you need to know is what happens with excess titrant.
This is actually really easy. Just take however much base you have past the final equivalence
point (we’ll do 5 ml), multiply by the concentration (.25 M) and divide by the total solution (now
65 mL). The new OH- concentration is .01923 M. So, the pOH is 1.72 and the pH is 12.28.
And now you know titration curves. Ta da!!
Buffers
 A buffer solution is a solution at equilibrium which contains both a weak acid and its
conjugate weak base, or a base and its conjugate acid e.g. HA ↔ H+ + A-.
 When an acid is added to this solution, the H+ concentration will initially increase, but the
increase will be buffered by the conjugate base, which will neutralize the H+ released by
the acid; an added base will similarly be buffered by the conjugate acid present in
solution i.e. it’ll throw off H+.
o Example: HCl is added to HA ↔ H+ + A-; the H+ ions released by HCl will be
neutralized by the A- ions, and Cl- is a spectator ion. This is an example of the
common ion effect; HCl and HA both share the H+ ion in their formulas, so HA
acted as a buffer.
[𝐴− ]
 Henderson-Hasselbalch equation: pH = pKa + 𝑙𝑜𝑔 [𝐻𝐴]. This equation calculates the pH of

a buffer solution.
The buffer capacity is generally based on amount of acid and base in solution; two 1-L
solutions of 1 M and .01 M acid/base have the same pH, but the former will have a
greater buffering capacity because the acid/base is simply present in larger amounts.
Designing a buffer solution:
 When designing a buffer solution, one will be given what the desired pH is, volume, and
what its buffering capacity will be.
o First using appendices given, select a weak acid or base with Ka given, after finding
[𝐴− ]
pKa, one can the know what the pH needs to be for the expression 𝑙𝑜𝑔 [𝐻𝐴].
o Using the buffering capacity required, one can determine the moles of A- and
setting up the equation for the previous pH needed and the A- concentration then
one can solve for the [HA]. Then it can be diluted to meet volume requirements.