Acid-Base Chemistry
The following flow-chart may be used to determine the type of acid base calculations required
in various situations. In a particular problem, it may be necessary to use this chart more than one time.
This chart does not list pH or pOH calculations, which may be done at any point once the [H+] or the
[OH-] has been calculated. In addition, the reaction of a weak acid with a weak base is not covered.
When using this chart an acid will be defined as an H+ donor or as a conjugate acid, and a base
will be defined as an OH- donor or as a conjugate base. Thus the following qualify as acids: HCl,
HSO4-, and NH4+. The following qualify as bases: NaOH, NH3, and C2H3O2-. Be careful when
looking for these so they are not missed in cases such as: NaHSO4, NH4Cl, and KC2H3O2. It is very
important to be able to identify the conjugates of weak acids or of weak bases. (Remember: a
conjugate acid - H+ = a conjugate base, and a conjugate base + H+ = a conjugate acid.) All salts
contain conjugate acids and conjugate bases, so separate salts into their ions, then classify the ions.
The ions from strong acids or strong bases are not important when predicting the acidity/basicity of the
solution (hydrolysis is negligible). The ions from weak acids or weak bases are usually important as
they will be involved in hydrolysis processes.
To use this chart begin at START and follow through each question taking the Y path if the
answer is yes and the N path if the answer is no. See the notes and the examples at the end for further
explanations. (Do not forget to convert grams or moles to molarity for most of the calculations.)
N
N
START ========> Is there an acid? ========> Is there a base? ========> STOP
Y
Y
Y
Is there a base? ======> Are the acid and base conjugates?
N
N
Is the acid strong?
Stoichiometry
(2, maybe 3)
Y
N
Stoichiometry (1)
(maybe 3)
Ka
Y
Is the base strong?
Y
Is either weak? Stoichiometry (1)
(maybe 3)
Y
N
Go to START
Buffer
with the "leftovers."
(Moles of acid, base,
and any conjugates.
There may be a new
volume.)
N
Kb
Stoichiometry (1)
(maybe 3)
The stoichiometry areas of the chart fall into three types (labeled as 1, 2, or 3 on the chart).
These refer to the distinct types of calculations that may be performed in these situations. The
procedures are:
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1. [H+] for a strong acid equals the molarity of the strong acid times the number of hydrogens it
donates.
[OH-] for a strong base equals the molarity of the strong base times the number of hydroxide
ions it donates.
2. There is a neutralization reaction. A balanced chemical equation is needed. The quantities
of both the acid and the base need to be converted to moles, and the limiting reagent
determined. The final number of moles of each of the reactants, and their conjugates are
needed. If there is a volume change step 3 is also required. Take the final list and go to the
START.
3. If the problem inquires about conditions after an acid and base have been mixed or after the
volume has been changed, then the final volume is needed. All concentrations need to be
expressed as the final moles divided by the final volume.
NOTES:
The strong acids are: HCl, HBr, HI, HNO3, H2SO4, HClO3, HClO4, plus any acid the problem
specifically labels as strong.
The strong bases are: (Li, Na, K, Rb, Cs)OH, (Ca, Sr, Ba)(OH)2, plus any base the problem
specifically labels as strong.
All acids or bases that are not strong require an equilibrium constant (K). A weak acid has a
Ka value, or if it is the conjugate acid of a weak base its Ka may be derived from the Kb of its
conjugate base. Similarly, a weak base has a Kb, or if it is the conjugate base of a weak acid its Ka
may be derived from the Kb of its conjugate base.
The relationship between the members of a conjugate acid-base pair may be expressed in two
ways:
Kw = Ka Kb = 1.00 x 10-14
pKw = pKa + pKb = 14.000
There is only ONE way to set-up a Ka problem:
CA ' H+(aq) + CB
Ka = [H+][CB]
[CA]
All coefficients and exponents MUST be 1 (ONE), any other set-up is not a Ka. (In the above
expressions CA = conjugate acid, and CB = conjugate base.) A polyprotic acid will require a repeat of
the Ka set-up for each of its hydrogen ions.
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In a similar manner, there is only ONE way to set-up a Kb problem:
CB + (H2O) ' OH- + CA
Kb = [OH-][CA]
[CB]
Again, all coefficients and exponents MUST be 1 (ONE), any other set-up is not a Kb. Note that the
H2O may or may not be present in the equilibrium expression, and is never present in the Kb
expression. This procedure will need to be repeated for a polybasic base until all of the separate Kb's
have been used.
There are ways of combining the Ka's of a polyprotic acid or the Kb's of a polybasic base.
Strictly speaking, the combined K's are no longer simple Ka or Kb situation. These combined K's may
be used as short cuts in calculating [H+] etc.
If the problem asks for the pH or the pOH of a solution this is done after all other calculations
using the definition of pH or pOH, the relationship between pH and pOH, or the HendersonHasselbalch equation (for buffers only). These relationships are:
pH = - log [H+]
pOH = - log [OH-]
pH + pOH = pKw = 14.000
Henderson-Hasselbalch:
pH = pKa + log ([CB] / [CA])
pKa = - log Ka
[CA] = conjugate acid
[CB] = conjugate base
pOH = pKb + log ([CA] / [CB])
pKb = - log Kb
© 2006 Sevagram Enterprises
EXAMPLES:
Each of the following examples will include a sample problem plus an indication of the path
through the chart to come to the type of calculation. The path through the chart will be abbreviated
with S = START, and either a Y or N for the path to follow after each question. Thus to get from the
beginning to the point labeled buffer the path would be: S-Y-Y-Y-Y. Similarly to get from the start to
the point labeled Kb: S-N-Y-N.
1. What is the pH of a 0.100 M HNO3 solution?
S-Y-N-Y
2. What is the pH of a 0.100 M HClO solution?
S-Y-N-N
(The Ka for HClO must be given or found in a table.)
3. What is the pH of a 0.100 M Ca(OH)2 solution?
S-N-Y-Y
4. What is the pH of a 0.100 M NH3 solution?
S-N-Y-N
(Kb needed)
5. What is the [H+] of a 0.100 M HF solution?
S-Y-N-N
6. What is the [H+] of a 0.100 M NH3 solution?
S-N-Y-N
7. What is the pH of a solution made by dissolving 0.25 grams of HF in 100. mL of solution?
S-Y-N-N
(Make sure grams and mL are converted to M.)
8. What is the [H+] of a 0.100 M NH4Cl solution?
S-Y-N-N
(If no Ka given, determine the value from the Kb for NH3.)
9. What is the pH of a solution made by dissolving 15. g of Ba(C2H3O2)2 in 250. mL of solution?
S-N-Y-N
(Make sure g and mL are converted to M. Also do not forget that
there are 2C2H3O2- for each Ba(C2H3O2)2, thus the acetate ion
concentration is double the barium acetate concentration.)
10. What are the concentrations of H+, H3PO4, H2PO4-, HPO42-, and PO43- in a 0.200M H3PO4
solution?
S-Y-N-N-S-Y-N-N-S-Y-N-N
(There are three Ka values for H3PO4.)
11. The pH of a solution of a weak base is 9.28, calculate Kb.
S-N-Y-N
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12. What is the pH of a solution that is 0.100 M in NH3 and 0.150 M in NH4Cl?
S-Y-Y-Y-Y
13. How many milliliters of 0.100M HCl are required to titrate 25.00mL of 0.100M Ca(OH)2?
S-Y-Y-N
14. If 45.00 mL of 0.1000 M NH3 were required to titrate 25.00 mL of HCl, what was the
concentration of the HCl solution?
S-Y-Y-N
15. What is the pH at the equivalence point in the titration of 25.00 mL of 0.1000 M Ba(OH)2 with
0.1000 M HClO?
S-Y-Y-N-S-N-Y-N
16. Calculate the pH after the addition of the following quantities of 0.150 M NaOH in the
titration of 30.00 mL of 0.100 M HF. a. 0.00 mL, b. 5.00 mL, c. 15.00 mL, d. 20.00mL,
e. 25.00mL, and f. 30.00 mL.
a. S-Y-N-N
b. S-Y-Y-N-S-Y-Y-Y-Y
Note all steps except (a.) require the sequence
S-Y-Y-N-S
c. S-Y-Y-N-S-Y-Y-Y-Y
d. S-Y-Y-N-S-N-Y-N
e. S-Y-Y-N-S-N-Y-Y
f. S-Y-Y-N-S-N-Y-Y
17. Calculate the concentration of C2O42- in a 0.100 M solution of H2C2O4.
S-Y-N-N-S-Y-N-N (Oxalic acid has two Ka values, so two cycles are required. The
K's may be combined as a short-cut with K = Ka1Ka2.)
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