THE MATHEMATICS IN A
TITRATION CURVE
(WITH A LITTLE BASE 10 AND
LOGARITHM ARITHMATIC ADDED)
INTRODUCTION OF BASIC CONCEPT
OF TITRATION CURVES
USING
TITRATION OF STRONG ACID WITH
STRONG BASE
How do calculations of pH where
the solution changes from acidic to
basic compare to the curve of the
titration graph?
Recall:
What is a mole? 6.022 x 10 23
molecules, atoms, or ions
# M represents the fraction ( # moles)/(1
liter of solution) and this value tells us
the concentration of the solution
pH = -- log [H+]
An example of a strong acid base
reaction is:
Add drops of NaOH
To a solution of HCL --Add an indicator which changes color
with pH
(phenalthalien turns from clear to pink)
Background Concepts: Strong Acid
titrated with a Strong Base
Consider the titration of a 50ml sample
of 1M HCl titrated with 1M NaOH
Add drops of NaOH
To 50 ml of 1M solution of HCL
3 important Calculations of pH
when the volume of NaOH added
is close to Equivalency Point:
ml of NaOH = 49.99 ml or 0.04999 liters
total volume in beaker is approximately 100
ml
There is 0.01 ml of H+ ions which have not
paired with OH- to form water.
[H+] = 10 ^ -4
pH = 4
When the ml of NaOH = 50 ml
then
total volume of added NaOH equals
volume of HCl
all H+ ions have combined with OH –
ions to form water
neutral pH = 7
note: K w = = [H+] [OH-] for this
+
volume of NaOH ---- [H ] =[OH ]
giving the equation K w =[H+] 2 and
[H+] = 10^-7.
This reviews laws of exponents –
taking square root means dividing
exponent by 2.
When ml of NaOH added = 50.01 ml
There is 0.01 ml of excess OH— ions in
solution.
[OH-]= 10^-4
Since [H+] [OH-] = 10^-14
then [H+] = 10^-14 / [OH-]
[H+] = 10^-14/10^-4
or [H+]= 10^-10 and
pH = 10
CONCLUSION: The pH changes 6
points when the volume of NaOH
changes 2 ml close to the
equivalency point for this mixture
of strong acid and strong base.
Relate this to the titration curve for
strong acids and strong bases.