Chemistry 12
UNIT 4
ACIDS AND BASES
PACKAGE #4
ACIDS
STRONG
monoprotic
CONSIDER
WEAK
polyprotic
polyprotic
H2SO4 ⇄ H+ + HSO4 -
If [H+]1 >> K2 ignore 2nd dissoc.
calculate in 2 stages since [H+]1 < K2 commonly
monoprotic
If [H+]1 < K2 do 2nd dissoc.
POLYPROTIC ACIDS:
SULPHURIC
H2SO4
OXALIC
H2C2O4
SULPHUROUS
H2SO3
PHOSPHORIC
H3PO4
CARBONIC
H2CO3
HYDROSULPHURIC
H2S
ARSENIC
H3AsO4
K1 = 6.0 x 10-3
K2 = 1.0 X 10-7
K3 = 3.0 X 10-12
MALONIC
H2C3H2O4
K1 = 1.5 x 10-3
K2 = 3.0 X 10-6
PHTHALIC
H2C8H4O4
K1 = 1.3 x 10-3
K2 = 3.9 X 10-6
MALEIC
H2C4H2O4
K1 = 1.4 x 10-2
K2 = 8.6 X 10-7
K2 = 1.3 X 10-13
DISSOCIATION MECHANISM for H3PO4:
K1: H3PO4 ⇄
K2: H2PO4- ⇄
K3: HPO42- ⇄
H+ + H2PO4 H+ + HPO4 2H+ + PO4 3-
--------------------------------------------------------------------------------------------------------------------H3PO4 ⇄ H+ + H2PO4 Ka = K1 = 7.5 x 10-3
I
R
E
1.0 M
-y
1.0-y
-+y
y
-+y
y
K 1=
H3PO4
⇄
H+ + H2PO4 -
7.5 x 10-3 =
y2
assume y<<1.0
1.0-y
-3
2
7.5 x 10 = y
y = 0.0866 = [H+]
pH = 1.06
--------------------------------------------------------------------------------------------------------------------H3PO4 ⇄ 2H+ + HPO4 2Ka = K1K2=
I
1.0 M
--(7.5 x 10-3)(6.2 x 10-8)
R
-y
+2y +y
=(4.65 x 10-10)
E
1.0-y
2y
y
K 2=
H2PO4- ⇄
H+ + HPO4 2-
4.65 x 10-10 = (y)(2y)2
assume y<<1.0
1.0-y
-10
4.65 x 10
= 4y3
y = 4.88 x 10-4 x 2= [H+] = 9.76 x 10-4 M
pH = 3.01
--------------------------------------------------------------------------------------------------------------------H3PO4<====>
3H+ + PO4 3Ka = K1K2K3=
I
1.0 M
--7.5 x 10-3)(6.2 x 10-8)(2.2 x 10-13)
R
-y
+3y +y
=(1.0 x 10-22)
E
1.0-y
3y
y
K 3=
HPO4-
⇄
H+ + PO4 3-
1.0 x 10-22 = (y)(3y)3
assume y<<1.0
1.0-y
1.0 x 10-22 = 27y4
y = 1.4 x 10-6 x 3= [H+] = 4.16 x 10-6 M
pH = 5.38
--------------------------------------------------------------------------------------------------------------------POLYPROTICS RELEASE HYDROGENS WITH INCREASING DIFFICULTY
K2<<K1 - the second hydrogen is more tightly bound
(in an anion versus a neutral molecule - because the radius is smaller - electrons held closer - more
electronegative central atom) The presence of the previous H+ has "common ion" effect!
Therefore K1 is most likely, K2 is less likely , and K3 is least likely
In reality, the pH of a 1.0 M phopshoric acid solution is ~1
So, when will subsequent (secondary, tertiary) proton release have an impact on our
solution pH???
[H+]1
K2
[H+]2
TOTAL [H+]
Secondary
effect
0.010 M
H2CO3
0.010 M
H2SO3
0.010 M
H3PO4
0.010 M
H2S
0.010 M
H2SO4
0.00010 M
H2SO4
NOTE:
If [H+]1 >> K2 then [H+]2 = K2 and there is no effect
If [H+]1 ~ K2 or [H+]1 < K2 then there is a significant effect
DO A TWO STAGE CALCULATION!
GENERALLY the second / third dissociations can be ignored for weak acids unless the solution is
extremely dilute (check [H+]1 versus K2 )
GENERALLY FOR H2SO4 a two stage problem must be done
IMPORTANT NOTE:
GENERALLY WEAK POLYPROTIC ACIDS WILL HAVE INSIGNIFICANT SECOND AND
THIRD DISSOCIATIONS, UNLESS THEY ARE EXTREMELY DILUTED.
eg. 0.0000010 M H3PO4 has [H+] = 10-5 M
Therefore assumptions such as y<< 0.0000010 M are no longer valid and therefore you must calculate
the second and if applicable third dissociations.
INDICATORS
Indicators are substances that must be added to a flask when doing a titration to bring about a colour
change at the endpoint of the titration. (i.e. when all of the H3O+ is neutralized by OH- or vice versa).
Transition point:
Equivalence point:
Indicators may be weak acids or bases whose conjugate has a different characteristic colour.
OR
Indicators may be weak acids or bases which dissociate in water and form ionic compounds which have
different colours.
If Indicators are weak acids - when they lose their H+ ion to form the conjugate base, they change
colour:
HIn +
H2O ⇄
H3O+ + Inweak acid
weak base
(one colour)
(different colour)
Ka expression for indicators:
KIn = [H+][In-]
[HIn]
Consider the titration of an acid with a base:
Inside the flask:
HX
+
HIn
+
H2O
⇄
H3O+ + X-
+
HIn
the colour does not change immediately
Add OH- drop by drop from the burette
The OH- will react with the strongest acid H3O+
H3O+ + OH- ⇄
2H2O
Start to use up the H3O+, the initial reaction shifts right
Eventually all of the HX dissociates and all of the H3O+ is consumed by the OHWe have reached the endpoint.
At the endpoint, Ka = [H+]
eg. If HX has a Ka = 10-7
then its endpoint occurs when [H+] = 10-7 or pH = 7
If you choose the indicator correctly, the endpoint (transition point) = the equivalence point (stoich)
NOTE: the total [H+] is from
1) the water dissociation
2) the acid dissolved
3) the indicator itself
The acid dissolved is by far the greatest contribution
On the next drop of OH-, the HIn now reacts (because the HX and H3O+ are totally consumed)
HIn +
H2O ⇄
H3O+ + InThere is a colour change as HIn changes to InConsider
[HIn]
>
[In-]
the acid colour dominates
HIn +
H 2O
⇄
H3O+ + In-
[HIn]
<
[In-]
the base colour dominates
HIn
H 2O
⇄
H 3O +
[HIn]
=
[In-]
intermediate colour occurs
The indicators endpoint
+
+
In-
At endpoint:
KIn = [H+][In-]
[HIn]
Ka = [H+]
since
[HIn]
= [In-]
Consider the titration of a base with an acid:
In a base the HIn is in the form of In- because the H from the HIn has been donated to form the
conjugate base InH3O+ +
In+
OH- ⇄
2H2O + Inweak
strong
base
base
As soon as all of the OH- is consumed, then on the next drop, the H3O+ reacts with the InH3O+ +
In⇄
HIn +
H2O
one colour
another colour
We have reached the endpoint.
UNIVERSAL INDICATOR
-an indicator solution which changes colour several times over several pH ranges
For example:
Mixing ethyl orange (which changes from red to yellow at pH = 4.1; HEth is red)
with bromthymol blue (which changes from yellow to blue at pH = 6.8; Brom- is blue)
yields a universal indicator.
When this mixed indicator is added to a solution of pH = 6, what colour will the indicator turn?
The indicator turns yellow (it is at the basic side of the HEth indicator - the yellow side, and on the
acidic side of the BBlue indicator - the yellow side)
SAMPLE QUESTIONS:
1.
An indicator, HIn, is red in acids and blue in bases. What is the colour of the
anion, In-?
Answer
2.
The indicator Clayton Yellow changes from yellow to amber at pH = 12.7.
What is the Ka for the inidicator?
Answer
3.
BLUE
pKa = pH = 12.7
Ka = 10-12.7 = 2x10-13
(Ka = [H+])
Given the following observations, determine the pH of the sample:
Indicator added
Methyl orange
Thymol blue
Bromthymol blue
Colour changed
yellow
yellow
yellow
Answer
pH is between 4.4 and 6.0
INDICATOR USE:
1)
Select an indicator which has an endpoint matching the range of the system studied
eg. Bromothymol blue in strong acid / strong base titration
- more to come on this later
2) Select an indicator with suitable colours
-don't use phenolphtalein in blood pH study
3) Use an appropriate (small) amount
Don't use so much that your test sample is changed wrt acidic vs. basic
-due to dissociation of the indicator HIn + H2O ⇄ H3O+
+ In-
SEE BOOKLET PAGE 7
pH ranges of indicators and colour changes
SEE "APPENDIX A" handout
Relative strengths of acids in aq solutions at RT
includes indicators and their Ka values
SEE 'INDICATORS' handout See chart of pH ranges and colour changes