1. LO (Learning Objectives)
World of Buffers
LO (Learning Objectives)
Understand the definition of a buffer
Understand buffer action
Calculate pH of buffers and new pH if acid or base added
Salt hydrolysis

2. SALT HYDROLYSIS
Many salts dissolve in water to produce solutions which are not neutral. This is because the ions formed react with the hydroxide and hydrogen ions formed when water dissociates.
KNOW:
Conjugate bases from strong acid, have essentially no ability to gain protons, so are not really bases at all.
There are four distinct systems.
All dissociated ions are aqueous ions.
Some Salts are neutral: NaCl  Na Cl Neutral
Some Salts form weak bases: NaF  Na F are bases
Some Salts form weak acids: NH4 Cl ——> NH4+ + Cl¯ are acids
Some salts form both acids and bases NH4 F ——> NH4+ + F¯ Acid and a base (biggest Ka or Kb wins)
Kb (F-): 1.4 x < Ka (NH 4 +) =: 5.6 x Thus it is acid

3. Salt ID Base from the Na+ is neutral but NO2¯ is a base
Identify each salt as neutral, acidic, or basic.
1) NH4I
Acid because of NH4+ I- from a conjugate base from strong acid
2) NaNO2
Base from the Na+ is neutral but NO2¯ is a base
3) KNO3
Neutral because K+ not acid or base, and I- from a conjugate base from strong acid
4) KC2H3O2
Base C2H3O2¯
5) NH4CH3COO (s)
Neutral: Ka (NH4+ ) = 5.6 x and Kb = 5.6 x 10-10

4. Predict whether the following salts will from a solution that is acidic, basic or neutral.
FeF
e. NaF b.
NH4ClO2
f. LiClO4 c.
K2CO3
g. C6H5NH3NO2 d.
CaBr2
h. NH4Br

5. What is a Buffer?
Contain weak acids or weak bases and their corresponding conjugate partners (common ions).
Resist changes in pH.
Buffering capacity depends on the amount of weak acid or weak base and their common ions.
Effective pH buffering range ~ pKa  1 . Depends on how much H3O+ or OH- the buffer can absorb so depends on the concentrations of HA and A־.

6. Calculating the pH of an acidic buffer solution
Calculate the pH of a buffer whose [HA] is 0.1 mol dm-3 and [A¯] of 0.1 mol dm-3.
Ka = [H+(aq)] [A¯(aq)]
[HA(aq)]

7. Calculating the pH of an acidic buffer solution
Calculate the pH of a buffer whose [HA] is 0.1 mol dm-3 and [A¯] of 0.1 mol dm-3.
Ka = [H+(aq)] [A¯(aq)]
[HA(aq)]
re-arrange [H+(aq)] = Ka x [HA(aq)]
[A¯(aq)]

8. Calculating the pH of an acidic buffer solution
Calculate the pH of a buffer whose [HA] is 0.1 mol dm-3 and [A¯] of 0.1 mol dm-3.
Ka = [H+(aq)] [A¯(aq)]
[HA(aq)]
re-arrange [H+(aq)] = [HA(aq)] x Ka
[A¯(aq)]
from information given [A¯] = 0.1 mol dm-3
[HA] = 0.1 mol dm-3

9. Calculating the pH of an acidic buffer solution
Calculate the pH of a buffer whose [HA] is 0.1 mol dm-3 and [A¯] of 0.1 mol dm-3.
Ka = [H+(aq)] [A¯(aq)]
[HA(aq)]
re-arrange [H+(aq)] = Ka [HA(aq)] [A¯(aq)]
from information given [A¯] = 0.1 mol dm-3
[HA] = 0.1 mol dm-3
If the Ka of the weak acid HA is 2 x 10-4 mol dm-3.
[H+(aq)] = x 2 x = 2 x 10-4 mol dm-3
0.1

10. Calculating the pH of an acidic buffer solution
Calculate the pH of a buffer whose [HA] is 0.1 mol dm-3 and [A¯] of 0.1 mol dm-3.
Ka = [H+(aq)] [A¯(aq)]
[HA(aq)]
re-arrange [H+(aq)] = [HA(aq)] x Ka
[A¯(aq)]
from information given [A¯] = 0.1 mol dm-3
[HA] = 0.1 mol dm-3
If the Ka of the weak acid HA is 2 x 10-4 mol dm-3.
[H+(aq)] = x 2 x = 2 x 10-4 mol dm-3
0.1
pH = - log10 [H+(aq)] =

11. Buffer solutions
It is essential to have a weak acid for an equilibrium to be present so that conjugate base ions are also present and they can go back and forth
CH3COOH(aq) CH3COO¯(aq) H+(aq)
A strong acid can’t be used as it is fully dissociated and cannot remove H+(aq), we have no effective conjugate base and no back and forth either.
HCl(aq) ——> Cl¯(aq) H+(aq)
Adding acid
Any H+ added is removed by reacting with CH3COO¯ ions to form CH3COOH
Adding alkali
This adds OH¯ ions which react with H+ ions
Removal of H+ according to Le Chatelier’s Principle we shift right 

12. Buffer Exercise
Indicate whether each of the following solution mixtures will make a buffer solution. Explain.
(a) 50.0 mL of 0.10 M NH mL of 0.10 M NH4NO3;
Yes
(b) 50.0 mL of 0.10 M NH mL of 0.10 M HNO3; NO
(c) 50.0 mL of 0.10 M NH mL of 0.10 M HNO3;
YES
(d) 50.0 mL of 0.10 M NH4NO mL of 0.10 M NaOH;
(e) mL of 0.10 M NH4NO mL of 0.10 M NaOH;

13. Strong Acid to Water
mol HCl only has 1.0 cm3 from 1.0 cm3 of 1.00 M originally)
Q.: What is the new pH after addition of mol HCl at 298K? (assume new volume is 100 cm3 )
[H3O+] =
Being added
0.0010
100/1000
= M
99 cm3 H2O +
1cm3 added
=100 cm3
pH water = 7.00
 pH = - log(0.010) = 2.00
 pH would decrease from 7 to 2 after adding mol HCl at 298K!

14. 0.0010 mol HCl cm3 from 1.0 cm3 of 1.00 mol/l originally
Strong Acid to Weak Acid
mol HCl cm3 from 1.0 cm3 of 1.00 mol/l originally
Q.: What is the new pH after addition of mol HCl at 298K? (assume no volume is 100 cm3 ) x2
0.099 – x
= 1.7610-5
x = 1.3310-3
pH = 2.88
Before addition of HCl:
99 cm3 0.10M CH3COOH Ka = 1.7610-5M
After addition of 1.0 cm3 HCl (100ml): CH3COO- is ALL destroyed
extra [H3O+] =
0.0010
100/1000
= M M
[H3O+] =
pH = 1.95
 pH would decrease from 2.88 to 1.95 after adding mol HCl at 298K!

15. Strong Acid to BUFFER 0.0010 mol HCl
Q.: What is the new pH after addition of mol HCl at 298K? (assume no volume change)
Acid 0.10
Base 0.10
X 1.7610-5 = H+
H+ = 1.7610-5
pH = 4.75
Before addition of HCl:
100 cm3 0.1M CH3COOH
& 0.1M CH3COO- Na+
n=CV
=0.100 x 0.100
Or moles of EACH
After addition of HCl: extra [H3O+] = M acid new increased to M
[CH3COO-] remained = – = M
acid
base
x 1.7610-5
H+ = 2.1510-5
pH = -log = 4.71
 pH would decrease from 4.75 to 4.67 after adding mol HCl at 298K!

16. 100cm3 0.1M CH3COOH and 0.1M CH3COO-Na+
Note the difference …
tends to resist the changes in pH when small amounts of acid is added
100cm3 H2O
100cm3 0.1M CH3COOH
100cm3 0.1M CH3COOH and 0.1M CH3COO-Na+
Initial pH 7
2.88
4.75
pH after adding mol HCl 2
1.99
4.67
i.e. it is a “BUFFER” solution!

17. 100 cm3 0.1M CH3COOH and 0.1M CH3COO- Na+
p.05
It tends to resist changes in pH when small amounts of acid or base are added, and their pH is not affected by dilution.
It is an ACIDIC BUFFER solution! (i.e. weak acid + conjugate base)
CH3COOH H2O
CH3COO- + H3O+
100 cm3 0.1M CH3COOH and 0.1M CH3COO- Na+
When acid is added, the additional H3O+ would combine with the conjugate base so that the eqm shifts LEFT.
i.e. [H3O+] does not change much, pH remains almost unchanged.
When base is added, some H3O+ ions are reacted. Some acid molecules dissociate to produce H3O+ so that the eqm shifts RIGHT
i.e. [H3O+] does not change much, pH remains almost unchanged.

18. NH3 + H2O NH4+ + OH- p.06 100 cm3 0.1M NH3 and 0.1M NH4Cl
It is an BASIC BUFFER solution! (i.e. weak base + conjugate acid)
NH H2O
NH OH-
When acid is added, some OH- are neutralized, some NH3 molecules would dissociate so that the eqm shifts FW.
100 cm3 0.1M NH3 and 0.1M NH4Cl
i.e. [OH-] does not change much, pH remains almost unchanged.
When base is added, additional OH- ions would combine with NH4+ so that the eqm shifts BW.
i.e. [OH-] does not change much, pH remains almost unchanged.

19. Calculating the pH of an acidic buffer solution
Calculate the pH of the solution formed when 500cm3 of 0.1 mol dm-3 of weak acid HX is mixed with 500cm3 of a 0.2 mol dm-3 solution of its salt NaX. Ka = 4 x 10-5 mol dm-3.
Ka = [H+(aq)] [X¯(aq)]
[HX(aq)]

20. Calculating the pH of an acidic buffer solution
Calculate the pH of the solution formed when 500cm3 of 0.1 mol dm-3 of weak acid HX is mixed with 500cm3 of a 0.2 mol dm-3 solution of its salt NaX. Ka = 4 x 10-5 mol dm-3.
Ka = [H+(aq)] [X¯(aq)]
[HX(aq)]
re-arrange [H+(aq)] = [HX(aq)] Ka
[X¯(aq)]

21. Calculating the pH of an acidic buffer solution
Calculate the pH of the solution formed when 500cm3 of 0.1 mol dm-3 of weak acid HX is mixed with 500cm3 of a 0.2 mol dm-3 solution of its salt NaX. Ka = 4 x 10-5 mol dm-3.
Ka = [H+(aq)] [X¯(aq)]
[HX(aq)]
re-arrange [H+(aq)] = [HX(aq)] Ka
[X¯(aq)]
The solutions have been mixed; the volume is now 1 dm3
therefore [HX] = mol dm-3 and
[X¯] = mol dm-3

22. Calculating the pH of an acidic buffer solution
Calculate the pH of the solution formed when 500cm3 of 0.1 mol dm-3 of weak acid HX is mixed with 500cm3 of a 0.2 mol dm-3 solution of its salt NaX. Ka = 4 x 10-5 mol dm-3.
[H+(aq)] = Ka [HX(aq)]
[X¯(aq)]
The solutions have been mixed; the volume is now 1 dm3
therefore [HX] = mol dm-3 and
[X¯] = mol dm-3
Substituting [H+(aq)] = 4 x = 2 x mol dm-3
0.1
pH = log10 [H+(aq)] =

23. Buffer with Acid Added
A mL buffer solution which is 2.00M of both HC2H3O2 and NaC2H3O2
It has a pH = What is the new pH after 2.00 mL of 6.00M HCl is added
Initial moles in buffer= (2.00mol/L)(0.500L) = 0.100mol (same for acid & base)
ADDED H+ moles=(6.00mol/L)( L) = HCl mol
The H+ ion reacts quantitatively with the conjugate base, C2H3O2- , form
of the buffer, changing the base to acid ratio and thus the pH.
 Equation H+
HC2H3O  C2H3O2
Initial moles
0.012
0.100
Change in moles
-0.012
+0.012
Final moles
0.000
0.088
0.112
New molarity
0.0
1.69
2.15