AP15NotesDoodle1.0Buffer

AP CHEMISTRY – Chapter 15 – Scotch Plains-Fanwood High School

Chapter 15 – Buffers, Titrations, and Solubility

Buffers – Solutions which resist change in pH Consist of: weak acid and conjugate base or

weak base and conjugate acid

Add Acid (H+) H+ + A- HA

Add Base (OH-) OH- + HA H2O + A-

HA/A-

Add Acid (H+) NH3 + H+ NH4+

Add Base (OH-) OH- + NH4+ H2O + NH3 NH3/NH4

+

Find the pH of a 0.75 M HC2H3O2/0.75 M NaC2H3O2 (Ka = 1.8(10)-5)

HC2H3O2 H+ + C2H3O2- Ka =

I 0.75 M 0 M 0.75 M

C

E

Find the pH of a 0.75 M NH3/0.75 M NH4+ (Kb = 1.8(10)-5)

NH3 + H2O OH- + NH4+ Kb =

I 0.75 M 0 M 0.75 M

C

E

Buffer Shortcut: (can use moles or Molarity)

[H+] = Ka A [OH-] = Kb B

B A


“3 Solution Scenarios”:

Strong Weak Buffer

Acid or Base Acid or Base Acidic or Basic

M = mol/L “ice” Ka,Kb Buffer Shortcut [H+],[OH-]

Use Molarity Use Molarity or moles

When you add something to a solution, ask “Does it react (ie – is it

an acid + base)?”

If “no,” then do appropriate “Solution Scenario.”

If “yes,” then do Stoichiometry in moles, and then appropriate

“Solution Scenario.”

Find the pH of 1.0 L of a solution of 0.25 M HCl and 0.50 M HC2H3O2

Find the pH of 1.0 L of a 0.50 M HC2H3O2/0.50 M NaC2H3O2 buffer

after adding 0.020 mol H+.


after adding 0.020 mol OH-.


Find the pH of 1.0 L of a 0.80 M NH3/0.70 M NH4+ buffer after adding

0.030 mol H+.

Find the pH of 1.0 L of a 0.80 M NH3/0.70 M NH4+ buffer after adding

0.030 mol OH-.

Buffer Capacity – compare with page 2 of notes.


after adding 0.020 mol H+.


Ways to prepare a buffer:

1. Add weak acid to conjugate base (0.10 mol HC2H3O2 and 0.10

mol NaC2H3O2 in 1.0 L of solution).

2. Add weak acid to lesser amount of strong base (0.10 mol

HC2H3O2 and 0.05 mol NaOH in 1.0 L of solution).

3. Add weak conjugate base to lesser amount of strong acid (0.10

mol NaC2H3O2 and 0.05 mol HCl in 1.0 L of solution).

4. Add weak base to conjugate acid (0.10 mol NH3 and 0.10 mol

NH4+ in 1.0 L of solution).

5. Add weak base to lesser amount of strong acid (0.10 mol NH3

and 0.05 mol HCl in 1.0 L of solution).

6. Add weak conjugate acid to lesser amount of strong base (0.10

mol NH4+ and 0.05 mol NaOH in 1.0 L of solution).


Titration Curves SA/SB WA/SB SA/WB WA/WB

pH pH pH pH

mL base added mL base added mL base added mL base added

For any titration:

1. Do Stoich in mol (Lim Reag).

Equivalence Point is when mol acid = mol base

2. Check what you have left in “Final Line” and use

appropriate Solution Scenario.

“3 Solution Scenarios”: Strong Weak Buffer

Acid or Base Acid or Base Acidic or Basic

M = mol/L “ice” Ka,Kb Buffer Shortcut [H+],[OH-]

Use Molarity Use Molarity or moles

@ ½ way point [H+] = Ka or [OH-] = Kb for WA or WB


Titration Curve – SA/SB 30.0 mL of 2.00 M HCl is titrated pH

with 4.00 M NaOH. Find the pH

after adding the following volumes of NaOH

4.00 M NaOH

2.00 M HCl = 0.060 mol mL NaOH added

0.030 L

mL NaOH added Stoich in moles “Solution Scenario”

0.00 mL NaOH No reaction yet Strong Acid

( Initial solution) [H+] = 2.00 M

pH = -0.301 H+

7.50 mL NaOH H+ + OH- H2O Strong Acid (1/2 way point) I 0.060 mol 4.00 M = 0.030 mol 0 mol [H+] = 0.030mol/0.0375L

0.0075 L = 0.800 M

H+ C -0.030 -0.030 +0.030 pH = 0.0969

F 0.030 0 0.030

15.0 mL NaOH H+ + OH- H2O Neutral Solution Equivalence pt I 0.060 mol 4.00 M = 0.060 mol 0 mol [H+] = 1.0(10)-7 M

0.0150 L

H2O C -0.060 -0.060 +0.060 pH = 7.00 M

F 0 0 0.060

16.0 mL NaOH H+ + OH- H2O Strong Base Past equivalence pt I 0.060 mol 4.00 M = 0.064 mol 0 mol [OH-] = 0.004 mol/0.0460L

0.0160 L = 0.0870 M

OH- C -0.060 -0.060 +0.060 pH = 12.85 M

F 0 0.004 0.060


Titration Curve – SA/SB – you try it 40.0 mL of 1.00 M HCl is titrated pH

with 4.00 M NaOH. Find the pH

after adding the following volumes of NaOH

4.00 M NaOH

1.00 M HCl = _______mol mL NaOH added

L


0.00 mL NaOH No reaction yet Strong Acid

( Initial solution) [H+] =

pH = H+

____ mL NaOH H+ + OH- H2O Strong Acid (1/2 way point) I ______ mol 4.00 M = ____ mol 0 mol [H+] =

L =

H+ C pH =

F

_____ mL NaOH H+ + OH- H2O Neutral Solution Equivalence pt I ______ mol 4.00 M = _____mol 0 mol [H+] =

L

H2O C pH =

F

_____ mL NaOH H+ + OH- H2O Strong Base Past equivalence pt I ______ mol 4.00 M = _____ mol 0 mol [OH-] =

L =

OH- C pH =

F


Titration Curve – WA/SB 50.0 mL of 0.10 M HC2H3O2 (Ka = 1.8(10)-5) is titrated pH

with 0.20 M NaOH. Find the pH at the following points:

0.20 M NaOH

0.10 M HC2H3O2 = 0.0050 mol 0.050 L mL NaOH added


0.00 mL NaOH No reaction yet Weak Acid

( Initial solution) HA H+ + A-

I 0.10 M 0 0 HC2H3O2 C -x +x +x

E 0.10-x x x

x = 2.7(10)-5 M = [H+], pH = 2.87 1.8(10)-5 = x2

(0.10 – x)

10.0 mL NaOH HA + OH- H2O + A- Buffer (buffer zone) I 0.0050 mol 0.20 M = 0.0020 mol 0 mol [H+] = 1.8(10)-5(0.0030)

0.0100 L (0.0020)

HC2H3O2 C -0.0020 -0.0020 +0.0020 [H+] = 2.7(10)-5 M

C2H3O2- F 0.0030 0 0.0020 pH = 4.57

12.5 mL NaOH HA + OH- H2O + A- Buffer (1/2 way point) I 0.0050 mol 0.20 M = 0.0025 mol 0 mol [H+] = 1.8(10)-5(0.0025)

0.0125 L (0.0025)

HC2H3O2 C -0.0025 -0.0025 +0.0025 [H+] = 1.8(10)-5 M

C2H3O2- E 0.0025 0 0.0025 pH = 4.74 [pH=pKa]

25.0 mL NaOH HA + OH- H2O + A- Weak Base (equivalence point) I 0.0050 mol 0.20 M = 0.0050 mol 0 mol A- + H2O HA + OH-

0.025 L I 0.0050mol/.075L 0 0

C -0.0050 -0.0050 +0.0050 C -x +x +x

C2H3O2- F 0 0 0.0050 E 0.067-x x x Kb = Kw/Ka=10-14/1.8(10)-5

x = 6.1(10)-6 M= [OH-], pH = 8.79 5.6(10)-10 = x2

(0.067 – x)

30.0 mL NaOH HA + OH- H2O + A- Strong Base (past equiv point) I 0.0050 mol 0.20 M = 0.0060 mol 0 mol [OH-] = 0.0010 mol

0.0300 L 0.0800 L

OH- C -0.0050 -0.0050 +0.0050 [OH-] = 0.0125 M

C2H3O2- F 0 0.0010 0.0050 pH = 12.10


Titration Curve – WA/SB – you try it 40.0 mL of 0.20 M HC2H3O2 (Ka = 1.8(10)-5) is titrated pH

with 0.10 M NaOH. Find the pH at the following points:

0.10 M NaOH

0.20 M HC2H3O2 = ________mol L mL NaOH added


0.00 mL NaOH No reaction yet Weak Acid

( Initial solution) HA H+ + A-

I HC2H3O2 C

E

x = _____________ M = [H+], pH =

_____ mL NaOH HA + OH- H2O + A- Buffer (buffer zone) I [H+] =

HC2H3O2 C [H+] =

C2H3O2- F pH =

_____ mL NaOH HA + OH- H2O + A- Buffer (1/2 way point) I [H+] =

HC2H3O2 C [H+] =

C2H3O2- E pH = [pH=pKa]

_____ mL NaOH HA + OH- H2O + A- Weak Base (equivalence point) I A- + H2O HA + OH-

I

C C

C2H3O2- F E Kb =

x = ___________ M= [OH-], pH =

_____ mL NaOH HA + OH- H2O + A- Strong Base (past equiv point) I [OH-] =

OH- C [OH-] =

C2H3O2- F pH =


Titration Curve – WB/SA 50.0 mL of 0.10 M NH3 (Kb = 1.8(10)-5) is titrated pH

with 0.20 M HCl. Find the pH at the following points:

0.20 M HCl

0.10 M NH3 = 0.0050 mol 0.050 L mL HCl added

mL HCl added Stoich in moles Solution Scenario

0.00 mL HCl No reaction yet Weak Base

( Initial solution) NH3+H2O NH4++ OH-

I 0.10 M 0 0 NH3 C -x +x +x

E 0.10-x x x

x = 2.7(10)-5 M = [OH-], pH = 11.13 1.8(10)-5 = x2

(0.10 – x)

10.0 mL HCl NH3 + H+ NH4+ Buffer (buffer zone) I 0.0050 mol 0.20 M = 0.0020 mol 0 mol [OH-] = 1.8(10)-5(0.0030)

0.0100 L (0.0020)

NH3 C -0.0020 -0.0020 +0.0020 [OH-] = 2.7(10)-5 M

NH4+ F 0.0030 0 0.0020 pH = 9.43

12.5 mL HCl NH3 + H+ NH4+ Buffer (1/2 way point) I 0.0050 mol 0.20 M = 0.0025 mol 0 mol [H-] = 1.8(10)-5(0.0025)

0.0125 L (0.0025)

NH3 C -0.0025 -0.0025 +0.0025 [H+] = 1.8(10)-5 M

NH4+ F 0.0025 0 0.0025 pH = 9.26 [pOH=pKb]

25.0 mL HCl NH3 + H+ NH4+ Weak Acid (equivalence point) I 0.0050 mol 0.20 M = 0.0050 mol 0 mol NH4+ H+ + NH3

0.025 L I 0.0050mol/.075L 0 0

C -0.0050 -0.0050 +0.0050 C -x +x +x

NH4+ F 0 0 0.0050 E 0.067-x x x Ka = Kw/Kb=10-14/1.8(10)-5

X = 6.1(10)-6 M= [H+], pH = 5.21 5.6(10)-10 = x2

(0.067 – x)

30.0 mL HCl NH3 + H+ NH4+ Strong Acid (past equiv point) I 0.0050 mol 0.20 M = 0.0060 mol 0 mol [H+] = 0.0010 mol

0.0300 L 0.0800 L

H+ C -0.0050 -0.0050 +0.0050 [H+] = 0.0125 M

NH4+ F 0 0.0010 0.0050 pH = 1.90


Titration Curve – WB/SA – you try it 60.0 mL of 0.30 M NH3 (Kb = 1.8(10)-5) is titrated pH

with 0.20 M HCl. Find the pH at the following points:

0.30 M HCl

0.10 M NH3 = _______ mol L mL HCl added

mL HCl added Stoich in moles Solution Scenario

0.00 mL HCl No reaction yet Weak Base

( Initial solution) NH3+H2O NH4++ OH-

I NH3 C

E

x = _______ M = [OH-], pH = 1.8(10)-5 =

_____mL HCl NH3 + H+ NH4+ Buffer (buffer zone) I [OH-] =

NH3 C [OH-] =

NH4+ F pH =

_____mL HCl NH3 + H+ NH4+ Buffer (1/2 way point) I [H-] =

NH3 C [H+] =

NH4+ F pH = [pOH=pKb]

_____mL HCl NH3 + H+ NH4+ Weak Acid (equivalence point) I NH4+ H+ + NH3

I

C C

NH4+ F E Ka =

x = ____________ M= [H+], pH =

_____mL HCl NH3 + H+ NH4+ Strong Acid (past equiv point) I [H+] =

H+ C [H+] =

NH4+ F pH =


Zone 1 Weak Acid alone

HA H+ + A-

I __ M 0 0

C -x +x +x

E __M-x x x

Ka = [H+][A-] = x2

[HA] (__M – x)

x = [H+]

Zone 2 Buffer

1st Stoich in mol

HA + OH- H2O + A-

I __ mol __ mol 0

C _____ _____ ____

F Acid 0 Base

[H+] = Ka Acid

Base

@ ½ way point, [H+] = Ka, pH = pKa

Zone 4 Excess Strong Base

1st Stoich in mol

HA + OH- H2O + A-

I __ mol __ mol 0

C _____ _____ ____

F 0 excess Base

[OH-] = excess OH-

Total L solution

4 Zones of a Titration of a Weak Acid (HA)

with a Strong Base (NaOH)

A- pH

OH-

½ way pt equiv pt A-

mL Strong Base Added HA

A-

HA

Zone 3 Weak Conjugate Base alone

1st Stoich in mol

HA + OH- H2O + A-

I __ mol __ mol 0

C _____ _____ ____

F 0 0 Base

A- + H2O HA + OH-

I __ M 0 0

C -x +x +x

E __M-x x x

Kb = [HA][OH-] = x2

[A-] (__M – x)

x = [OH-]


Zone 1 Weak Base alone

NH3 + HOH NH4+ + OH-

I __ M 0 0

C -x +x +x

E __M-x x x

Kb = [NH4+][OH-]= x2

[NH3] (__M – x)

x = [OH-]

Zone 3 Weak Conjugate Acid alone

1st Stoich in mol

NH3 + H+ NH4+

I __ mol __ mol 0

C _____ _____ ____

F 0 0 Acid

NH4+ + H+ + NH3

I __ M 0 0

C -x +x +x

E __M-x x x

Kb = [H+][ NH3] = x2

[NH4+] (__M – x)

x = [OH-]

4 Zones of a Titration of a Weak Base (NH3)

with a Strong Acid (HCl)

H+

NH4+

pH

½ way pt equiv pt NH3

mL Strong Base Added NH4+

NH4+

NH3

Zone 2 Buffer

1st Stoich in mol

NH3 + H+ NH4+

I __ mol __ mol 0

C _____ _____ ____

F Base 0 Acid

[OH-] = Kb Base

Acid

@ ½ way point, [OH-]=Kb, pOH=pKb

Zone 4 Excess Strong Acid 1st Stoich in mol

NH3 + H+ NH4+

I __ mol __ mol 0

C _____ _____ ____

F 0 excess Acid

[H+] = excess H+

Total L solution


Common Ion Effect Find the [H+] of 1.0 M HF (Ka = 7.2(10)-4)…

a) …in water

HF H+ + F- Ka = [H+][F-]

I 1.0 M 0 0 [HF]

C -x +x +x 7.2(10)-4 = x2

E 1.0-x x x (1.0-x)

H2O [H+] = x = 0.027M

b) …in 0.50 M NaF

HF H+ + F- Ka = [H+][F-]

I 1.0 M 0 0.50 M [HF]

C -x +x +x 7.2(10)-4 = x(0.50+x)

E 1.0-x x 0.50+x (1.0-x)

F- [H+] = x = 0.0014 M

Less [H+] in part (b) because less HF dissociated due to the

presence of F- in solution (Le Châtelier’s Principle).

Increased [F-] in the flask shifts the original equilibrium to

the left.


Solubility Product (Ksp) – see p. 718 for Ksp values @25°C

Find the solubility of MgF2(s) in H2O if Ksp = 6.4(10)-9

MgF2(s) Mg+2(aq) + 2F-(aq) Ksp = [Mg+2][ F-]2

I solid 0 M 0 M 6.4(10)-9 = (x)(2x)2

C -x +x +2x 6.4(10)-9 = 4x3

E solid-x x 2x solubility = x = 0.0012 M

0.0012 M 0.0024M

H2O

Common Ion Effect

Find the solubility of MgF2(s) if Ksp = 6.4(10)-9 in a 0.10M

MgCl2 solution.


I solid 0.10 M 0 M 6.4(10)-9 = (0.10+x)(2x)2

C -x +x +2x 6.4(10)-9 = (0.010)4x2

E solid-x 0.10+x 2x solubility = x = 1.6(10)-8 M

0.10 M 1.6(10)-8 M

Note that the MgF2 is much less soluble in the 0.10 M MgCl2

solution (Le Châtelier’s Principle – shifts left). Mg+2


Precipitation of Insoluble Salts

Will a precipitate from when 100. mL of a 0.10 M

MgCl2 is mixed with 100.0 mL of 0.10 M NaF?

Ksp of MgF2 = 6.4(10)-9

Mg+2 Cl- Na+ F-

Cl- Na+ Mg+2

0.10 M =0.010 mol 0.010M = 0.010 mol MgF2 F-

0.100 L 0.100 L

0.200 L total volume

To test for precipitation perform a “Q” calculation to compare w/Ksp

The insoluble solid in question is MgF2 where MgF2(s)Mg+2(aq)+2F-(aq)

Q = [Mg+2] [F-]2

Q = (0.010 mol) (0.010 mol)2 1.3(10)-4 > 6.4(10)-9

0.200 L (0.200 L)2 Q > Ksp

Q = 1.3(10)-4 YES, solid MgF2 will precipitate

How much solid MgF2 precipitates?

1st make ALL possible MgF2 precipitate out

STOICH in mol Lim Reag

Mg+2 + 2F- MgF2

I 0.010 mol 0.010 mol 0

C -0.005 mol -0.010 mol + 0.005 mol

F 0.005 mol 0 mol 0.005 mol

0.005 mol MgF2 · 62.3g/mol = 0.31 g MgF2(s) are formed


Calculate the concentration of ALL IONS in the final

solution above.

2nd make MgF2 redissolve a little bit using Ksp

0.005 mol 0 mol

0.200 L 0.200 L


I solid 0.025 M 0 M 6.4(10)-9 = (0.025+x)(2x)2

C -x +x +2x 6.4(10)-9 = (0.025)4x2

E solid-x 0.025+x 2x solubility = x = 2.5(10)-4 M

0.025 M 2.5(10)-4 M

Final Concentrations of all Ions in Solution:

[Mg+2] = 0.025 M

[F-] = 2.5(10)-4 M

[Cl-] = 2(0.0100 mol) = 0.10 M

0.200L

[Na+] = 0.0100 mol = 0.050 M

0.200L Mg+2 F- Cl- Na+

MgF2

0.31 g MgF2(s) were formed as ppt


Precipitation of Insoluble Salts – you try it

Will a precipitate from when 750. mL of a 0.00400 M

AgNO3 is mixed with 300.0 mL of 0.0200 M NaCl?

Ksp of AgCl = 1.6(10)-10

Ag+ NO3- Na+ Cl-

Ag+ NO3- Na+

M = _______ M = _______ AgCl Cl-

______ L total volume

To test for precipitation perform a “Q” calculation to compare w/Ksp

The insoluble solid in question is AgCl where AgCl(s)Ag+(aq)+Cl-(aq)

Q =

How much solid AgCl precipitates?

1st make ALL possible AgCl precipitate out

STOICH in mol Lim Reag

_____ + _____ _____

I _______ _______ _______

C _______ _______ _______

F _______ _______ _______

= _____g AgCl(s) are formed


Calculate the concentration of ALL IONS in the final

solution above.

2nd make AgCl redissolve a little bit using Ksp

____ mol ____mol

L L

____(s) ____(aq) + ____(aq) Ksp =

I _____ _____ _____ =

C _____ _____ _____ =

E _____ _____ ____solubility = x = ______ M

Final Concentrations of all Ions in Solution:

[Ag+] = __________ M

[Cl-] = __________ M

[Na+] = mol = M

L

[NO3-] = mol = M

L Ag+ Cl- Na+ NO3-

AgCl

_______ g AgCl(s) were formed as ppt


Indicators:

Indicators themselves are weak acids or bases. An

appropriate indicator for an acid/base titration will have

a pKa close to the pH of the equivalence point.

Qualitative Analysis

Use appropriate flame tests and solubility rules (selective

precipitation) to ID unknown ions.

Acid-Base properties of Oxides

Metal Oxide + H2O base Na2O + H2O 2NaOH

CaO + H2O Ca(OH)2

Nonmetal Oxide + H2O acid SO3 + H2O H2SO4

(no redox) N2O3 + H2O 2HNO2

Complex Ions

Comprised of a metal ion (Lewis acid) which accepts e-

pairs from a ligand (Lewis base).

Examples of ligands: CN-, SCN-, OH-, NH3, halogens

Several complex ion formulas can be predicted by

doubling the charge of the metal ion to determine the

number of ligands that will attach.

Ex. Ag+(aq) + 2NH3(aq) [Ag(NH3)2]+(aq)

Al+3(aq) + 6OH-(aq) [Al(OH)6]-3(aq)

Co+2(aq) + 4Cl-(aq) CoCl4-2(aq)

pH and Solubility

Mg(OH)2(s) Mg+2(aq) + 2OH-(aq) dissolves better in acid

b/c H+ + OH- H2O (shifts equilibrium to the right by removing OH-)

AP15NotesDoodle1.0Buffer

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