Acid-Base Equilibrium II.pptx

Acid-Base Equilibrium II

15.1 Common Ion Effect

• _{The reduction in solubility of an ionic solid}

(salt) in solution, because one of the ions in the salt is already present in solution.

• _{According to common ion effect, if one of}

the ions is already present in solution, then less salt is capable of dissociating, so the

Example Problem #1

• _{Calculate the [H}+_{] , pH and the percent}

dissociation of HF in a solution containing 1.0M HF (Ka=7.2x10-4_{) and 1.0M NaF.}

• _{Write the equilibrium equation for the}

dissociation of HF.

• _{Set up your ice chart, except this time [F}-_{] is}

Example Problem #1 Con’t

HF H₂O _ H₃O+ _F

-1.0M 0 1.0

-x +x +x

1.0-x x 1.0+x

7.2x10-4 _{= (x)(1.0+x)}

1.0-x

Again, because the dissociation of HF is very small, we can

assume that x will not be a significant

change compared to 1.0M .

7.2x10-4 _{= (x)(1.0)}

1.0

[H₃O+_{]= x= 7.2 x10}-4_M

Example Problem #1 Con’t

• Percent dissociation = [H₃O+_{] x 100}

[HA]₀

7.2x10-4 _{x 100 = 0.072%}

You Try…

• _{Calculate the [H}+_{], pH and the percent}

dissociation of HCN in a solution containing 0.50M M HCN (Ka=6.2x10-10_{) and 0.25M KCN.}

• [H₃O+_{]= 1.2x10}-9_M

15.2 Buffer Solutions

• _{A buffer solution is one in which the pH of the}

solution is "resistant" to small additions of either a strong acid or strong base.

• _{Buffers usually consist of a weak acid and its}

15.2 Buffer Solutions

• _{Calculations are based on the equation for the}

ionization of the weak acid in water forming the

hydronium ion and the conjugate base of the acid.

HA(aq) + H₂O(l) --> H₃O+_{(aq) + A}-_(aq)

K_a = [H₃O+_][A-_]

15.2 Buffer Solutions

• _{A buffer system can be made by mixing a}

soluble compound that contains the conjugate base with a solution of the acid

– _{such as sodium acetate with acetic acid or ammonia}

with ammonium chloride.

• By knowing the K_a of the acid, the amount of

acid, and the amount of conjugate base, the pH of the buffer system can be calculated.

Example #1

• _{A buffer solution was made by dissolving 10.0}

grams of sodium acetate in 200.0 mL of 1.0 M acetic acid. Assuming the change in volume when the sodium acetate is added is not

significant, estimate the pH of the acetic acid/ sodium acetate buffer solution. The K_a for

Example #1 Con’t

HC₂H₃O₂ H₂O _ H₃O+ _C

2H3O2

-1.0M 0 0.610M

-x +x +x

1.0-x X 0.610+x

1.8x10-5 _{= (x)(0.610+x)}

1.0-x

x will be very small so we can use our

approximation.

1.8x10-5 _{= (x)(0.610)}

1.0

You Try…

• _{A buffered solution contains 0.50M acetic acid}

and 0.50M sodium acetate. Calculate the pH of this solution. (Ka= 1.8 x10-5₎

Henderson-Hasselbalch Equation

• _{Helps us find the pH of a buffer solution}

without using an ice chart.

Example #2

• _{Calculate the pH of a solution containing 0.75M}

lactic acid (Ka=1.4x10-4_{) and 0.25M sodium lactate.}

• _{pH = 3.85 + log(.25/.75)}

• _{pH= 3.37}

• _{Bases work the same way only you will find the}

You try…

• A buffered solution contains 0.25M NH₃

(K_b=1.8x10-5_{) and 0.40M ammonium chloride.}

Calculate the pH of this solution.

– _{Be careful!!! This is a base!!}

Example #3

• _{Calculate the ratio of ammonium chloride to}

ammonia that is required to make a buffer solution with a pH of 9.00. The Kb for

ammonia is 1.8 x 10-5_.

• _{First we must find the [OH}-_]

• 1.8x10-5 _{= 1x10}-5 _[NH 4+]

[NH₃]

You Try…

• _{Calculate the ratio of lithium nitrite to nitrous}

that is required to make a buffer solution with a pH of 3.00. Ka for nitrous acid is 7.1 x 10-4

Calculation of the pH of a Buffer Solution

after Addition of a Small Amount of Acid

• When a strong acid (H₃O+_{) is added to a buffer}

solution the conjugate base present in the buffer consumes the hydronium ion converting it into water and the weak acid of the conjugate base.

A-_{(aq) + H}

3O+(aq) --> H2O(l) + HA(aq) • _{This results in a decrease in the amount of}

Calculation of the pH of a Buffer Solution

after Addition of a Small Amount of Acid

• _{The pH of the buffer solution decreases by a}

very small amount because of this ( a lot less than if the buffer system was not present).

• _{An "ICE" chart is useful in determining the pH}

Example #4

• _{Calculate the pH of the solution that results when}

0.10M HCl is added to 1.0L of a buffered solution composed of 0.25M NH₃ and 0.4M NH₄Cl.

• _{Write the reaction that occurs}

NH₃ + H+  _NH 4+

• _{Set up the ICE Chart, but because this reaction can}

be assumed to go to completion we use

Example #4 con’t

NH3 H+ _ NH4+

0.25 mol 0.1 mol 0.4 mol

-0.1mol -0.1mol +0.1mol

0.15mol 0.5mol

Now that the reaction is complete, we can use the

Henderson-Hasselbalch equation to determine the pH pOH = 4.74+ log (0.5M/0.15M)

pH = 14-5.26

pOH = 5.26

Comparison…

• _{Compare the pH from Example #4 to the pH in}

Practice Problem #2.

• _{The pH did not change much because of the}

You Try…

• _{Calculate the pH of a solution when 0.100mol}

of HCl is added to 1.0L of a buffer solution consisting of 5.00M HC₂H₃O₂ and 5.00M NaC₂H₃O₂. Ka= 1.8x10-5_.

You Try again…

• _{Calculate the pH of a solution when 0.100mol}

of HCl is added to 1.0L of a buffer solution consisting of 0.050M HC₂H₃O₂ and 0.050M NaC₂H₃O₂. Ka= 1.8x10-5_.

15.3 Buffering Capacity

• _{The buffering capacity represents the amount}

of protons or hydroxide ions the buffer can absorb with a significant change in pH.

• _{A buffer with a large capacity contains large}

concentrations of buffering components.

• _{The capacity of a buffered solution is}

15.3 Buffer Capacity

• _{Since large changes in the ratio of [A-]/[HA]}

will produce large changes in pH we want to avoid this for effective buffering.

• _{Optimum [A-]/[HA] = 1 because it will be most}

Example #5: Preparing a Buffer

• _{A chemist needs a solution buffered at pH=4.30}

and can choose from the following acids and their sodium salts. Calculate the ratio of

[HA]/[A-] required for each system to yield a pH of 4.30. Which system will work best?

• _{Chloroacetic acid (Ka= 1.35x10}-3₎

• _{Propanoic acid (Ka= 1.3x10}-5₎

• _{Benzoic acid (Ka= 6.4x10}-5₎

15.4 Titrations and pH Curves

• _{Titration is a process of controlled addition of}

a known concentration acid/base to an unknown acid/base.

• _{Use to determine the concentration of a}

solution.

• _{The stoichiometric (equivalence) point is often}

Strong Acid-Strong Base Titrations

– H+ _{(aq) + OH}- _(aq) _H

2O (l)

• _{The addition of the acid to the base neutralizes}

the pH of the basic solution.

• _{We need to know the number of moles of acid}

added and the concentration of the [H+] to determine the pH.

• _{Millimoles (mmol) is commonly used because we}

Strong Acid-Strong Base Titration

• _{At the equivalence}

point the number of mmol of acid is equal to the mmol of base.

Weak Acid- Strong Base Titration

• _{How will this titration be}

different?

– _{The equivalence point}

occurs at a pH higher than 7.

– _{There is a leveling off the} pH just before the

equivalence point.

Weak Base – Strong Acid Titration

• _{How will this}

titration be

different from WA/ SB?

– _{The equivalence}

Example # 7

• _{Calculate the pH of solution at the indicated}

amounts of 0.100M NaOH added to 50.0mL of 0.200M HNO₃.

a) 0.00mL b) 10.0mL c) 20.0mL d) 50.0 mL e) 100.0mL f) 150.0mL g) 200.0 mL

**This is just a stoichiometry problem***

a) 0.00 mL NaOH added

• _{How do we determine the pH?}

b) 10.0 mL of 0.100M NaOH is added

H+ _OH- _{ H}

2O

Before rxn 50.0 mL x

0.200M = 10.0 mmol

10.0mL x 0.100M = 1.00 mmol

After rxn

10.0mmol-1.00 mmol = 9.0mmol

1.00mmol-1.00mmol=

0.00mmol

[H+]= mmol H+ left mL of solution

[H+]= 9.0 mmol

C) 20.0 mL of 0.100 M NaOH added

H+ _OH- _{ H}

2O

Before rxn 10.0 mmol 20.0mL x

0.100M = 2.00 mmol

After rxn 10.0-2.00=

8.0 mmol 2.00-2.00= 0.00 mmol

[H+]= 8.0 mmol

d) 50.0 mL of 0.100M NaOH added

H+ _OH- _{ H}

2O

Before rxn 10.0 mmol 50.0mL x

0.100M = 5.00 mmol

After rxn 10.0-5.00=

5.0 mmol 5.00-5.00= 0.00 mmol

[H+]= 5.0 mmol

e) 100.0mL of NaOH added

H+ _OH- _{ H}

2O

Before rxn 50.0 mL x

0.200M = 10.0 mmol 100.0mL x 0.100M = 10.00 mmol

After rxn 10.0-10.0 =

0.0mmol 10.0-10.0= 0.00mmol

f) 150.0 mL of NaOH added

H+ _OH- _{ H}

2O

Before rxn 50.0 mL x

0.200M = 10.0 mmol 150.0mL x 0.100M = 15.00 mmol

After rxn 10.0-10.0 =

0.0mmol 15.0-10.0= 5.0mmol

[OH-] = 5.0 mmol = 0.025M 200.0mL

g) 200.0 mL of NaOH added

Example #8

• _{Consider the titration of 50.0mL of 0.10M}

acetic acid with 0.10M NaOH. As before, we will calculate the pH at various points.

• _{Ka = 1.8x10}-5

• _{To determine the pH you must first complete}

a) 0.00mL of NaOH added

• _{This is a typical weak acid calculation…}

b) 10.0mL of 0.10 M NaOH added

HC₂H₃O₂ OH- _{ C}

2H3O2- H2O

Before

rxn 50.0 mL x 0.100M = 5.0 mmol 10.0mL x 0.100M = 1.00 mmol 0

After rxn 5.0-1.0 =

4.0mmol 10.0= 15.0-5.0mmol

1.0mmol formed

by rxn

HC₂H₃O_{2 } C₂H₃O₂- _H+

I 4.0mmol/

60.0mL = 0.067M

1.0mmol/ 60.0mL =

0.017M

0

C -x +x +x

E 0.067-x 0.17+x x

1.8x10-5 _{= (0.17+x) (x)}

c) 25.0mL of 0.10M NaOH added

d) 40.0mL of 0.10M NaOH added

• _{pH= 5.35}

e) 50.0mL 0f 0.10M NaOH

f) 60.0mL of NaOH added

g) 75.0mL of 0.10M NaOH

You Try…

• _{Hydrogen Cyanide gas (HCN), a powerful}

respiratory inhibitor, is highly toxic. It is a very

weak acid (Ka=6.2x10-10_{) when dissolved in water.} If a 50.0mL sample of 0.100M HCN is titrated

with 0.100M NaOH, calculate the pH of the solution

a) after 8.00mL of 0.100M NaOH has been added b) At the halfway point of the titration

Answers

Example #9: Calculating Ka

• _{A chemist has synthesized a monoprotic weak}

acid and wants to determine its Ka value. To do so, the chemist dissolves 2.00mmol of the solid acid in 100.0mL of water and titrates the

resulting solution with 0.0500M NaOH. After 20.0mL of NaOH has been added, the pH is 6.00. What is the Ka value for the acid?

15.5 Acid-Base Indicators

• _{There are two methods for determining the}

pH at the equivalence point.

– _{1. use a pH meter to monitor the pH and plot the}

titration curve. The vertical region of the curve represents the equivalence point.

– _{2. Use an indicator which marks the end point by}

Indicators

• _{We can determine the pH of the color change}

in an acid using the equation

• _{pH= pKa + log ([In-]/[HIn])}

• _{Where the ratio of [In-]/[HIn] = 1/10 because}

that’s the ratio required for color change.

Indicators

• _{We can determine the pH of the color change}

in a base using the equation

• _{pH= pKa + log ([In-]/[HIn])}

• _{Where the ratio of [In-]/[HIn] = 10/1 because}

that’s the ratio required for color change.

Example #10

• _{The Ka for bromothymol blue is 1.0x10}-7_{. It is}

yellow in acid (HIn) form and blue in base (In-) form. Suppose we put a few drops of this

indicator in a strongly acidic solution. If the

solution is then titrated with NaOH, at what pH will the indicator color change first be visible?

Example #11

• _{What would be the pH of the color change, if}

you put bromothyol blue in a base and then titrated it with an acid?

15.6 Solubility Equilibria and the Solubility

Product

• _{When solids dissolve there are two competing}

processes occurring:

– _{The dissolution of the solid into ions} – _{Reforming of the solid from ions}

• _{Ultimately, a dynamic equilibrium is}

established.

Solubility Product

AB(s) A+_{(aq) + B}-_(aq)

• _{To write the equilibrium expression (solubility}

product), remember we do not include solids. K_sp= [A+_][B-_]

• _{Solubility product is an equilibrium constant and}

has one value for a given solid at a given temperature.

Example #12

• _{Copper (I) bromide has a measured solubility}

of 2.0x10-4_{M at 25}◦_{C. Calculate its K} sp.

CuBr(s)  Cu+1_{(aq) + Br}-1_(aq)

K_sp= [Cu+1_][Br-1_]

K_sp= (2.0x10-4_)(2.0x10-4₎

Example #13

• Calculate the K_sp value for Bismuth sulfide (Bi₂S₃) which has a solubility of 1.0x10-15_M

Bi₂S₃(s)  2Bi+3 _{+ 3S}-2

K_sp= [Bi+3_]2_[S-2_]3

K_sp=[(2) (1.0x10-15_)]2_[(3)(1.0x10-15_)]3

Example #14

• The Ksp for copper (II) iodate (Cu(IO₃)₂) is 1.4x10-7 _{at 25}◦_{C. Calculate its solubility at}

25◦C.

Cu(IO₃)₂(s)  Cu+2 _{+ 2IO} 3-1

K_sp= [Cu+2_][IO

3-1]2

1.4x10-7_{= (x)(2x)}2

Example #15

• Calculate the solubility of solid CaF₂

(Ksp=4.0x10-11_{) in 0.025M NaF solution.}

• _{How do we solve this?}

• _{Just like common ion problems before…}

15.7 Precipitation & Qualitative Analysis

• _{We can use Q to determine if precipitation will}

occur…

Example #16

• _{A solution is prepared by adding 750.0mL of}

4.00x10-3_{M Ce(NO}

3) 3 to 300.0mL of 2.00x10-2

M KIO₃. Will Ce(IO₃) ₃ (K_sp=1.9x10-10₎

precipitate from this solution?

• First: calculate the [Ce+3_{] and [IO}

3-1] in the

mixed solution.

• Then calculate Q, same as you would K_sp.

Con’t

• _[Ce+3_{] = (750.0mL)(4.00x10}-3_{M) = 2.86x10}-3_M

(750.0mL + 300.0mL)

• [IO₃-1_{] = (300.0mL)(2.00x10}-2_{M) = 5.71x10}-3_M

(750.0mL + 300.0mL)

• Q= [Ce+3_{] [IO}

3-1]3 = (2.86x10-3)(5.71x10-3)

• Q= 5.32x10-10 _{> K}

sp=1.9x10-10