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It is Wednesday. The first homework is due. Place it on the bench in front. Acids and bases of varying strengths. Acids and bases of varying strengths. Strong acid = 100% ionization. Strong acid = 100% donation of acidic proton.

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It is Wednesday.

The first homework is due.

Place it on the bench in front.



Acids and bases of varying strengths.

Strong acid = 100% ionization

Strong acid = 100% donation of acidic

proton.


HCl(aq) + H2O(l) H3O+(aq) + Cl-(aq)


HCl(aq) + H2O(l) H3O+(aq) + Cl-(aq)

[H3O+][Cl-]

K =

[HCl]


HCl(aq) + H2O(l) H3O+(aq) + Cl-(aq)

[H3O+][Cl-]

large

K =

=

[HCl]


Generic acid

HA(aq) + H2O(l) H3O+(aq) + A-(aq)


Generic acid

HA(aq) + H2O(l) H3O+(aq) + A-(aq)

[H3O+][A-]

= acidity

constant

= Ka

[HA]


Generic acid

HA(aq) + H2O(l) H3O+(aq) + A-(aq)

[H3O+][A-]

= acidity

constant

= Ka

[HA]

-log10 Ka = pKa


Acid Ka pKa

HI  1011-11

HCl  107  -7

H2SO4  102  -2

CH3COOH 1.8 x 10-5 4.74

Table page 332



Base strength

Inversely related to strength of

conjugate acid.


Base strength

Inversely related to strength of

conjugate acid.

H2O(l) + B(aq) HB+(aq) + OH-(aq)

conjugate acid


H2O(l) + B(aq) HB+(aq) + OH-(aq)

[HB+][OH-]

= Kb

= basicity

constant

[B]


[HB+][OH-]

= Kb

= basicity

constant

[B]

[H3O+][B]

= acidity

constant

= Ka

[HB+]

[H3O+][OH-] = Kw


[HB+][OH-]

= Kb

= basicity

constant

[B]

[H3O+][B]

= acidity

constant

= Ka

[HB+]

Conjugate acid

[H3O+][OH-] = Kw

KbKa = Kw


[HB+][OH-]

= Kb

= basicity

constant

[B]

[H3O+][B]

= acidity

constant

= Ka

[HB+]

[H3O+][OH-] = Kw

KbKa = Kw

pKb + pKa = pKw


KbKa = Kw

pKb + pKa = pKw

Expressions can be used for any

conjugate acid-base pair in water.


Indicators :

Usually a weak organic acid that has a

color different from its conjugate base.


Indicators :

Usually a weak organic acid that has a

color different from its conjugate base.

HA + H2O H3O+ + A-



Indicators :

Usually a weak organic acid that has a

color different from its conjugate base.

+ H2O H3O+ + A-




Equilibria with weak acids

and weak bases.

Weak acid:


Equilibria with weak acids

and weak bases.

Weak acid: Ka < 1


Equilibria with weak acids

and weak bases.

Weak acid: Ka < 1

[H3O+][A-]

Ka =

[HA]


Equilibria with weak acids

and weak bases.

Weak acid: Ka < 1

Ka (H3O+) = 1


Equilibria with weak acids

and weak bases.

Weak acid: Ka < 1

Ka (H3O+) = 1

Why is a weak acid weak?


HCl Ka 107 strong

HF Ka = 6.6 x 10-4 weak


HCl Ka 107 strong

HF Ka = 6.6 x 10-4 weak

HF more ionic than HCl


HCl Ka 107 strong

HF Ka = 6.6 x 10-4 weak

HF more ionic than HCl

Relative Kas show that HF holds

proton more strongly than HCl


HCl Ka 107 strong

HF Ka = 6.6 x 10-4 weak

HF more ionic than HCl

Relative Kas show that HF holds

proton more strongly than HCl

Electrostatic attraction for H+ stronger

for F- than for Cl-.


Organic acids

Ka

CH3COOH 1.8 x 10-5

CH3CH2COOH 1.3 x 10-5

CCl3COOH 2 x 10-1


Organic acids

Ka

CH3COOH 1.8 x 10-5

CH3CH2COOH 1.3 x 10-5

CCl3COOH 2 x 10-1


Ka

CH3COOH 1.8 x 10-5

CH3CH2COOH 1.3 x 10-5

CCl3COOH 2 x 10-1

The more stable the conjugate base,

the stronger the acid.


Ka

CH3COOH 1.8 x 10-5

CH3CH2COOH 1.3 x 10-5

CCl3COOH 2 x 10-1


Ka

CH3COOH 1.8 x 10-5

CH3CH2COOH 1.3 x 10-5

CCl3COOH 2 x 10-1


Ka

CH3COOH 1.8 x 10-5

CH3CH2COOH 1.3 x 10-5

CCl3COOH 2 x 10-1


Ka

CH3COOH 1.8 x 10-5

CH3CH2COOH 1.3 x 10-5

CCl3COOH 2 x 10-1


Ka

CH3COOH 1.8 x 10-5

CH3CH2COOH 1.3 x 10-5

CCl3COOH 2 x 10-1


Equilibria with weak acids

and weak bases.

HA(aq) + H2O(l) H3O+(aq) + A-(aq)

[H3O+][A-]

= Ka

[HA]


HA(aq) + H2O(l) H3O+(aq) + A-(aq)

[H3O+][A-]

= Ka

[HA]


HA(aq) + H2O(l) H3O+(aq) + A-(aq)

[H3O+][A-]

= Ka

[HA]

CH3COOH(aq) + H2O(l)

H3O+(aq) + CH3COO-(aq)


HA(aq) + H2O(l) H3O+(aq) + A-(aq)

[H3O+][A-]

= Ka

[HA]

CH3COOH(aq) + H2O(l)

H3O+(aq) + CH3COO-(aq)

Ka = 1.8 x 10-5


CH3COOH(aq) + H2O(l)

H3O+(aq) + CH3COO-(aq)

Ka = 1.8 x 10-5

1 mol CH3COOH

1 L water solution


CH3COOH(aq) + H2O(l)

H3O+(aq) + CH3COO-(aq)

Ka = 1.8 x 10-5

1 mol CH3COOH

Calculate pH

1 L water solution


[CH3COOH] [H3O+] [CH3COO-]

[M]

1.0 0 0

initial

change

Eq.


[CH3COOH] [H3O+] [CH3COO-]

[M]

1.0 0 0

initial

change

-y +y +y

Eq.


[CH3COOH] [H3O+] [CH3COO-]

[M]

1.0 0 0

initial

change

-y +y +y

Eq.

1.0 -y y y


[CH3COOH] [H3O+] [CH3COO-]

[M]

1.0 0 0

initial

change

-y +y +y

Eq.

1.0 -y y y

[H3O+][A-]

= Ka

[HA]


[CH3COOH] [H3O+] [CH3COO-]

[M]

1.0 0 0

initial

change

-y +y +y

Eq.

1.0 -y y y

[H3O+][CH3COO-]

[H3O+][A-]

= Ka

[HA]

[CH3COOH]


[CH3COOH] [H3O+] [CH3COO-]

[M]

1.0 0 0

initial

change

-y +y +y

Eq.

1.0 -y y y

[H3O+][CH3COO-]

= 1.8 x 10-5

[CH3COOH]


[CH3COOH] [H3O+] [CH3COO-]

[M]

1.0 0 0

initial

change

-y +y +y

Eq.

1.0 -y y y

(y)(y)

[H3O+][CH3COO-]

=

= 1.8 x 10-5

(1-y)

[CH3COOH]


[CH3COOH] [H3O+] [CH3COO-]

[M]

1.0 0 0

initial

change

-y +y +y

Eq.

1.0 -y y y

(y)(y)

(y2)

=

=

1.8 x 10-5

(1-y)

(1-y)


[CH3COOH] [H3O+] [CH3COO-]

[M]

1.0 0 0

initial

change

-y +y +y

Eq.

1.0 -y y y

(y)(y)

(y2)

=

=

1.8 x 10-5

(1-y)

(1-y)

y << 1


[CH3COOH] [H3O+] [CH3COO-]

[M]

1.0 0 0

initial

change

-y +y +y

Eq.

1.0 -y y y

(y)(y)

(y2)

=

=

1.8 x 10-5

(1-y)

(1-y)

1-y  1

y << 1


[CH3COOH] [H3O+] [CH3COO-]

[M]

1.0 0 0

initial

change

-y +y +y

Eq.

1.0 -y y y

(y)(y)

(y2)

=

=

= y2

1.8 x 10-5

(1-y)

(1-y)

1-y  1

y << 1


[CH3COOH] [H3O+] [CH3COO-]

[M]

1.0 0 0

initial

change

-y +y +y

Eq.

1.0 -y y y

=

1.8 x 10-5

y2

y = 4.24 x 10-3


[CH3COOH] [H3O+] [CH3COO-]

[M]

1.0 0 0

initial

change

-y +y +y

Eq.

1.0 -y y y

=

1.8 x 10-5

y2

y = 4.24 x 10-3

-log10 4.24 x 10-3 =


[CH3COOH] [H3O+] [CH3COO-]

[M]

1.0 0 0

initial

change

-y +y +y

Eq.

1.0 -y y y

=

1.8 x 10-5

y2

y = 4.24 x 10-3

-log10 4.24 x 10-3 = 2.37 = pH


[CH3COOH] [H3O+] [CH3COO-]

[M]

1.0 0 0

initial

change

-y +y +y

Eq.

1.0 -y y y

y = 4.24 x 10-3

What is the fraction of acid ionized?


[CH3COOH] [H3O+] [CH3COO-]

[M]

1.0 0 0

initial

change

-y +y +y

Eq.

1.0 -y y y

y = 4.24 x 10-3

What is the fraction of acid ionized?

[A-]

[HA]


[CH3COOH] [H3O+] [CH3COO-]

[M]

1.0 0 0

initial

change

-y +y +y

Eq.

1.0 -y y y

y = 4.24 x 10-3

What is the fraction of acid ionized?

[A-]

4.24 x 10-3

=

[HA]

1.0


[CH3COOH] [H3O+] [CH3COO-]

[M]

1.0 0 0

initial

change

-y +y +y

Eq.

1.0 -y y y

y = 4.24 x 10-3

What is the percent of acid ionized?

[A-]

4.24 x 10-3

X 100% = 0.42%

=

[HA]

1.0


[CH3COOH] [H3O+] [CH3COO-]

[M]

0.000100 0

initial

change

-y +y +y

Eq.

0.00010 -y y y

Calculate % ionization for 0.00010 M


[CH3COOH] [H3O+] [CH3COO-]

[M]

0.000100 0

initial

change

-y +y +y

Eq.

0.00010 -y y y

(y2)

(y)(y)

=

=

1.8 x 10-5

(0.0001-y)

(0.0001-y)

[A-]

[HA]


[CH3COOH] [H3O+] [CH3COO-]

[M]

0.000100 0

initial

change

-y +y +y

Eq.

0.00010 -y y y

(y2)

(y)(y)

=

=

1.8 x 10-5

(0.0001)

(0.0001)

[A-]

[HA]


[CH3COOH] [H3O+] [CH3COO-]

[M]

0.000100 0

initial

change

-y +y +y

Eq.

0.00010 -y y y

(y2)

(y)(y)

=

=

1.8 x 10-5

(0.0001)

(0.0001)

[A-]

y2 = 1.8 x 10-9

[HA]


[CH3COOH] [H3O+] [CH3COO-]

[M]

0.000100 0

initial

change

-y +y +y

Eq.

0.00010 -y y y

(y2)

(y)(y)

=

=

1.8 x 10-5

(0.0001)

(0.0001)

[A-]

y2 = 1.8 x 10-9

[HA]

y = 4.2 x 10-5


[CH3COOH] [H3O+] [CH3COO-]

[M]

0.000100 0

initial

change

-y +y +y

Eq.

0.00010 -y y y

(y2)

(y)(y)

=

=

1.8 x 10-5

(0.0001)

(0.0001)

[A-]

4.2 x 10-5

y = 4.2 x 10-5

=

1 x 10-4

42%

[HA]


1 x 10-4 and 4.2 x 10-5

do not differ enough to

ignore y.


[CH3COOH] [H3O+] [CH3COO-]

[M]

0.000100 0

initial

change

-y +y +y

Eq.

0.00010 -y y y

(y2)

(y)(y)

=

=

1.8 x 10-5

(0.0001-y)

(0.0001-y)

34%

Full solution: pg 342


1.0 M CH3COOH 0.42% ionized

0.0001 M CH3COOH 34% ionized


1.0 M CH3COOH 0.42% ionized

0.0001 M CH3COOH 34% ionized

[A-]

=

[HA]


1.0 M CH3COOH 0.42% ionized

[A-] = 4.2 x 10-3 M

0.0001 M CH3COOH 34% ionized

[A-] = 3.4 x 10-5 M

[A-]

=

[HA]


1.0 M CH3COOH 0.42% ionized

[A-] = 4.2 x 10-3 M

0.0001 M CH3COOH 34% ionized

[A-] = 3.4 x 10-5 M

[A-]

[A-]  10-2

=

[HA]

[HA]  10-4


Problem 29:

Calculate pH of 0.35 M solution of propionic

acid.


Problem 29:

Calculate pH of 0.35 M solution of propionic

acid.

Ka = 1.3 x 10-5

Table 8-2


Problem 29:

Calculate pH of 0.35 M solution of propionic

acid.

Ka = 1.3 x 10-5

Table 8-2

HA(aq) + H2O(l) H3O+(aq) + A-(aq)


Problem 29:

Calculate pH of 0.35 M solution of propionic

acid.

Ka = 1.3 x 10-5

Table 8-2

HA(aq) + H2O(l) H3O+(aq) + A-(aq)

[HA] = 0.35 M


Calculate pH of 0.35 M solution of propionic

acid.

Ka = 1.3 x 10-5

Table 8-2

HA(aq) + H2O(l) H3O+(aq) + A-(aq)

[HA] = 0.35 M

[H3O+][A-]

= Ka

[HA]


Calculate pH of 0.35 M solution of propionic

acid.

Ka = 1.3 x 10-5

Table 8-2

HA(aq) + H2O(l) H3O+(aq) + A-(aq)

[HA] = 0.35 M

[H3O+][A-]

(x2)

= Ka

=

[HA]

[HA]


Calculate pH of 0.35 M solution of propionic

acid.

Ka = 1.3 x 10-5

Table 8-2

HA(aq) + H2O(l) H3O+(aq) + A-(aq)

[HA] = 0.35 M

x2 = [HA] Ka


Calculate pH of 0.35 M solution of propionic

acid.

Ka = 1.3 x 10-5

Table 8-2

HA(aq) + H2O(l) H3O+(aq) + A-(aq)

[HA] = 0.35 M

x2 = [HA] Ka

= 0.35 x 1.3 x 10-5

=


Calculate pH of 0.35 M solution of propionic

acid.

Ka = 1.3 x 10-5

Table 8-2

HA(aq) + H2O(l) H3O+(aq) + A-(aq)

[HA] = 0.35 M

x2 = [HA] Ka

= 0.35 x 1.3 x 10-5

=

4.55 x 10-4


Calculate pH of 0.35 M solution of propionic

acid.

Ka = 1.3 x 10-5

Table 8-2

HA(aq) + H2O(l) H3O+(aq) + A-(aq)

x2 = [HA] Ka

= 0.35 x 1.3 x 10-5

=

4.55 x 10-4

4.55 x 10-4

x =


Calculate pH of 0.35 M solution of propionic

acid.

Ka = 1.3 x 10-5

Table 8-2

HA(aq) + H2O(l) H3O+(aq) + A-(aq)

x2 = [HA] Ka

= 0.35 x 1.3 x 10-5

=

4.55 x 10-4

= 2.13 x 10-3

4.55 x 10-4

x =


Calculate pH of 0.35 M solution of propionic

acid.

Ka = 1.3 x 10-5

Table 8-2

HA(aq) + H2O(l) H3O+(aq) + A-(aq)

x2 = [HA] Ka

= 0.35 x 1.3 x 10-5

=

4.55 x 10-4

= 2.13 x 10-3

4.55 x 10-4

x =

-log10 2.13 x 10-3

=


Calculate pH of 0.35 M solution of propionic

acid.

Ka = 1.3 x 10-5

Table 8-2

HA(aq) + H2O(l) H3O+(aq) + A-(aq)

x2 = [HA] Ka

= 0.35 x 1.3 x 10-5

=

4.55 x 10-4

= 2.13 x 10-3

4.55 x 10-4

x =

-log10 2.13 x 10-3

= 2.67



Equilibria involving weak bases

H2O(l) + B(aq) OH-(aq) + HB+(aq)


Equilibria involving weak bases

H2O(l) + B(aq) OH-(aq) + HB+(aq)

Acid1 base2 base1 acid2


Equilibria involving weak bases

H2O(l) + B(aq) OH-(aq) + HB+(aq)

Acid1 base2 base1 acid2

[OH-][HB+]

= Kb

[B]


Equilibria involving weak bases

H2O(l) + B(aq) OH-(aq) + HB+(aq)

Acid1 base2 base1 acid2

[OH-][HB+]

= Kb

[B]

Weak base: Kb < 1


Equilibria involving weak bases

H2O(l) + NH3(aq) OH-(aq) + NH4+(aq)

Acid1 base2 base1 acid2

Kb = ?


Equilibria involving weak bases

H2O(l) + NH3(aq) OH-(aq) + NH4+(aq)

Acid1 base2 base1 acid2

Kb = ?

KaKb = Kw


Equilibria involving weak bases

H2O(l) + NH3(aq) OH-(aq) + NH4+(aq)

Acid1 base2 base1 acid2

Kw

Kb =

Ka


Equilibria involving weak bases

H2O(l) + NH3(aq) OH-(aq) + NH4+(aq)

Acid1 base2 base1 acid2

Kw

Kb =

Ka


Equilibria involving weak bases

H2O(l) + NH3(aq) OH-(aq) + NH4+(aq)

Acid1 base2 base1 acid2

Kw

Kb =

Ka

Table page 332: Ka = 5.6 x 10-10


Equilibria involving weak bases

H2O(l) + NH3(aq) OH-(aq) + NH4+(aq)

Acid1 base2 base1 acid2

1 x 10-14

Kw

=

Kb =

5.6 x 10-10

Ka


Equilibria involving weak bases

H2O(l) + NH3(aq) OH-(aq) + NH4+(aq)

Acid1 base2 base1 acid2

1 x 10-14

Kw

=

= 1.8 x 10-5

Kb =

5.6 x 10-10

Ka



Hydrolysis

The reaction of water with an ion

resulting in a change in the [H3O+]-

[OH-] balance.


Hydrolysis

The reaction of water with an ion

resulting in a change in the [H3O+]-

[OH-] balance.

Hydrolysis of anions raises pH.


Hydrolysis

The reaction of water with an ion

resulting in a change in the [H3O+]-

[OH-] balance.

Hydrolysis of anions raises pH.

Hydrolysis of cations lowers pH.


Hydrolysis

H2O(l) + F-(aq) HF(aq) + OH-(aq)


Hydrolysis

H2O(l) + F-(aq) HF(aq) + OH-(aq)

Ka (HF) = 6.6 x 10-4


Hydrolysis

H2O(l) + F-(aq) HF(aq) + OH-(aq)

Ka (HF) = 6.6 x 10-4

[HF][OH-]

Kb =

[F-]


Hydrolysis

H2O(l) + F-(aq) HF(aq) + OH-(aq)

Ka (HF) = 6.6 x 10-4

[HF][OH-]

Kb =

[F-]

pH 0.095 M NaF


Hydrolysis

H2O(l) + F-(aq) HF(aq) + OH-(aq)

Ka (HF) = 6.6 x 10-4

Kw

[HF][OH-]

=

Kb =

Ka

[F-]

pH 0.095 M NaF


Hydrolysis

H2O(l) + F-(aq) HF(aq) + OH-(aq)

Ka (HF) = 6.6 x 10-4

[F-]Kw

Kw

[HF][OH-]

=

=

[OH-]2

Ka

[F-]

Ka

pH 0.095 M NaF


Hydrolysis

H2O(l) + F-(aq) HF(aq) + OH-(aq)

Ka (HF) = 6.6 x 10-4

(0.095)(1 x 10-14)

[F-]Kw

=

=

[OH-]2

(6.6 x 10-4)

Ka

pH 0.095 M NaF


Hydrolysis

H2O(l) + F-(aq) HF(aq) + OH-(aq)

(0.095)(1 x 10-14)

[F-]Kw

=

=

=

[OH-]2

(6.6 x 10-4)

Ka


Hydrolysis

H2O(l) + F-(aq) HF(aq) + OH-(aq)

(0.095)(1 x 10-14)

[F-]Kw

=

=

=

[OH-]2

(6.6 x 10-4)

Ka

1.44 x 10-12

=

[OH-]2


Hydrolysis

H2O(l) + F-(aq) HF(aq) + OH-(aq)

(0.095)(1 x 10-14)

[F-]Kw

=

=

=

[OH-]2

(6.6 x 10-4)

Ka

1.44 x 10-12

=

[OH-] = 1.2 x 10-6

[OH-]2


Hydrolysis

H2O(l) + F-(aq) HF(aq) + OH-(aq)

[OH-] = 1.2 x 10-6

1 x 10-14

=

[H3O+] =

1.2 x 10-6


Hydrolysis

H2O(l) + F-(aq) HF(aq) + OH-(aq)

[OH-] = 1.2 x 10-6

1 x 10-14

8.34 x 10-9

=

[H3O+] =

1.2 x 10-6


Hydrolysis

H2O(l) + F-(aq) HF(aq) + OH-(aq)

[OH-] = 1.2 x 10-6

1 x 10-14

8.34 x 10-9

=

[H3O+] =

1.2 x 10-6

-log10

(8.34 x 10-9) = 8.08


Hydrolysis

H2O(l) + F-(aq) HF(aq) + OH-(aq)

pH = 8.08


It is Wednesday.

The first homework is due.

Place it on the bench in front.


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