ACIDS AND BASES 3 WEAK ACID BASE...

ACIDS AND BASES – 3WEAK ACID BASE

EQUILIBRIUMDr. Sapna Gupta

ACID BASE EQUILIBRIUM

Acid base equilibrium occurs when we deal with weak acids and bases.

We can write an acid equilibrium reaction for the generic acid, HA.

HA(aq) + H2O(l) H3O+(aq) + A−(aq)

Acetic acid is a weak acid. It reacts with water as follows (HA = HC2H3O2)

HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2−(aq)

Equilibrium constant, Kc, can be calculated as in any equilibria (products over reactants); however here Kc is referred to as Ka as it is specific for acids. Liquid water is not part of the equilibrium as water has no concentration.

Note: the higher the Ka value – the more the concentration of hydronium ion and the stronger the acid.

Dr. Sapna Gupta/Acids-Bases - Equilibrium 2

HA

A OH3a

K

FOR SOME ACIDS


Ka

EXAMPLE: QUICK CHECK: FOR SOME ACIDS

You have prepared dilute solutions of equal molar concentrations of HC2H3O2 (acetic acid), HNO2, HF, and HCN. Rank the solutions from the highest pH to the lowest pH. Use data table given before or your text book.

Solution:

Acid Ka pH

HF 6.8 x 10-4

HNO2 4.5 x 10-4

HC2H3O2 1.7 x 10-5

HCN 4.9 x 10-10


Ka

highest

lowest

CALCULATIONS USING

• Given the Ka values and concentrations of acids, one can calculate the concentrations of all ions at equilibrium, and then pH of the acid.

• Remember for weak acids and bases, the ionization is not complete, so not all the hydronium ions will be in solution.

• If the pH of the solution is given then x can be calculated from it.

• These setup is similar to equilibrium setup.

• For the equilibrium conc. of C1 (C1-x); x can usually be ignored because in weak acids and bases “x” is very small. To know when to ignore x use the equation shown. If the percent is less than 5% then ignore x.


HA(aq) H2O(l) H3O+(aq) A−(aq)

Initial C1 0 0

Change –x +x +x

Equilibrium C1 – x x x

x1005%iC

x

Ka

EXAMPLE: CALCULATING

Calculate the pH of a 0.18 M solution of a weak acid that has Ka = 9.2 x 106.

Solution: HA(aq) H+ (aq) + A-(aq)

Initial 0.18 M 0 0

Change -x +x +x

At Eq. 0.18-x x x

Ignore –x for equilibrium conc.

Check % ionization


xM

xx

0.1810x9.2 6 ))((

M

x

0.1810x9.2 6

2

xM 310x1.3

0.72%1000.18

10x1.3 3

xM

M5%0.72%

( . x ) . 3pH log 1 3 10 2 89M

pH

EXAMPLE: CALCULATING

Para−hydroxybenzoic acid is used to make certain dyes. What are the concentrations of this acid, of hydronium ion, and of para−hydroxybenzoateion in a 0.200 M aqueous solution at 25°C? What is the pH of the solution and the degree of ionization of the acid? The Ka of this acid is 2.6 × 10−5.

Solution: (use HA as a generic acid)

HA + H2O H3O+ + A−


HA

A OH3

a

K

252.6 10

0.200

x

x

5

0.200 1000, so 0.200 0.200

2.6 10x

3

3pH log [H O ] log (2.3 10 )

2610 5.2 x

0.200

2

a

xK

]O[H 10 2.3 3

3 x

2.64 pH

pH

EXAMPLE: CALCULATING , % IONIZATION

Sore−throat medications sometimes contain the weak acid phenol, HC6H5O. A 0.10 M solution of phenol has a pH of 5.43 at 25°C.

a) What is the acid−ionization constant, Ka, for phenol at 25°C?

b) What is the degree of ionization?

Solution:

If pH = 5.43 then [H3O+] = 3.7 × 10−6 M = x = [C6H5O−].

[HC6H5O] = 0.10 – x = 0.10 M. (x is a very small amount so it’s subtraction from the original conc. will give almost the same original conc.)


HC6H5O(aq) + H2O(l) H3O+(aq) + C6H5O−(aq)

Initial 0.10 0 0

Change –x +x +x

Equilibrium 0.10 – x x x

Ka

EXAMPLE…CONTD…

HC6H5O + H2O H3O+ + C6H5O−

[HC6H5O] = 0.10 M [H3O+] = [C6H5O−] = 3.7 × 10−6 MOnce we have all the values we can calculate Ka.

Degree of ionization is ratio of ionized to original concentrations.

Percent ionization = 3.7 10−3% or 0.0037%


OHHC

OHC OH

56

563a

K

0.10

)10 (3.7 26

a

K

10a 1.4 10K

6

5

3.7 10Degree of ionization

0.10 0.10

Degree of ionization 3.7 10

x

CALCULATING FOR WEAK BASES

• The degree to which a weak base ionizes depends on

• the concentration of the base.

• the equilibrium constant for the ionization called the base ionization constant, Kb.

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

• Solving for x gives the hydroxide ion conc. – not the hydronium ion conc., so one more calculation is needed to get the pH – calculate [H3O+] using Kw.

• Example of some weak base equations:

Ammonia becomes ammonium ion: NH3+ H2O NH4+ + OH−

Ethylamine becomes ethyl ammonium ion: C2H5NH2 + H2O C2H5NH3+ + OH−


[B]

]][OH[HBb

K

pH

SOME FOR WEAK BASES


Kb

EXAMPLE: USING TO CALCULATE

Determine the Kb of a weak base if a 0.50 M solution of the base has a pH of 9.59 at 25°C.

Solution:

pOH = 14.00-9.59 = 4.41

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

Initial 0.050 M

Change -3.89 x 10-5 M +3.89 x 10-5 M +3.89 x 10-5 M

Eq. conc 0.050* 3.89 x 10-5 M 3.89 x 10-5 M

* 3.89 x 10-5 M is too small – so it does not have to be subtracted


.[ ] . x 4 41 5OH 10 3 89 10 M

[B]

]][OH[HBb

K

( . ).

.

59

b

3 89x10 23 0 x 10

0 50

MK

M

pH Kb

POLYPROTIC ACIDS

A polyprotic acid has more than one acidic proton—for example, H2SO4, H2SO3, H2CO3, H3PO4.

These acids have successive ionization reactions with Ka1, Ka2, . . .

The calculations of the pH includes ionization of all the protons and then adding all the concentrations to calculate the final pH.


SOME S FOR POLYPROTIC ACIDS


Ka

EXAMPLE: POLYPROTIC ACID

Tartaric acid, H2C4H4O6, is a diprotic acid used in food products. What is the pH of a 0.10 M solution? What is the concentration of the C4H4O6

2 ion in the same solution? Ka1 = 9.2 104; Ka2 = 4.3 105.

Solution: Find the first and second ionizations…

Cannot ignore “x”


AH

HA OH

2

3

1a

K

x

x

-0.1010 9.2

24

pH

EXAMPLE:…CONTD….1

• Use the quadratic equation.

• At the end of the first acid ionization equilibrium, the concentrations are • [H3O+] = 9.1 10−3 M

• [HA−] = 9.1 10−3 M

• [H2A] = 9.0 10−4 M

• Use these to calculate the 2nd conc. of hydronium ion.


4 51 9.2 10 9.2 10a b c a

acbbx

2

42

10 9.210 9.2 245 xx

24 )(0.1010 9.2 xx

2 4 5 9.2 10 9.2 10 0x x

4 4 2 5(9.2 10 ) (9.2 10 ) 4(1)( 9.2 10 )

2(1)x

3424

10 9.6 10 4.62

)10 (1.92)10 (9.2

x

3 2 9.1 10 and 1.0 10x x

EXAMPLE:….CONTD….2

• So x = 4.3 x 10-5 M (4.3 x 10-5 /.0091 x 100% = 0.47%) so x is insignificant

• Calculating the final pH.

• Use this for pH


HA−(aq) + H2O(l) H3O+(aq) + A2−(aq)

Initial 0.0091 0.0091 0

Change –x +x +x

Equilibrium 0.0091 – x 0.0091 + x x

23

a2

H O A

HAK

5 (0.0091 )4.3 10

(0.0091 )

x x

x

We can assume that

0.0091 0.0091 and 0.0091 0.0091x x

2 54 4 6[C H O ] 4.3 10 M

34 4 6[HC H O ] 9.1 10 M

M 10 9.1 ]O[H 3

3

M 10 9.0 ]OHC[H 4

6442

)10 (9.1 logpH 3

2.04pH

KEY CONCEPTS

• Calculation of pH and pOH of weak acids and bases

• Calculation of Ka and Kb of weak acids and bases

• Calculation of pH for diprotic acids.


ACIDS AND BASES 3 WEAK ACID BASE...

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