(8.2) Weak Acids & Bases: Ionization Constants

Percent Ionization for Weak Acids

Most weak acids ionize < 50% Percent ionization (p)

General Weak Acid: HA(aq) + H2O(l) H3O+ (aq) + A-

(aq)

• p varies depending on concentration: increase [HA ] decreases p

• This is caused by Le Chatelier’s Principle

• Remember, for strong acids we assume complete ionization (100%)

100xsoluteacidofionconcentrat

ionizedacidofionconcentratp

100x]HA[]H[

p)aq(

)aq(

Percent Ionization Calculation

Ex. 1

The pH of a 0.10mol/L methanoic acid (HCOOH) solution is 2.38. Calculate the percent ionization of methanoic acid.

Ans: 4.2%

Ionization Constants

Same as K, but represents acids (Ka) and bases (Kb)

Weak Acids

A weak acid is a weak electrolyte that doesn’t completely ionize when mixed with water to form H+

General formula: HA (aq) + H2O (l) H3O+ (aq) + A-

(aq)

Ka is the acid ionization constant for weak acids (don’t include pure states or units, just as with K)

The stronger the acid the larger the Ka

Ka values are found in Appendix C9 p 803

][HA

][A]O[HK

(aq)

(aq)(aq)3a

Calculating Ka

Ex. 2

A 0.24 mol/L solution of H2CO3 has a pH of 3.49.

Determine the Ka for H2CO3 when one H+ is donated.

Step 1: Write out balance chemical equation, determine we know/what we

want to know

Step 2: Determine formula for calculating Ka

Step 3: Complete an ICE table

Step 4: Determine [H3O+]

Step 5: Substitute [H3O+] into Ka equation

Ans: 4.4x10-7 (let’s compare to Appendix C Table p803 – Ka1)

Recall: B-L acids can also be polyprotic-capable of losing

(donating) more than one proton. Polyprotic acids lose their protons in separate steps or

reactions.

Ex:

H2CO3(aq) + H2O(l) HCO3-(aq) + H3O+

(aq) (Ka1)

HCO3-(aq) + H2O(l) CO3

2-(aq) + H3O+

(aq) (Ka2)

Polyprotic Acids (Ka values)

1st

H+

lost

2nd

H+

lost

Calculating Ka from Percent Ionization

Ex.3

Calculate the acid ionization constant (Ka )of acetic acid if a 0.1000mol/L solution at equilibrium at SATP has a percent ionization of 1.3% (Hint: ICE table)

Ans: 1.7x10-5

Weak Bases

Do not react completely in an aqueous solution Are molecules or ions that remove a proton from H2O and therefore

produce OH-

General Formula: B:(aq) + H2O (l) OH-(aq) + HB+

(aq)

To act as a weak B-L base a compound must have an atom with lone pairs of electrons since this is where the H+ is accepted from water

A substance can undergo two chemical changes, one can represent a strong base while another weak

For example: Na3PO4(aq) 3Na+(aq) + PO4

3-(aq) (strong)

PO43-

(aq) + H2O(l) OH- aq) + HPO4

2-(aq) (weak)

Weak Bases

Base ionization constant (Kb)

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

(aq)

In the above equation H2O acts as a B-L acid, and B as a B-L base. H20 and OH- are conjugate base pairs, as are B: and HB+

Kb values are also found in Appendix C9 pg. 803

][B

][HB][OHK

(aq)

(aq)(aq)b

Relationship Between Ka and Kb for Conjugate Base Pairs

• Recall: Conjugate Pairs – an acid and base that differ by one hydrogen

• Lets consider the hypothetical weak acid, HX, and its conjugate base, X -

HX (aq) + H2O (l) H3O+ (aq) + X- (aq)

][HX

][X]O[HK

(aq)

(aq)(aq)3a

Relationship Between Ka and Kb for Conjugate Base Pairs

• Now consider the hypothetical weak base, X -

X- (aq) + H2O (l) HX (aq) + OH- (aq)

• Now let’s put that together

HX (aq) + H2O (l) H3O+ (aq) + X- (aq) Ka

X- (aq) + H2O (l) HX (aq) + OH- (aq) Kb

H2O (l) H+ (aq) + OH-

(aq) Kw

][X

][OH][HXK

(aq)-

(aq)-

(aq)b

Relationship Between Ka and Kb for Conjugate Base Pairs

][HX

][X]O[HK

(aq)

(aq)(aq)3a

][X

][OH][HXK

(aq)-

(aq)-

(aq)b

KxK ba

][X][OH][HX

x][HX

][X]O[H(aq)

-

(aq)-

(aq)

(aq)

(aq)-

(aq)3

][OH]O[H (aq)(aq)3 Kw

Recall: Autoionization of water H2O(l) + H2O(l) H3O+

(aq) + OH-(aq)

Kw=1.00x10-14 **must remember this value**

Relationship Between Ionization Constants for Conjugate Base Pairs

For acids and bases whose chemical formulas differ by only one hydrogen (conjugate pairs) the following apply:

Kw = KaKb Kb =Kw/Ka Ka = Kw/Kb

• Therefore if only the Ka value is available in the table, we can determine the conjugate pairs Kb by using the above equations

Note: these equations show the larger the Ka the smaller the Kb

Important Concepts Between Conjugate Pairs

Summary : Text p562

Important Concepts Between Conjugate Pairs

Summary : Text p562

Using Kw to Calculate Ka & Kb

Ex.4

What is the value of the base ionization constant (Kb) for the acetate ion, C2H3O2

- (aq)

at SATP?

Ans: 5.6x10-10

Example Calculations

Ex.5

Calculate the percent ionization of propanoic acid, HC3H5O2(aq), if a 0.050 mol/L solution has a pH of 2.78.

Ans. 3.3%