Chapter 8, Problem 81P

Interpretation Introduction

Interpretation:

Why the setup for $K_b$ values for bases and in a similar manner the $K_{\text{a}}$ values for acids is considered unnecessary should be explained.

Concept Introduction:

There are associated scales within the chemistry utilized to find out how basic and acidic a solution is, as well as the strength of bases and acids. Though most familiar is pH scale, K_a, pKa, K_b and pK_b are the common calculation which offer understanding into acid-base reactions.

Answer to Problem 81P

Scale of K_b values for bases is unnecessary because the strength of two bases can be compared from the strength of their conjugate acids.

Explanation of Solution

The acid ionization constant, K_a, is the equilibrium constant for acid’s ionization. For the ionization of an acid, say HA, in an aqueous solution we get.

$\begin{array}{l}{\text{HA + H}}_{\text{2}}\text{O }\rightleftharpoons {\text{ H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{ + A}}^{\text{-}}\\ {\text{K}}_{\text{a}}\text{ = }\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}\end{array}$

The square brackets denote the concentration. The concentration of water is not considered as water is exist in excess.

The K_a equation displays that the greater the dissociation of acid, the greater the value of K_a.

If there is a stringer acid, it will dissociate to a greater extent. Therefore, the numerator within the equation of K_a is larger, and thus, K_a will have the greater value.

Therefore, the weak acid’s strength is expressed in terms of K_a.

As per the Bronsted-Lowry concept of bases and acids, an acid is known as the proton donor, and a base is the proton acceptor.

When an acid donates a proton to another ion or molecule, it is converted to the conjugate base. For illustration,

$\begin{array}{l}{\text{CH}}_3\text{COOH}(\text{a}q){\text{ + H}}_2\text{O}(\text{l}\text{) }\rightleftarrows {\text{ CH}}_{\text{3}}{\text{COO}}^{-}(\text{a}q){\text{ + H}}_3{\text{O}}^{+}\text{(aq)}\\ \text{Acetic acid Acetate ion Hydronium ion}\\ \text{ (acid) (Conjugate base}\\ \text{ of acid) }\end{array}$

When a base accepts a proton, it is converted to the conjugate acid. For illustration.

$\begin{array}{l}{\text{NH}}_3(\text{a}q){\text{ + H}}_2\text{O}(\text{l})\rightleftarrows {\text{ NH}}_4{}^{+}(\text{a}q){\text{ + OH}}^{-}(\text{a}q)\\ \text{Ammonia water Ammomium ion Hydroxide ion}\\ \text{ (base) (Conjugate acid}\\ \text{ of base) }\end{array}$

As per the Bronsted-Lowry definitions any molecules or ions pairs which might be interconverted by the transfer of proton is known as conjugate acid-base pair. Therefore, CH₃ COOH and CH₃ COO- is a conjugate acid-base pair and NH₃ andNH₄ + is a conjugate acid-base pair.

There is an inverse relationship amid the strength of acid and its conjugate base. The stronger the acid, the weaker is the conjugate base. The acid ionization constant, K_a, expresses the weak acid’s strength.

The below table shows values of k_a for weak acids:

K_a values of weak acids.

Formula	Name	Ka
H3 PO4	Phosphoric acid	$\text{7}\text{.5}\times {\text{10}}^{-3}$
HCOOH	Formic acid	$\text{1}\text{.8}\times {\text{10}}^{-4}$
CH3 CH(OH)COOH	Lactic acid	$\text{1}\text{.4}\times {\text{10}}^{-4}$
CH3 COOH	Acetic acid	$\text{1}\text{.8}\times {\text{10}}^{-5}$
H2 CO3	Carbonic acid	$\text{4}\text{.3}\times {\text{10}}^{-7}$
H2 PO4 -	Dihydrogen phosphate ion	$\text{6}\text{.2}\times {\text{10}}^{-8}$
H3 BO3	Boric acid	$\text{7}\text{.3}\times {\text{10}}^{-10}$
NH4 +	Ammonium ion	$\text{5}\text{.6}\times {\text{10}}^{-10}$
HCN	Hydrocyanic acid	$\text{4}\text{.9}\times {\text{10}}^{-10}$
C6 H5 OH	Phenol	$\text{1}\text{.3}\times {\text{10}}^{-10}$
HCO3 -	Bicarbonate ion	$\text{5}\text{.6}\times {\text{10}}^{-11}$
HPO42 -	Hydrogen phosphate ion	$\text{2}\text{.2}\times {\text{10}}^{-13}$

If we want to compare the strength of two bases, we can compare the strength of their conjugate acids.

For illustration, let us compare the strength of two bases, phenoxide ion C₆ H₅ O- and bicarbonate ion HCO₃^-.

The conjugate acid of phenoxide ion i.e. C₆ H₅ O- is phenol C₆ H₅ OH.

K_a of phenol = $\text{1}\text{.3}\times {\text{10}}^{-10}$

The conjugate acid of bicarbonate ion HCO₃ - is carbonic acid H₂ CO₃.

K_a of carbonic acid = $\text{4}\text{.3}\times {\text{10}}^{-7}$

Therefore, K_a of the carbonic acid is greater than K_a of phenol. Hence, the carbonic acid is stronger than phenol. The stronger the acid, the weaker is the conjugate base. Therefore, a phenoxide ion is stronger base than a bicarbonate ion.

Therefore, we could compare the strength of weak bases from the K_a scale of acids. Hence, a scale of K_b values for bases is unncessary.

Conclusion

Since, the strength of two bases can be compared from the strength of their conjugate acids thus, scale of K_b values for bases is unnecessary.