Advance discussion of acids bases

- Explain conjugate acids of bases.

- Evaluate
*K*_{a}of the conjugate acid of a base.

- Treat the conjugate acid of a base as an acid in numerical calculations.

- Reverse the role of acid and base for the previous skills.

The conjugation of acids and bases have been discussed earlier. After losing a proton, the acid species becomes the conjugate base. A base and its protonated partner also form a conjugated acid-base pair. These relationships have been represented by

Acid = H

HAc = H

Let BH^{+} be the conjugate acid of a base, then the expression
for the acidic constant *K*_{a} for the conjugate acid:

[B] [HThus,^{+}]K_{a}= --------- [BH^{+}] [B] [H^{+}] [OH^{-}] = --------- ------ [BH^{+}] [OH^{-}] [B] = ----------- [H^{+}] [OH^{-}] [BH^{+}] [OH^{-}] 1 = ---K_{w}K_{b}

and at 298 K, we have

**
Examples 1
**

*Solution*

The conjugate base is CO_{3}^{2-}.

*K*_{b} = (1E-14)/(4.7E-11)

= 2.1E-4

*DIscussion*

The *K*_{b} so calculated is for the reaction,

COThe anion CO_{3}^{2-}+ H_{2}O = HCO_{3}^{-}+ OH^{-}[HCO_{3}^{-}] [OH^{-}]K_{b}= ------------- [CO_{3}^{2-}]

**
Example 2
**

*Solution*

The *K*_{a} for oxalic acid is

*K*_{a} = (1E-14) / (1.8E-10)

= 5.6E-5

*DIscussion*

The calculation regarding *K*_{a} and *K*_{b}
conversion is simple, but understanding what problems require
this type of conversion is difficult. The concept is rather useful,
and it further broadens the concept of acid and base.

Let A^{-} be the conjugate base of an acid HA, then the expression
for the equilibrium constant for the reaction:

can be written as

[HA] [OHMultiplying the numerator and denominator with [H^{-}]K_{b}= ---------- [A^{-}]

[HA] [OHThus,^{-}] [H^{+}]K_{b}= ---------- ---- [A^{-}] [H^{+}]Rearrangement gives

[HA]

K_{b}= ---------- [OH^{-}] [H^{+}] [A^{-}] [H^{+}][HA] = ---------

K_{w}[A^{-}] [H^{+}]

K_{w}= ----K_{a}

and this formula is the same as the one derived for the conjugate acid of a base. Again, at 298 K, we have

and the value for *K*_{w} is larger than 1E-14 at higher temperatures.
*K*_{w} is smaller at temperature less than 298 K.

**Calculate***K*_{b}for the acetate ion from the*K*_{a}for acetic acid of 1.8E-5.**Answer***5.6E-10*

**Consider...**

*K*_{b}= (1e-14)/(1.8e-5) = 5.6E-10 If*K*_{b}for the acetate ion is 5.6E-10, what is*K*_{a}for acetic acid?**The***K*_{a}for trimethylammonium ion (CH_{3})_{3}NH^{+}is 1.6E-10. Calculate*K*_{b}for its conjugate base.**Answer***6.25E-5*

**Consider...**

*K*_{a}= (1e-14)/(1.6e-10) = 6.25E-5 You know all about conjugate acid-base pairs now. Learning is a pleasure.**At some temperature,***K*_{w}= 1e-13. Calculate the*K*_{b}value for the acetate ion. (*K*_{a}for acetic acid is 9.5E-5 at the same temperature).**Answer***1.05e-9*

**Consider...**

*K*_{b}= 1e-13/9.5e-5 The acidic constants are dependent on temperature.