Acid and Base Equilibrium III

polyprotic acids

weak bases, K_b

=[OH^-] and pH of a weak base solution

types of weak bases, NX₃ and anions of weak acids

K_a and K_b

Polyprotic Acids

Several Acids have more than one H⁺ that can react. For example for carbonic acid

H₂CO₃ = H⁺ + HCO₃^- K_a1 = 4.3 X 10^-7

HCO₃^- = H⁺ + CO₃^-2 K_a2 = 5.6 X 10^-11

The second ionization constant is much smaller than the first because it is easier to remove a H⁺ from a neutral molecule than from an anion. In sample exercise 16.13 they calculate the [H⁺] from the first ionization of 3.7 X 10^-3M carbonic acid to be 4.0 X 10^-5M and the [H⁺] from the second ionization (y = CO₃^-2 = 2^nd H⁺) = 5.6 X 10^-11. So in calculating the pH from a polyprotic acid, it is usually safe to assume that the second and higher ionization steps will produce a negligible amount of H⁺. The text mentions that as long as successive K_a values differ by a factor of 10³ or more we can use K_a1 to determine the pH.

Weak Bases

Consider the weak base ammonia, NH₃, in water.

NH₃ + H₂O = NH₄⁺ + OH^-

K_c = [NH₄⁺] [OH^-] / [H₂O] [NH₃ ]

If we leave out the water concentration since it is constant, we get K_b the base-dissociation constant.

K_b= [NH₄⁺] [OH^-] / [NH₃ ]

It just happens that K_b for ammonia = K_a for acetic acid = 1.8 X 10^-5

Earlier we calculated the pH of a 1.0 X 10^{-2 M} acetic acid solution to be 3.37 (compared to HCl’s 2.00).

What is the pH of a 1.0 X 10^-2M NH₃ solution? (compare to 1.0 X 10^-2M NaOH which had a pH of 12.00).

NH₃ + H₂O = NH₄⁺ + OH^-

Initial	1.0 X 10^-2	0	0*
Change	-x	+x	+x
Equilibrium	1.0 X 10^-2 – x	X	X

*We are neglecting the [OH^-] from the autoionization of water because it is probably way to small to contribute much.

K_a = x² / (1.0 x 10^-2 –x) ~ x² / 1.0 X 10^-2

(Lets see if we can simplify by neglecting x compared to 0.0010 M. Solve for x= [OH^-] = 4.2 X 10^-4M (which is indeed very small compared to 1.0X10^-2M.) Since we want pH we need to solve for [H⁺]

[H⁺] = K_w / [OH^-] = 1.0 X 10^-14 / 4.2 X 10^-4M = 2.4 X 10^-11M

pH= -Log (2.4 X 10^-11) = 10.62 (compared to [NaOH] giving a pH of 12.00).

Types of Weak Bases

Amines are a type of organic compound where one or more of the N-H bonds in NH₃ have been replaced with C-N bonds. The lone pair of electrons on the nitrogen can accept a H⁺ .

The second type of weak base are anions of weak acids. Consider sodium acetate, NaC₂H₃O₂. When dissolved in water the acetate ion can react with water as follows

C₂H₃O₂^- + H₂O = HC₂H₃O₂ + OH^-

K_b = [HC₂H₃O₂] [OH^-]

[C₂H₃O₂^-]

Remember K_a for the ionization of acetic acid

HC₂H₃O₂ = H⁺ + C₂H₃O₂^-

K_a = [H⁺] [C₂H₃O₂^-]

[HC₂H₃O₂]

K_a K_b = [H⁺] [C₂H₃O₂^-] X [HC₂H₃O₂] [OH^-] = [H⁺] [OH^-] = K_w

[HC₂H₃O₂] [C₂H₃O₂^-]

So we get that an acid dissociation constant multiplied times its conjugate base dissociation constant equals the water ion-product constant. A practical consequence of this is that if we know one constant, we can calculate its’ conjugate constant. Ionization constants are typically listed only for the neutral acid or base molecule. The base dissociation constant of an anion of a weak acid can be calculated as shown above. And the acid dissociation constant of the cation of a weak base can similarly be calculated.

And, as K_a decreases, K_b increases. And vice versa. So, as an acid get weaker, (K_a decreases), the conjugate base get stronger (K_b increases). An as an acid gets stronger, (K_a increases), the conjugate base gets weaker (K_b decreases).

So, calculate the K_b of the acetate ion

K_w = K_a K_b

K_w / K_a = K_b

1.0 X 10^-14 / 1.8 X 10^-5 = 5.6 X 10^-10

What is the pH of a 1.0 X 10^-2 M NaC₂H₃O₂ solution?

C₂H₃O₂^- + H₂O = HC₂H₃O₂ + OH^-

Initial	1.0 X 10^-2	0	0*
Change	-x	+x	+x
Equilibrium	1.0 X 10^-2 – x	X	X

Since [ C₂H₃O₂^- ]_initial > 400 X K_b, we can neglect x compared to 1.0 X 10^-2

5.6 X 10^-10 = x² / 1.0 X 10^-2

x = [OH^-] = 2.4 X 10^-6 M => [H⁺] = K_w / [OH^-] = 4.2 X 10^-9

pH = - Log 4.2 X 10^-9 = 8.38 (you see, I told you it was a base).