Is the Conjugate Base of a Weak Acid Always a Weak Base?

[dissolver]

If the conjugate base of a weak acid ends up being determined as a weak base by dividing the acid dissociation constant by the ion product of water, how can this be?

My book had an example, but goes on to generalize that if a salt's anion is the conjugate base of a weak acid, then the salt will cause the water to be basic. However, it was shown that a weak acid's conjugate base can be a weak base. Are these just the rare exceptions?

 

[mrjeffy321]

Is there a specific example the book gives to illustrate the idea?

The conjugate base of a weak acid should turn the solution alkaline.
For example, The conjugate base of acetic acid is C2H3O2-1, when this is in solution, it tends to "Steal" H+ ions from the water to form HC2H3O2. Now that the water is lacking H+ ions, it has an excess of OH- and becomes basic.

The conjugate acid of a weak base should turn the solution acidic.
For example, the conjugate acid of the weak base Ammonia is NH4+. In solution, this ion will steal OH- ions from the water and form Ammonium Hydroxide, NH4OH, leaving the solution acidic.

When I say "steal" ___ ions from the water I mean it will take the ions which are already present due to the fact water will naturally break up into H+ and OH- ions,
H2O (l) --> H+ (aq) + OH- (aq)
and since the weak acids / weak bases which are formed with these ions do not dissassociate completely, the H+ and OH- ions form the undissassociated compounds and don't effect the pH.

 

[GCT]

To the OP. You need to cite specific examples. Both weak acids and weak bases have pKa to be considered, that is, the equilibrium is important in understanding the ratios between the conjugate acid and base which is comparable; this is part of the basis for the essential buffer properties. For a strong base, the reaction goes towards completion...the same with a strong acid.

 

[dissolver]

mrjeffy321, would a conjugate acid of a strong base also turn an aqueous solution acidic?

Also the example give was for HCN.

For example given a weak acid whose Ka is 10^-8, and then the conjugate base should be a strong one. But dividing the water ion product 10^-14 by 10^-8, we find Kb is actually on 10^-5, so the base is weak. How can this be? The book said it was because the base reaction had the conjugate acid and OH- competing for the H+ while the acid reaction had the conjugate base and H2O competing for the H+.

 

[mrjeffy321]

The conjugate acid/base of a strong base/acid would not affect the solution's pH.
Take the strong acid HCl. The conjugate base of Hydrochloric acid is Cl-, a fairly common ion, in table salt for example. But then why does table salt not effect the pH of the solution, after all, it has the Cl- ion and water will supply the H+ (forming the strong acid HCl)...or the other way around, Na+ from the salt can form the strong base NaOH with the OH- from the water. The reason is that the acids/bases which are formed are strong acids/bases, they will disassociate completely into ions in solution. If the acid formed with the strong acid's conjugate base disassociates completely, then no ions are "stolen" from the water and the solution remains neutral.

HCN is a weak acid with the conjugate base of CN-. When a CN- salt (like KCN) is dissolved in soltution, the CN- ion will "see" the H+ ion from the water and it will form HCN. Some of that HCN will disassociate back into solution, but most of it will come to equilibrium in the undissassociated state HCN (aq) instead of H+ (aq) + CN- (aq).
Since the water now 'lacks' H+ ions to balance out the OH- ions, the CN- solution should be alkaline.

 

[Renge Ishyo]

dissolver, please consider that the notion of "strong" and "weak" is simply relative. For example, in pure water the acid species and base species are considered of equal "strength."

Kw = (Ka)(Kb)

For water, Ka and Kb are equal (1x10^-7) with neither the acid (ka) or base (kb) term dominating over the other. It goes to show that for a acid or base other than water, one value (either ka or kb) will be stronger than the other as their product is equal to 1x10^-14. However, keep in mind that the *closer the value of ka and kb get to 10^-7* the smaller the difference in strength between an acid and it's conjugate base.

Here are three examples (and I may be fudging on the numbers a bit due to laziness so forgive me if I am a bit off in numerical details ^^):

#1 Pure Water
kw = (ka)(kb) ----> 1x10^-14 = (1x10^-7)(1x10^-7)

The solution is neutral. Neither dominates over the other (they are of equal strength).

#2 HCN (in water)
kw = (ka)(kb) ----> 1x10^-14 = (1x10^-8)(1x10^-6)

The solution is slightly basic. The reason why is that the base term is only slightly stronger than the acid (compare to water equation, both sides are only off by one order of 10 from the solution being completely nuetral. Therefore, both species are weak because their constants are so close to water that not much of a change happens in the solution. However, note that the base is STILL stronger than the acid so it still will be a slightly more basic solution than nuetral water. It can be said that both HCN and CN- are weak because they differ so little in pka and pkb values from completely neutral water.

#3 HCL (in water)

kw = (ka)(kb) ---> 1x10^-14 = (1x10^7)(1x10^-21)

This solution is strongly acidic. The reason is that the difference in pka and pkb terms are much farther away from 10^-7 as in the water and HCN examples which are either at 10^-7 or very close to it. Because pka is clearly the dominant term almost all of HCL forms acid. It can be said that HCL is a very strong acid and CL^- is a very weak base. This is because of the large difference in their pka and pkb values.

Hope this helped.