"<...> a weak mountain is one that only rarely dissociates in water <...>. Likewise, because the conjugate base is a weak base, <...>"

which seems to was standing in dispute with my assumption a) above. So, what is correct?

Furthermore, if b) is correct, isn"t any solution of a weak acid systems a buffer, since any type of weak mountain in water renders an equilibrium of the form

$HA ext (weak acid) leftrightarrow H^+ + A^- ext (conjugate base)$

and is therefore a systems of a weak acid and also its base?

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inquiry Jul 16 "16 at 15:58

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Your declare a) isn"t constantly true.

You are watching: Is the conjugate base of a weak acid a strong base

Water dissociation is represented by:

$$ceH2O + H2O H3O+ + OH-$$

$$ K_mathrm w=cdot = 1 imes 10^-14 ( extrmat 25^circ ~mathrm C) $$

Note 1: we don"t create the $ceH_2O$ activity, because it have the right to usually it is in rounded to 1 and the ions tasks can it is in rounded to your concentrations.

Note 2: This value was obtained experimentally, considering the concentration that $ceH_3O^+$ and $ceOH^-$ in the medium was the same and also measuring the ionization that water.

On a weak acid dissociation:

$$ceHA + H2O A- + H3O+$$

$$K_mathrm a=fraccdot $$

On that is conjugate base:

$$ceA- + H2O HA + OH-$$

$$K_mathrm b=fraccdot $$

If we "add" both reactions we mean that both equilibriums will happen in the mixture, so we have:

$$ceHA + H2O A- + H3O+ (K_mathrm a)$$


$$ceA- + H2O HA + OH- (K_mathrm b)$$


$$ceH2O + H2O H3O+ + OH- (K_mathrm w)$$

$$K_mathrm bcdot K_mathrm a= cdot cdot (cdot )/( cdot ) = cdot = K_mathrm w$$

$$K_mathrm b=K_mathrm w/K_mathrm a~~~~~~ K_mathrm b=10^-14/K_mathrm a extrmat 25^circ ~mathrm C$$

Another means to create this:

eginalign-log(K_mathrm b)&=-log(10^-14K_mathrm a) \ implies mathrm pK_mathrm b &= -log(10^⁻14) - (-log(K_mathrm a))\ implies mathrm pK_mathrm a+mathrm pK_mathrm b &=14;.endalign

This is the relation in between a conjugate basic strength and also its mountain strength. A very strong acid has actually a weak conjugate base, but a weak mountain doesn"t necessarily have a very strong base. Speak you have actually a $mathrm pK_mathrm a=5$, i beg your pardon is a weak acid, through $K_mathrm a=1 imes 10^-5$. The conjugate base would have a $mathrm pK_mathrm b=14-5=9$ or a $K_mathrm b=1 imes 10^-9$, which is no a strong base.

However, if we have a strong acid, prefer $ceHCl$ with a $mathrm pK_mathrm a=-6.3$ and also $K_mathrm a=10^6.3$. Its conjugate base would have actually a $mathrm pK_mathrm b=14-(-6.3)=20.3$ and also $K_mathrm b=10^-20.3$ i m sorry is a yes, really weak base.

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Hopefully i didn"t do it even more confusing for you! It"s mostly an reaction equilibrium issue.