CH3OH(aq) + OH-(aq) ⇌ CH3O-(aq) + H2O(l)
If methanol has pKa = 18.3, what is pKb for its conjugate base?
pKa + pKb = pKw and pKw = 14 @ 25 C.
Thank you!
To find the pKb for the conjugate base of methanol, we need to use the relationship between pKa and pKb.
The pKa of a weak acid is defined as:
pKa = -log(Ka)
where Ka is the acid dissociation constant for the weak acid. This constant describes the strength of the acid and its ability to donate a proton.
Now, for a conjugate base, the dissociation constant is called Kb, the base dissociation constant. It describes the strength of the base and its ability to accept a proton.
The relationship between pKb and pKa is given by:
pKw = pKa + pKb
where Kw is the ionization constant of water (Kw = 1.0 x 10^-14 at 25°C).
Since we know the pKa of methanol (18.3), we can rearrange the equation to solve for pKb:
pKw = pKa + pKb
pKb = pKw - pKa
Substituting for pKw (1.0 x 10^-14) and pKa (18.3):
pKb = 1.0 x 10^-14 - 18.3
Calculating this expression, we find:
pKb ≈ -18.3
Therefore, the pKb for the conjugate base of methanol is approximately -18.3.