Henderson-Hasselbalch Equation: Solving for acid

[Krishil_Gandhi_1D]

Hello,
I'm sorry if this question sounds dumb but I am a little confused on how we could solve the Henderson-Hasselbalch equation for the concentration of a conjugate acid given the pH, pka, and the concentration of the base? Thanks

Krishil

[Michael Lesgart 1H]

The Henderson-Hasselbach Equation is
pH=pKa + log ([A-]/[HA])
[HA] is the molar concentration of the weak acid, [A⁻] is the molar concentration of this acid's conjugate base, and pKa is −log10 Ka where Ka is the acid dissociation constant.

So if we are given the pH, pKa, and the concentration of [HA], we could solve for [A-]

example: pH=4 , pKa=10^-2 , [HA]=4M
So we put in these constants in the equation:

4=10^-2 + log ([A-]/4)
We know that log (A/B)=logA-logB
SO, we have 4=10^-2 + log[A-] - log [4]
from here we put everything on one side except the log [A-]
and we get
-95.39=log[A-]
then we base 10 each side to get rid of the log and we find that [A-]=10^-95.39
(I just chose random numbers so I'm not sure if these concentrations are physically possible or not. Hope this helped)

[Krishil_Gandhi_1D]

Thank you so much.
Krishil