# Deriving the Henderson-Hasselbalch equation

Huzaifa
Im not able to understand the derivation equations and all please.
\begin{aligned} \mathrm{HA}+& \mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{A}^{-}+\mathrm{H}_{3} \mathrm{O}^{+} \\ K_{\mathrm{a}} &=\frac{\left[\mathrm{A}^{-}\right]\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}{[\mathrm{HA}]} \\ K_{\mathrm{a}} &=\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] \times \frac{\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]} \\ K_{\mathrm{a}} &=\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] \times \frac{[\text { conjugate base }]}{[\text { acid }]} \\ -\log _{10} \mathrm{~K}_{\mathrm{a}} &=-\log _{10}\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]-\text {log }_{10} \frac{[\text { conjugate base }]}{[\text { acid }]} \\ \mathrm{pK}_{\mathrm{a}} &=\mathrm{pH}-\log _{10} \frac{[\text { conjugate base }]}{[\text { acid }]} \\ \mathrm{pH} &=\mathrm{pK}_{\mathrm{a}}+\log _{10} \frac{[\text { conjugate base }]}{[\text { acid }]} \end{aligned}

## Answers and Replies

Gold Member
Which steps are you having trouble with? Are you familiar with the definitions of ##K_a## and ##\text{pH}##?

Huzaifa
Which steps are you having trouble with? Are you familiar with the definitions of ##K_a## and ##\text{pH}##?
I think I know those definitions, I am having trouble with conjugate acid base thing and the positive and negative signs.

Gold Member
Can you write out the definitions of ##\text{pH}## and ##\text{pK}_\text{a}##? Can you tell me what a conjugate acid or base is? For example, if I give you ##HCl## as an acid, can you tell me its conjugate base?

Last edited:
jim mcnamara
Homework Helper
Gold Member
A = Anion

HA —> H+ + A-

Ionic compound —> Cation + Anion

DrJohn
You are missing a line in your set of equations
simplest example I can think of:
28 = 4 x 7
so log 28 = log4 + log7 <=== this is the equivalent missing line
so -log28 = -log4 - log7 - changing the signs on each side for each part of the equation

Your text is the equivalent of going straight from 28 = 4x7 to -log28 = -log4 - log7, so there are two changes on a single line which always leads to confusion, yet mathematicians seem to do this all the time to save writing lots of lines that other experts can instantly see are obvious

The switch to logs is because for weak acids (there are only four or six strong acids - where acids means forms H+ ions) we end up dealing with numbers like 2.309 x (10 to power of -5) ( which can be written as 2.309e-5 or in normal life as 0.00002309)
But with logs we get a simple number -4.63657606708
And then for some reason which I can't remember just now, we use negative logs for the pKa (and for pH)
This gives us 4.63657606708, which we then normally only write to one or two decimal places
ie the pKa = 4.6 or pKa = 4.63

Does this help a bit?

Last edited:
Huzaifa
Huzaifa
You are missing a line in your set of equations
simplest example I can think of:
28 = 4 x 7
so log 28 = log4 + log7 <=== this is the equivalent missing line
so -log28 = -log4 - log7 - changing the signs on each side for each part of the equation

Your text is the equivalent of going straight from 28 = 4x7 to -log28 = -log4 - log7, so there are two changes on a single line which always leads to confusion, yet mathematicians seem to do this all the time to save writing lots of lines that other experts can instantly see are obvious

The switch to logs is because for weak acids (there are only four or six strong acids - where acids means forms H+ ions) we end up dealing with numbers like 2.309 x (10 to power of -5) ( which can be written as 2.309e-5 or in normal life as 0.00002309)
But with logs we get a simple number -4.63657606708
And then for some reason which I can't remember just now, we use negative logs for the pKa (and for pH)
This gives us 4.63657606708, which we then normally only write to one or two decimal places
ie the pKa = 4.6 or pKa = 4.63

Does this help a bit?
Yes sir, It was helpful!