Deriving the Henderson-Hasselbalch equation

In summary: Thank you. In summary, the equations say that if I have an ionic compound, like HCl, then I need to add a negative log of the acid to the log of the conjugate base to find the pH. The conjugate base is the anion, and the pH is the amount of acid that is needed to make the conjugate base completely dissociated.
  • #1
Huzaifa
40
2
Im not able to understand the derivation equations and all please.
$$
\begin{aligned}
\mathrm{HA}+& \mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{A}^{-}+\mathrm{H}_{3} \mathrm{O}^{+} \\
K_{\mathrm{a}} &=\frac{\left[\mathrm{A}^{-}\right]\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}{[\mathrm{HA}]} \\
K_{\mathrm{a}} &=\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] \times \frac{\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]} \\
K_{\mathrm{a}} &=\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] \times \frac{[\text { conjugate base }]}{[\text { acid }]} \\
-\log _{10} \mathrm{~K}_{\mathrm{a}} &=-\log _{10}\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]-\text {log }_{10} \frac{[\text { conjugate base }]}{[\text { acid }]} \\
\mathrm{pK}_{\mathrm{a}} &=\mathrm{pH}-\log _{10} \frac{[\text { conjugate base }]}{[\text { acid }]} \\
\mathrm{pH} &=\mathrm{pK}_{\mathrm{a}}+\log _{10} \frac{[\text { conjugate base }]}{[\text { acid }]}
\end{aligned}
$$
 
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  • #2
Which steps are you having trouble with? Are you familiar with the definitions of ##K_a## and ##\text{pH}##?
 
  • #3
TeethWhitener said:
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.
 
  • #4
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?
 
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Likes jim mcnamara
  • #5
A = Anion

HA —> H+ + A-

Ionic compound —> Cation + Anion
 
  • #6
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:
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  • #7
DrJohn said:
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!
 

FAQ: Deriving the Henderson-Hasselbalch equation

1. What is the Henderson-Hasselbalch equation?

The Henderson-Hasselbalch equation is a mathematical formula used to calculate the pH of a solution based on the concentration of an acid and its conjugate base.

2. How is the Henderson-Hasselbalch equation derived?

The Henderson-Hasselbalch equation is derived from the equilibrium constant expression for a weak acid, using the principles of equilibrium and conservation of mass.

3. What are the assumptions made in deriving the Henderson-Hasselbalch equation?

The Henderson-Hasselbalch equation assumes that the acid is a weak acid, the solution is dilute, and the acid dissociation constant (Ka) is known.

4. What is the significance of the Henderson-Hasselbalch equation?

The Henderson-Hasselbalch equation is important in biochemistry and pharmaceutical sciences, as it allows for the calculation of the pH of a solution containing a weak acid and its conjugate base. It is also used in acid-base titrations and buffer solutions.

5. Can the Henderson-Hasselbalch equation be used for strong acids and bases?

No, the Henderson-Hasselbalch equation is only applicable to weak acids and their conjugate bases. For strong acids and bases, the pH can be calculated using the concentration of the acid or base and the dissociation constant.

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