# Why does pH of a buffer solution change according to Henderson–Hasselbalch equation? [closed]

If I add a little amount of acid or base to a buffer solution, the pH of the buffer won't change. That's what the diagramatic mechanism tells us.

But according to Henderson–Hasselbalch equation, the $$\mathrm{pH}$$ changes!

For example, a buffer solution of $$\ce{CH3COOH}$$ and $$\ce{CH3COONa}$$:

$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log\left(\frac{[\ce{CH3COONa}]}{[\ce{CH3COOH}]}\right)$$

Now I add a little amount of acid $$(\pu{0.01M},\;\pu{5 mL}).$$ It disturbs $$\ce{CH3COONa}$$. Reaction:

$$\ce{CH3COONa + HCl -> CH3COOH + H2O}$$

It is noticed here that a small amount of $$\ce{CH3COOH}$$ is produced and a little amount of $$\ce{CH3COONa}$$ is spared. So simply $$[\ce{CH3COOH}]$$ and $$[\ce{CH3COONa}]$$ go high and low, respectively. As a result, the ratio between $$\ce{CH3COONa}$$ and $$\ce{CH3COOH}$$ decreases. As the $$\mathrm{pH}$$ depends on the ratio, the $$\mathrm{pH}$$ changes as well.

• First, Henderson–Hasselbalch equation is: $\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \left(\frac{[\text{base}]}{[\text{acid}]}\right)$. Second, $\frac{[\text{base}]}{[\text{acid}]}$ ratio is in log scale so numerical value of change is minimal at $\mathrm{pH}$ around $\mathrm{p}K_\mathrm{a}$. – Mathew Mahindaratne Jun 19 '19 at 18:52
• so Log ([base]/[acid]) is minimal but not 0. Still a little change at pH but according to mechanism there shouldn't be any change in pH, shouldn't it? Don't these two ideas contradict each other? – Mohammad Mizanur Rahaman Jun 19 '19 at 19:11
• Do your calculations with suggested change (adding $\ce{HCl}$), and see what happens. – Mathew Mahindaratne Jun 19 '19 at 19:17
• Yes I did before asking this question and got a little change around 1.8*10^(-3). If add more acid, the change is more – Mohammad Mizanur Rahaman Jun 19 '19 at 19:38
• Your premise re: "If I add a little amount of acid or base to a buffer solution, the pH of the buffer won't change." is wrong. The pH does change, just not by much. The gist is that the pH of a buffer solution changes much less than the pH of pure water would. Also look up "buffer capacity." – MaxW Jun 19 '19 at 21:31

## 1 Answer

Suppose you have a $$\pu{1.0 L}$$ of acetate buffer solution made by $$\pu{500.0 mL}$$ of $$\pu{1.0 M}$$ $$\ce{CH3COONa}$$ solution adding to $$\pu{500.0 mL}$$ of $$\pu{1.0 M}$$ $$\ce{CH3COOH}$$ solution. Thus, $$[\ce{CH3COOH}] = [\ce{CH3COONa}] = \pu{0.5 M}$$. Therefore, according to Henderson–hasselbalch equation, $$\mathrm{pH} = \mathrm{p}K_\mathrm{a} = 4.75$$ ($$\mathrm{p}K_\mathrm{a} = 4.75$$ for acetic acid).

Now you add a little amount of acid ($$\pu{0.01M}, \; \pu{5 mL}$$). It disturbs the equilibrium and change both $$[\ce{CH3COONa}]$$ and $$[\ce{CH3COOH}]$$, according to the reaction:

$$\ce{CH3COONa + HCl -> CH3COOH + H2O}$$

Now, the amount of $$[\ce{CH3COONa}]$$ is reduced by $$\pu{0.01M} \times \pu{0.005 L} = \pu{0.00005 mol}$$ and, similarly, $$[\ce{CH3COOH}]$$ is increased by $$\pu{0.00005 mol}$$.

Hence, new $$[\ce{CH3COONa}]$$ is $$\pu{0.49995 mol/L}$$ and $$[\ce{CH3COOH}]$$ is $$\pu{0.50005 mol/L}$$ (neglecting the volume change for convenience). Thus, new $$\mathrm{pH}$$:

$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \left(\frac{[\ce{CH3COONa}]}{[\ce{CH3COOH}]}\right)= 4.75 + \log \left(\frac{\pu{0.49995 mol/L}}{\pu{0.50005 mol/L}}\right) = 4.7499 \approx 4.75$$

You see, it is a very minimal change. It is true that if you add more acid or base, change could be significant. However, all buffers have buffer capacity (usually, $$\mathrm{p}K_\mathrm{a} \pm 1$$). Nobody ever said $$\mathrm{pH}$$ of buffer is constant. It fluctuate withing the capacity. If you add too much acid or base, its $$\mathrm{pH}$$ would drastically change.

• You have just avoided my question. I acknowledge it's a minimal change but not 0 . The mechanism shows all H+ added get neutralised , no extra H+ left. Then how and why does this minimal change happen? – Mohammad Mizanur Rahaman Jun 19 '19 at 20:18
• See my last paragraph. – Mathew Mahindaratne Jun 19 '19 at 20:21
• The mechanism also includes that the system has reached a new equilibrium state in line with the minimal change in pH – Adnan AL-Amleh Jun 19 '19 at 21:50
• I am a bit confused at this point that "How does the mechanism include minimal change in pH?" – Mohammad Mizanur Rahaman Jun 20 '19 at 19:00
• Your buffer has $\ce{CH3CO2H}$ and $\ce{CH3CO2Na}$ in equilibrium with water. Their concentrations will adjust according to the changes following Le Chatelier principle. – Mathew Mahindaratne Jun 20 '19 at 19:26