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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. enter image description here

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closed as unclear what you're asking by Zhe, Todd Minehardt, airhuff, ashu, Mithoron Jun 21 at 19:39

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    $\begingroup$ 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 $\mathrm{p}K_\mathrm{a}$. $\endgroup$ – Mathew Mahindaratne Jun 19 at 18:52
  • $\begingroup$ 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? $\endgroup$ – Mohammad Mizanur Rahaman Jun 19 at 19:11
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    $\begingroup$ Do your calculations with suggested change (adding $\ce{HCl}$), and see what happens. $\endgroup$ – Mathew Mahindaratne Jun 19 at 19:17
  • $\begingroup$ 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 $\endgroup$ – Mohammad Mizanur Rahaman Jun 19 at 19:38
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    $\begingroup$ 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." $\endgroup$ – MaxW Jun 19 at 21:31
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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.

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    $\begingroup$ 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? $\endgroup$ – Mohammad Mizanur Rahaman Jun 19 at 20:18
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    $\begingroup$ See my last paragraph. $\endgroup$ – Mathew Mahindaratne Jun 19 at 20:21
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    $\begingroup$ The mechanism also includes that the system has reached a new equilibrium state in line with the minimal change in pH $\endgroup$ – Adnan AL-Amleh Jun 19 at 21:50
  • $\begingroup$ I am a bit confused at this point that "How does the mechanism include minimal change in pH?" $\endgroup$ – Mohammad Mizanur Rahaman Jun 20 at 19:00
  • $\begingroup$ Your buffer has $\ce{CH3CO2H}$ and $\ce{CH3CO2Na}$ in equilibrium with water. Their concentrations will adjust according to the changes following Le Chatelier principle. $\endgroup$ – Mathew Mahindaratne Jun 20 at 19:26

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