How do buffers work? and why only solutions of a weak acid and its salt with a strong base or vice versa work as a buffer [closed]








**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
 
 






---








 **Homework questions** must demonstrate some effort to understand the underlying concepts. For help asking a good homework question, see: [How do I ask homework questions on Chemistry Stack Exchange?](https://chemistry.meta.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange)




Closed 2 years ago.








[Improve this question](/posts/156087/edit) 



Please note that the definition of a buffer in my textbook is given as



> 
> The solutions which resist change in pH on dilution or with the addition of small amounts of acid and alkali are called Buffer Solutions
> 
> 
> 


Is this the right explanation of a buffer?
If not what is the right explanation?




pH basically means "how much of $\ce{H+}$ there is in the solution". And the job of the buffer is to keep this concentration more or less constant: if someone "consumed" $\ce{H+}$ our buffer has to generate new $\ce{H+}$; and if someone produced $\ce{H+}$, our buffer needs to "consume" the extra amount.


Suppose you have a solution with $\mathrm{pH=4}$ and you add a weak acid (let's call it $\ce{HA}$) with approximately the same $\mathrm{pK\_a=4}$. Because $\mathrm{pH=pK\_a}$ half of that $\ce{HA}$ substance will be dissociated into anion ($\ce{A-}$) and $\ce{H+}$ and the other half will stay intact as $\ce{HA}$. But if something changed the $\mathrm{pH}$ of the solution, then the proportions of Dissociated:Intact versions of the molecule would shift.


So if we then try adding another acid which is stronger ($\mathrm{pK\_a=2}$):


1. We'll get an excess of $\ce{H+}$ from that new acid
2. And those extra $\ce{H+}$ will be consumed by the original $\ce{A-}$ turning it back to $\ce{HA}$. Therefore $\mathrm{pH}$ of the solution will stay approximately the same.


If we add a base ($\mathrm{pK\_a=6}$):


1. Those $\ce{H+}$ that left the original $\ce{HA}$ will be consumed by that base
2. Which means the concentration of $\ce{H+}$ drops, but this in turn leads to our original $\ce{HA}$ dissociating further producing more $\ce{A-}$ and $\ce{H+}$


Therefore $\mathrm{pH}$ of the solution again stays more or less the same. Here are some video explanations of [Acid-Base equilibrium](https://www.youtube.com/watch?v=l5fk7HPmo5g) and [Buffer Solutions](https://www.youtube.com/watch?v=rIvEvwViJGk).


 1