5
$\begingroup$

I read this today:

If you mix equal amounts of $1 M$ $\ce{H_3PO_4}$ and $1 M$ $\ce{NaH_2PO_4}$ you get a buffer at $pH<6$.

The reason given is that a weak acid plus a salt containing the conjugate base forms a acidic buffer.

If you mix equal amounts of $1 M$ $\ce{NH_3}$ and $1 M$ $\ce{NH_4Cl}$, you get a buffer at $pH>8$.

The reason given is that a weak base plus a salt containing the conjugate acid forms a basic buffer.

I don't understand why this is true. Let's take the first statement for example. If you have equal amounts of a weak acid and its conjugate base, then by definition you have a buffer. Whether the buffer is acidic or basic shouldn't depend on whether the salt contains the conjugate acid or the conjugate base, right?

$\endgroup$
  • 1
    $\begingroup$ Try to find pH by yourself :) The first acidity to $\ce{H_3PO_4}$ is a strong acidity $\ce{pKa_1}=2,15$ and $\ce{pKa_2}=7,20$ and the third $\ce{pKa_3}=12,42$ $\endgroup$ – Hexacoordinate-C Apr 27 '15 at 7:36
  • $\begingroup$ Do you know the Henderson-Hasselbalch equation? The answer comes from a straightforward analysis. $\endgroup$ – Nicolau Saker Neto Apr 27 '15 at 15:21
  • $\begingroup$ Yes I know the Henderson-Hasselbach equation, it gives that the pH equals the pKa which for an acid is low, but this is just a mathematical trick and I want a conceptual explanation. $\endgroup$ – Joshua Benabou Apr 27 '15 at 19:57
3
$\begingroup$

The statement "a weak acid plus a salt containing the conjugate base forms an acidic buffer." is not correct. More properly, a mixture of equal amounts of a weak acid (but not too weak: with a pKa < 7) with its conjugate base (salt) does indeed give an acidic buffer. To get a basic buffer you would use equal amounts of an even weaker acid (with a pKa > 7) and its conjugate base (salt). Similar considerations apply to formation of buffers from a weak base plus a conjugate acid salt.

$\endgroup$
  • $\begingroup$ This answer is correct. Not sure why you got a downvote. $\endgroup$ – Nicolau Saker Neto Apr 28 '15 at 18:42
2
$\begingroup$

Your suspicion is correct: the buffer pH depends on the $\rm{pK_a/pK_b}$ of the acid/base in question, not whether the buffer salt contains a weak conjugate acid or conjugate base. We can test this out via the application of the Henderson-Hasselbalch equation.

Consider acetic acid ($\rm{pK_a = 4.76}$) mixed with the same concentration of sodium acetate. Via the Henderson-Hasselbalch equation, $\rm{pH = pK_a + log \frac{[Ac^{-}]}{[HAc]}}=pK_a + log\ 1 = pK_a = 4.76$. Thus, the acid/conjugate base buffer is acidic.

Now take hydrocyanic acid ($\rm{pK_a = 9.21}$) mixed with the same concentration of sodium cyanide. Again, via the Henderson-Hasselbalch equation, $\rm{pH = pK_a + log \frac{[CN^{-}]}{[HCN]}}=pK_a + log\ 1 = pK_a = 9.21$. The acid/conjugate base buffer is now basic. A similar analysis can be made for bases mixed with their conjugate acid salts.

So why does this happen? It's quite simple. If an acid has a $\rm{pK_a < 7}$, then it is more acidic than its conjugate base is basic ($\rm{pK_b > 7}$), so when mixing an equal amount of the acid and its conjugate base, the acidic character wins and the buffer is acidic. If the acid has a $\rm{pK_a > 7}$, then it is less acidic than the conjugate base is basic ($\rm{pK_b < 7}$). Mix an equal amount of both, and as you might expect, the basic character wins, and the buffer ends up alkaline. Again, a similar analysis can be made for buffers made with bases and their conjugate acid salts.

$\endgroup$
1
$\begingroup$

I am a pretty elementary level student myself, so please excuse and ignore if you find me saying something wrong.

By definition a buffer solution resists the sudden change in $\mathrm{pH}$ on addition of a small amount of strong acid or strong base or due to dilution. One possible way for this to occur (in case of acid buffer) is for the strong acid introduced into the system (buffer) to be converted to a weaker acid (that is why we require both a weak acid and its salt), same for a strong base.

Now, consider ethanoic acid and sodium ethanoate:

$$\ce{H+ + CH3COONa -> CH3COOH}$$

where $\ce{H+}$ – proton from strong acid; $\ce{CH3COOH}$ – a weaker acid than a one introduced, hence no considerable change in $\mathrm{pH}$.

Also,

$$\ce{CH3COOH + OH- -> CH3COO- + H2O}$$

where $\ce{OH-}$ is from introduced base (mildly reversible reaction).

The acetate here lacks a hydroxyl group as opposed to the species you introduced.

For a basic buffer

$$\ce{NH4OH + H+ -> NH4+ + H2O}$$

where $\ce{H+}$ is from acid; $\ce{NH4+}$ is weakly acidic; and

$$\ce{NH4Cl + OH- -> NH4OH + Cl-}$$

where $\ce{OH-}$ is from base; $\ce{NH4OH}$ – weak base.

However, both basic and acidic buffers serve the same purpose and in nearly the same way.. they just have different compositions and consequently different $\mathrm{pH}$ and $\mathrm{pOH}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.