Buffer Solution with Strong Acid?

I was thinking: Could we could make a strong acid-strong base buffer solution? Take, for instance, a mixture of 1 L $\ce{HCl}$ (0.1 M) and 1 L of aqueous salt, $\ce{Na_2SO_4}$ (also 0.1 M), note that $k_2=0.012$ for the polyprotic acid. How can the pH of the mixture be calculated? Additionally, would it be "stable" like a buffer?

Buffers can be made two ways.

Buffers

Weak Acid/Base + Salt of the Conjugate Base/Acid

Let’s take acetic acid ($\ce{HC2H3O2}$) and sodium acetate ($\ce{NaC2H3O2}$).

$$\ce{HC2H3O2 + H2O <=> H3O+ + C2H3O2-~~~~~~~~~~K_a=1.8 x 10^{-5}}$$

Acetic acid has a small $\ce{K_a}$ value, meaning that it does not dissociate much. Conversely, sodium acetate readily dissociates as sodium is an alkali metal. By having equimolar concentrations of both weak acid and conjugate base, the addition of an acid adds hydronium, which reacts with the acetate ion to shift the equilibrium to the left, and the addition of a base absorbs hydronium to shift the equilibrium to the right. Either way, the hydronium concentration, which determines the pH, remains close to the same.

Weak Acid/Base + Strong Base/Acid

This relies on the same principle as above but supplies the conjugate base in a different manner. Continuing with our acetic acid example, here, this would be added, alone. Then you would add half of the moles of acid in strong base, such as $\ce{NaOH}$. This would neutralize any available hydronium, shifting the equilbrium to the right until there was the same half-and-half of acid and conjugate base.

Strong Acid?

Unfortunately, strong acids do not make for good buffers as they fully dissociate -- the equilibrium is so far to the right that it is very unstable.

Salt of the Conjugate Base?

$\ce{Na2SO4}$ shares no common ion with $\ce{HCl}$, so it provides no conjugate base ($\ce{{SO4}^2-}$ is the conjugate base of the second dissociation of $\ce{H2SO4}$). Even if you used $\ce{NaCl}$, it still would not work because of the reason outlined above.

pH?

If you're still interested, the pH of the solution above is easily calculated. As $\ce{HCl}$ fully dissociates, its concentration (0.1 M, as you say above) is equal to the hydronium concentration. However, the addition of another liter of neutral solution halves the hydronium concentration to 0.05 M. Therefore,

$$\ce{-log(0.05)=pH=1.3}$$

Hope this helps!

• Even if we are given the fact that $K_2=0.012$ for sulfuric acid???? – riemannium Jun 20 '18 at 20:54
• @riemannium Unfortunately, yes. Due to the fact that sulfuric acid's first dissociation Ka is large, as explained in the article to which I linked, it is a strong acid. If you found a way to isolate the bisulfate ion (HSO4-), it might work, but probably not. Any other questions? – Shady Puck Jun 20 '18 at 21:01

My reasoning was different...The above salt is "neutral", but as it is a solution in water, you double the amount of solvent, so I thought in the following: 1 liter HCl 0.1M equals to [H⁺]=0.1 of course, and thus 0.1 mole. Then you get another liter of aqueous salt...So a more precise answer, it seems to me, would be pH(mixture)=-log(0.05)=1.3> 1. It was obvious for me that the mixture should raise a little bit the pH, even if the salt was neutral and from a strong acid/strong base...If $K_2(H_2SO_4)=0.012$, then $K_h=K_w/k_2≃10^{-12}$. So, $\alpha=10^{-6}$ and pH=1.45 (not far from 1.3).

• Nice catch on my missing the dilution's effect on the pH -- I edited my answer. However, the solution still would not buffer. – Shady Puck Jun 20 '18 at 21:22
• Do you have any other questions I can help with? If not, would you consider accepting my answer? I’d really appreciate it! – Shady Puck Jun 21 '18 at 20:15
• Read my extra answer...I only have the doubt about $K_2$...Since, I wonder if it would matter...For instance, imagine instead of $Na_2SO_4$ had been $Na_2CO_3$, and its $K_2=10^{-11}$... – riemannium Jun 21 '18 at 20:31
• In my experience, $\ce{K_2}$ has little impact on acid dissociations (relevant only to me when calculating the pH of a polyprotic acid to a number of significant figures). If you wanted to use $\ce{H2CO3}$ (carbonic acid) and $\ce{NaHCO3}$ (sodium bicarbonate), I believe that would create a reasonable buffer. Just remember: the acid/base and salt MUST share a common ion. Hope this helps! – Shady Puck Jun 21 '18 at 20:42