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From these three salts, which will be best able to buffer the addition of $\ce{HCl}$?

  1. ammonium chloride

  2. sodium carbonate

  3. ammonium acetate.

I think sodium carbonate is the right choice because when it dissociates in water forms a basic solution, so when $\ce{HCl}$ mixes with sodium carbonate it will be the best buffer, but it's a weak base so I'm confused. So can someone explain which is best?

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To pick the best salt in this case, we should pick the least acidic cation and the most basic anion.

While sodium cannot dissociate into $\ce{H+}$ ions, ammonium can, so the ammonium salts will be more acidic than their sodium counterparts.

In addition to being the most basic ($\mathrm{p}K_\mathrm{b}(\ce{CO3^{2-}})=3.67$, $\mathrm{p}K_\mathrm{b}(\ce{C2H3O2-})=9.24$, $\mathrm{p}K_\mathrm{b}(\ce{Cl-})=20.3$), carbonate is also a diacidic base whose conjugate acid is also a stronger base than the other two anions ($\mathrm{p}K_\mathrm{b}(\ce{HCO3-})=7.63$).

So, sodium carbonate is the best choice.

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  • $\begingroup$ From the first line, can you tell me the reason for choosing most basic anion? I am struggling to understand this concept of buffer. Thanks alot. $\endgroup$ – samjoe Nov 17 '17 at 13:04
  • $\begingroup$ Is it because the more basic anion will abstract $\ce{H+}$ from solution given by $\ce{HCl}$ ? $\endgroup$ – samjoe Nov 17 '17 at 13:05
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Of the top of my head...

  • Ammonium chloride

Definitely not this salt. There will be little free ammonia from the solvation of the $\ce{NH4^+}$ cation. The $\ce{Cl^-}$ is a spectator anion.

  • Sodium carbonate

The $\ce{Na^+}$ is a spectator cation. The $\ce{CO3^{2-}}$ will be mostly $\ce{CO3^{2-}}$ with a tiny bit of $\ce{HCO3^-}$. So the ratio of $\frac{\ce{HCO3^-}}{\ce{CO3^{2-}}}$ (and hence the pH) will change rapidly with the addition of $\ce{H+}$.

  • Ammonium acetate

Some of ammonium will protonate the acetate anion. So you end up with a mixture of $\ce{NH4^+}$, $\ce{NH3}$, $\ce{OAc^{-}}$ and $\ce{HOAc}$. I think this solution would have the most buffer capacity for acid.

It seems you need to solve all of these for $\dfrac{d\text{pH}}{d\ce{H+}}$ -- well the last two anyway...

SOLUTION

My calculus-foo is lost on this, so let's just brute force an answer. Let's assume we have 1.00 liters of 0.1 molar solutions of each of the salts. We'll calculate the pH for the salt, add $1.00\times10^{-3}$ moles of $\ce{H+}$ and calculate the new pH. We can then calculate:

$\dfrac{d\text{pH}}{d\ce{H+}} = \dfrac{\text{pH}_2 - \text{pH}_1}{1.00\times10^{-3}}$

  • Ammonium chloride

$\ce{NH4+ <=> H+ + NH3 }$

$\text{K}_\text{a} = 5.75\times10^{-10} = \dfrac{\ce{[H+][NH3]}}{\ce{[NH4+]}}$

Salt alone

assume $\ce{[H+]=[NH3]}$ and $\ce{[NH4+]=0.1}$ molar then simplify

$\ce{[H+]} = \sqrt{\text{K}_\text{a}\times\ce{[NH4+]}} = 7.56\times10^{-6}$

pH = 5.121

we can now solve for

$\dfrac{\ce{[NH3]}}{\ce{[NH4+]}} = \dfrac{\text{K}_\text{a}}{\ce{[H+]}} = \dfrac{5.75\times10^{-10}}{7.56\times10^{-6}} = 7.61\times 10^{-5} $

$\ce{[NH3] = 7.61\times10^{-6}}$

Salt plus 0.001 moles acid

So there isn't any significant amount of $\ce{NH3}$ to protonate and the acid added can be added to the initial concentration of $\ce{H+}$ from the salt.

final $\ce{[H+]} = 7.56\times10^{-6} + 1.00\times10^{-3} = 1.008\times10^{-3}$

pH = 2.997

$\dfrac{d\text{pH}}{d\ce{H+}} = \dfrac{\text{pH}_2 - \text{pH}_1}{1.00\times10^{-3}} = \dfrac{2.997 - 5.121 }{1.00\times10^{-3}} = -2124$

  • Sodium carbonate

$\ce{CO3^{2-} + H2O <=> OH^- + HCO3^{-}}$

$\text{K}_\text{b} = 2.14\times10^{-4} = \dfrac{\ce{[OH^-][HCO3^-]}}{\ce{[CO3^{2-}]}}$

Salt alone

assume $x = \ce{[HCO3^-] = [OH^-]}$ and $\ce{[CO3^{2-}]} = 0.100 -x$ molar then simplify

$0 = x^2 + 2.14\times10^{-4}x - 2.14\times10^{-5}$
$x = 4.52\times10^{-3}$

$\ce{[OH-]} = 4.52\times10^{-3}$

$\ce{[H+]} = \dfrac{K_\rm{w}}{\ce{[OH-]}} =2.21\times10^{-12}$

pH = 11.655

Salt plus 0.001 moles acid

Now if we add 0.001 moles of $\ce{H+}$.

We'll essentially neutralize some $\ce{OH-}$ and make some $\ce{HCO3-}$, but we know that

$\ce{[HCO3-] - [OH-]} = 0.001$ or $\ce{[HCO3-]} = 0.001 + \ce{[OH-]}$

Let $x = \ce{[OH-]}$

$2.14\times10^{-4} = \dfrac{\ce{[OH^-][HCO3^-]}}{\ce{[CO3^{2-}]}}$

$2.14\times10^{-4} = \dfrac{(x)(0.001+x)}{0.1 - (0.001+x)} = \dfrac{x^2 + 0.001x}{0.099-x}$

$0 = x^2 + 1.214\times10^{-3}x - 2.1186\times10^{-4}$
$x = 0.004036$

$\ce{[OH^-]} = 0.004036$

$\ce{[H+]} = \dfrac{1.00\times10^{-14}}{\ce{[OH-]}} = 2.48\times10^{-12}$

pH = 11.606

$\dfrac{d\text{pH}}{d\ce{H+}} = \dfrac{\text{pH}_2 - \text{pH}_1}{1.00\times10^{-3}} = \dfrac{11.655 - 11.606}{1.00\times10^{-3}} = 49$

  • Ammonium Acetate

Salt alone

Salt plus 0.001 moles acid

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  • $\begingroup$ You have not added your analysis for Ammonium acetate, but nevertheless a good and detailed answer. $\endgroup$ – samjoe Nov 17 '17 at 13:09
  • $\begingroup$ $\ce{CO3^- <=> OH^- + HCO3^{-}}$ that does not appear right, the charges and atoms are not balanced. I can't make sense of it. $\endgroup$ – Martin - マーチン Jul 31 '18 at 10:10
  • $\begingroup$ @Martin-マーチン - No new Nobel prize chemistry just plain sloppiness. I fixed it. $\endgroup$ – MaxW Jul 31 '18 at 15:10
  • $\begingroup$ And user ringo no doubt deduced the right answer using chemical logic not math. $\endgroup$ – MaxW Jul 31 '18 at 15:43

protected by Community Jul 30 '18 at 15:51

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