Chemistry Stack Exchange is a question and answer site for scientists, academics, teachers, and students in the field of chemistry. It only takes a minute to sign up.
Sign up to join this community
From these three salts, which will be best able to buffer the addition of $\ce{HCl}$?
ammonium chloride
sodium carbonate
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?
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 ($\mathrm{p}K_\mathrm{b}(\ce{CO3^{2-}})=3.67$, $\mathrm{p}K_\mathrm{b}(\ce{CH3COO^{-}})=9.24$, $\mathrm{p}K_\mathrm{b}(\ce{Cl-})=20.3$), 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.
Of the top of my head...
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.
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 $\ce{H^+}$.
Some of ammonium will protonate the acetate anion. So you end up with a mixture of $\ce{NH4^+}$, $\ce{NH3}$, $\ce{CH3COO^-}$ and $\ce{CH3COOH}$. 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:
$\dfrac{d\text{pH}}{d\ce{H^+}} = \dfrac{\text{pH}_2 - \text{pH}_1}{1.00\times10^{-3}}$
$\ce{NH4^+ <=> H^+ + NH3 }$
$\text{K}_\text{a} = 5.75\times10^{-10} = \dfrac{\ce{[H^+][NH3]}}{\ce{[NH4^+]}}$
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
$\dfrac{d\text{pH}}{d\ce{H^+}} = \dfrac{\text{pH}_2 - \text{pH}_1}{1.00\times10^{-3}} = \dfrac{2.997 - 5.121 }{1.00\times10^{-3}} = -2124$
$\ce{CO3^{2-} + H2O <=> OH^- + HCO3^-}$
$\text{K}_\text{b} = 2.14\times10^{-4} = \dfrac{\ce{[OH^-][HCO3^-]}}{\ce{[CO3^{2-}]}}$
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
$\dfrac{d\text{pH}}{d\ce{H^+}} = \dfrac{\text{pH}_2 - \text{pH}_1}{1.00\times10^{-3}} = \dfrac{11.655 - 11.606}{1.00\times10^{-3}} = 49$
Salt alone
Salt plus 0.001 moles acid
