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Problem:

A buffer was made by mixing $\pu{500 mL}$ of $\pu{1 M}$ acetic acid and $\pu{500 mL}$ of $\pu{0.5 M}$ calcium acetate.

What are the resulting concentrations of acetic acid (henceforth called $\ce{AcH}$), $\ce{Ca^2+}$ and acetate (henceforth called $\ce{Ac-}$, and what is the resulting $\mathrm{pH}$? $K_\mathrm{a}$(acetic acid) = $1.75\times10^{-5}.$

My attempted solution:

This problem gives me two equilibrium reactions to take into account.

\begin{align} \ce{AcH + H2O &<=> Ac- + H3O+}\tag{1}\\ \ce{Ac- + H2O &<=> AcH + OH- }\tag{2}\\ \end{align}

In (2), I assume that calcium acetate completely dissociates in water. I also assume that the concentration of $\ce{OH- }$ from the auto ionization of water is negligible.

First I calculate the concentrations of $\ce{AcH}$, $\ce{Ac- }$ and $\ce{H3O+}$ in (1), ignoring what I assume to be minuscule amounts of $\ce{H+}$ from the auto ionization of water. To do this, I use the equation for the acid dissociation constant:

$$K_\mathrm{a}=\frac{[\ce{Ac- }][\ce{H3O+}]}{[\ce{AcH}]}\tag{3}$$

Initial values for $\ce{AcH}$, $\ce{Ac- }$ and $\ce{H3O+}$ are set to 1, 0 and 0 M, respectively. Equilibrium concentrations will then be $1-x$, $x$ and $x$ M.

Plugging this into (3), I get $x = 4.17\times10^{-3}.$

I have now found (or can at least easily calculate) the concentrations of the compounds in (1) at equilibrium.

On to (2), I know that $K_\mathrm{a}K_\mathrm{b}=K_\mathrm{w}$, and I have the values for $K_\mathrm{a}$ and $K_\mathrm{w}$, which gives me the value $5.71\times10^{-10}$ for $K_\mathrm{b}.$

Initial concentrations in (2) for $\ce{Ac- }$, $\ce{AcH}$ and $\ce{OH- }$ are 1 (because 0.5 moles of calcium acetate gives $2\times0.5=1$ moles of acetate), 0 and 0 M, respectively. Equilibrium concentrations will then be $1-x$, $x$ and $x$. Plugging this into (3) (though using the equation for the base dissociation constant instead), I get $x = 2.39\times10^{-5}.$

I have now found all the values I need to calculate the final concentrations and pH when adding the two solutions to each other, and up till this point I am fairly sure that all my calculations are correct.

Questions:

  1. How do I write the chemical equation for the mixing of the two solutions? I tried the following; $\ce{AcH + Ac- <=> Ac- + AcH},$ but this doesn’t seem to help at all.

  2. From my younger days, I seem to remember a much simpler and quicker method of doing these calculations than what I've shown here; does such a method exist? Please note that if it does exist I’d like to learn about it, but I would also like answers based on the method I've used.

  3. Let’s say you mix one weak acid with one weak base, and they aren't conjugate pairs. How would the method differ from the one in this problem? I am specifically wondering about whether both the dissociation constants must be provided to be able to solve, and also how to write the chemical equation, but other observations are more than welcome as well.

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  1. In general: $\ce{acid + H2O <=> base + H_3O^+}$
  2. It is called the Henderson-Hasselbalch equation $\ce{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{\ce{[base]}}{\ce{[acid]}}$
  3. I believe the equilibrium constant for a such a neutralisation would look like $K_n = K_\mathrm{a}K_\mathrm{b}\frac{1}{K_\mathrm{w}}$
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Problem: A buffer was made by mixing 500 mL of 1 M (mol/L) acetic acid and 500 mL of 0.5 M calcium acetate. What are the resulting concentrations of acetic acid, Ca2+ and acetate and what is the resulting pH? Ka(acetic acid) = 1.75×10−5

Final volume = $500+500$ mL = 1L
Concentration of Acetic Acid = $\frac{\rm moles}{\rm volume}=\frac{0.5*1}{1}=0.5$ M
Concentration of Calcium Acetate = $\frac{\rm moles}{\rm volume}=\frac{0.5*0.5}{1}=0.25$ M
So using Henderson equation: $$pH=pKa+\log\frac{0.25*2}{0.5}=4.76$$ Rest is easy.

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