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I'd start off by saying it's been many years since I was in chemistry lessons and have forgotten much, but recently I have had to relearn a lot of basic chemistry. The following doesn't really impede me in my work, but it irritates me that I can't figure out what is basically two sides of the same coin.

Let's imagine I want to calculate the pH of a 0.1M acetic acid ($c_\ce{CH3COOH} = \pu{0.1M}$)

Acetic acid is a weak acid and dissociates in water as follows: $\ce{CH_3COOH + H_2O <<=> CH_3COO^- + H_3O^+}$. Expressing this in terms of acid dissociation constant (and ignoring the water) yields $K_\mathrm{a} = {\large{\ce{\frac{[CH_3COO^-][H^+]}{[CH_3COOH]}}}}$

We can look up the $K_\mathrm{a}$ of acetic acid, which is $1.8 \times 10^{-5}$. If we assume that the amount of $\ce{CH_3COOH}$ change is negligible (due to it being a weak acid and only slightly dissociating), and that $\ce{CH_3COO^-}$ and $\ce{H^+}$ concentrations are equal, we come up with the following equation:

$1.8 \times 10^{-5} = \large{\frac{[\ce{H^+}]^2}{\pu{0.1M}}}$

Rearranging and solving for $\ce{[H^+]}$ gives us $\sqrt{1.8 \times 10^{-5} \times \pu{0.1M}} = \pu{1.34E-3 M}$

Since $\mathrm{pH} = -\log([\ce{H+}])$ (or $\mathrm{pH} = \log(1/[\ce{H+}])$, whichever you prefer), $\mathrm{pH} = -\log(1.34 \times 10^{-3}) \approx 2.88$

I've checked this and I know it's correct. But I don't know how to do it from the $K_\mathrm{b}$. Or rather, I get the answer, but I don't fully understand the process.

So the opposite reaction, $\ce{CH_3COO^- + H_2O <=>> CH_3COOH + OH^-}$, also has a equilibrium constant which is called $K_\mathrm{b}$, defined as follows (again ignoring the water):

${\normalsize K_\mathrm{b}} = {\large \ce{\frac{[CH_3COOH][OH^-]}{[CH_3COO^-]}}}$

We can look it up or calculate it from the relationship that $K_\mathrm{a} \times K_\mathrm{b} = 1\times10^{-14}$

$K_\mathrm{b} = {\large \frac{1\times10^{-14}}{K_\mathrm{a}}}$

In either case, $K_\mathrm{b} = 5.6 \times 10^{-10}$

Now, I don't know what to do with the $\ce{[OH^-]}$. My basic assumption is that the change in the amount of $\ce{CH_3COOH}$ is negligible, so that stay as 0.1M. But unlike previously where $\ce{CH_3COO^-}$ and $\ce{[H^+]}$ were formed in equal amounts due to the dissociation of the acid, how much of $\ce{[OH^-]}$ is there? My (wrong) logic was to assume that $\ce{[CH_3COO^-] = -[OH^-]}$, but then they cancel out, leaving me with $K_\mathrm{b} = \ce{[CH3COOH]}$, which can't be right. So yeah, I'm lost.

I solved this by substituting $\ce{[OH^-]}$ with $\large{\frac{K_\mathrm{w}}{[\ce{H+}]}}$ (because $\ce{[H^+][OH^-]} = K_\mathrm{w}$)

This yields ${\normalsize K_\mathrm{b}} = {\large \ce{\frac{[CH_3COOH] K_w}{[CH_3COO^-][H^+]}}}$. Again assuming that $\ce{[CH_3COO-]} = \ce{[H+]}$, and knowing that $K_w = 1 \times 10^{-14}$, I come up with ${5.6 \times 10^{-10}} = {\large \ce{\frac{0.1M \times 1 \times 10^{-14}}{[H^+]^2}}}$

Solving for $\ce{[H^+]}$ gives us $\sqrt{{\large \frac{\pu{0.1M} \times 1 \times 10^{-14}}{5.6 \times 10^{-10}}}} = \pu{1.33 \times 10^{-3}M}$

This is the same concentration of $\ce{[H^+]}$ that I got using the $K_\mathrm{a}$ above, making the $\mathrm{pH} = -\log(1.33 \times 10^{-3}) \approx 2.88$

So I can get to the same answer ignoring (or rather substituting) the $\ce{[OH^-]}$, but I'd really like to know how to do it using that. This is probably a really simple answer but after having spent many hours staring my many pieces of paper, my mind is coming up with a blank.

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  • $\begingroup$ In this answer chemistry.stackexchange.com/questions/60068/… there is a general method to solve problems of this type that might help to answer some of your queries. $\endgroup$ – porphyrin Jun 21 '18 at 14:17
  • $\begingroup$ @porphyrin Thanks, I'll work through it and come back if I have questions! $\endgroup$ – Mazuriel Jun 21 '18 at 17:20

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