Find the $\mathrm{pH}$ of a $\pu{0.500 M}$ solution of $\ce{KC2H3O2}$.

I am preparing for an exam on equilibrium and I have two questions:

  1. How would I go about calculating this?
  2. Is the $\ce{H3O}$ in the middle of $\ce{KC2H3O2}$ one hydroxide ion?

Here is what I have attempted to do: find the conjugate acid/bases, write out a balanced equation, then take the $-\log{[\ce{H+}]}$ of the conjugate acid. But I am a little stuck on the best course of action to find the conjugate acid/base.

  • 1
    $\begingroup$ Welcome to Chemistry Stack Exchange! Please add what you have attempted towards solving the problem into the body of your question. For more information, see the site's homework policy for how to ask homework questions. Thanks! $\endgroup$ – jonsca May 15 '14 at 22:28
  • $\begingroup$ Thanks for the reply. I have added more info and context to the question. $\endgroup$ – Jake Chasan May 15 '14 at 22:30

First of all let us introduce the given compound in a more bonding descriptive way: $\ce{H3C-COOK}$. This compound is commonly referred to as potassium acetate:

structure of potassium acetate

It reacts with water in the following fashion: $$\ce{H3C-COOK_{(aq)} + H2O <=> H3C-COOH_{(aq)} + K+_{(aq)} + {}^{-}OH_{(aq)}}$$

Then the mass action law can be applied in the following way: $$K_c=\frac{\ce{[H3C-COOH]}\ce{[K+]}\ce{[{}^{-}OH]}}{\ce{[H3C-COOK]}\ce{[H2O]}}$$

And you can then rearrange to form the base dissociation constant: $$K_b=\frac{\ce{[H3C-COOH]}\ce{[{}^{-}OH]}}{\ce{[H3C-COOK]}}$$

The $\mathrm{p}K_b$ for potassium should be around $4.7$.

As dissenter pointed out, you do not need the Henderson-Hasselbalch equation to calculate $\mathrm{p}\ce{H}$ values for a pure solution of potassium acetate. You can have a look at wikipedia for an example calculation.

For your second question:

  • Hydroxide Ion: $\ce{{}^{-}OH}$ or $\ce{OH-}$ or $\ce{HO-}$ ...
  • Hydronium Ion: $\ce{H3+O}$ or $\ce{H3O+}$ or $\ce{H+OH2}$ ...
  • $\begingroup$ I like how you wrote out all the different ways of expressing hydroxide ion or hydronium ion (some more correct than others). However, I do not see why the OP needs to use the Henderson-Hasselbalch equation to solve this problem. There is essentially no conjugate acid/base pair in this system. $\endgroup$ – Dissenter May 16 '14 at 4:41
  • $\begingroup$ @Dissenter You are right, for this problem you don't need it - fixed that. However, there is a conjugate acid-base pair in this system. (Incomplete dissolution - you have all species in the first equation) $\endgroup$ – Martin - マーチン May 16 '14 at 4:53

You need the $K_\mathrm{a}$ value of the solute of interest. That would be $\ce{[C2H3O2]-}$ as $\ce{K+}$ has highly limited Brønsted-Lowry acid/base properties in water.

  • $\begingroup$ This is more like a comment than an answer. Please consider elaborating a little more. Also I disagree on the acid-base properties. It is one of the most commonly used buffer systems. $\endgroup$ – Martin - マーチン May 16 '14 at 4:37
  • $\begingroup$ Potassium as a buffer? Please elaborate. You are right however I should have probably expanded on my answer but when I posted it appeared that the OP had not given much thought to the question so I merely wished to nudge the OP in the right direction. $\endgroup$ – Dissenter May 16 '14 at 4:39
  • $\begingroup$ Potassium acetate has the same buffer applicability as sodium acetate just add some acid and your done. $\endgroup$ – Martin - マーチン May 16 '14 at 4:45
  • $\begingroup$ Right, I meant potassium by itself. $\endgroup$ – Dissenter May 16 '14 at 6:07
  • $\begingroup$ I don't understand what you are talking about. Are you talking about elemental potassium, then I agree not much of acid base properties - strung reducing agent though. But the question was about something entirely different: the weak base potassium acetate. (I also corrected your charges.) $\endgroup$ – Martin - マーチン May 16 '14 at 6:17

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