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Having read the following was wondering something: $$\mathrm {p}K_\mathrm{a} = -\log\frac{[\ce{H+}]_\mathrm{eq}[\ce{A-}]_\mathrm{eq}}{[\ce{HA}]_\mathrm{eq}} $$ $$\ce{HA <=> H+ + A-}$$ $$\mathrm{pH = p}K\mathrm{_{a}} +\mathrm{log\frac{[\ce{A-}]}{[\ce{HA}]}} $$ This rmgd $$ \mathrm{[\ce{A-}] = [salt] }$$ when acid is weak.

Now my question is like this: take a beaker of $\ce{CH3COOH}$. Now add $\ce{CH3COONa}$. Now here using the a over formula as salt concentration increases $\mathrm{pH}$ should decrease when take in $\ce{A-}$ as salt concentration. However this same $\ce{A-}$ was there in the $\mathrm{p}K_\mathrm{a}$ formula so shouldn't the $\mathrm{p}K_\mathrm{a}$ change to compensate? If so then there will be no $\mathrm{pH}$ change but I have read that there is $\mathrm{pH}$ change on adding $\ce{CH3COONa}$ in a solution. What is wrong with this?

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    $\begingroup$ The first equation is wrong : it’s Ka or not pKa $\endgroup$ – Nicolas Mar 1 at 13:16
  • $\begingroup$ It is now editted . $\endgroup$ – ask Mar 1 at 13:23
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    $\begingroup$ The first equation only works if the system is in equilibrium. I added to subscripts to indicate that. $\endgroup$ – Karsten Theis Mar 1 at 19:17
  • $\begingroup$ indeed @KarstenTheis, I corrected my answer as well $\endgroup$ – Nicolas Mar 1 at 20:39
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  1. the second relation is wrong to : $pK_{a} = -\log\frac{[\ce{H+}]_{eq}[\ce{A-}]_{eq}}{[\ce{HA}]_{eq}}=-\log[\ce{H+}]_{eq} -\log\frac{[\ce{A-}]_{eq}}{[\ce{HA}]_{eq}}$ so $pH=pK_a+\log\frac{[\ce{A-}]_{eq}}{[\ce{HA}]_{eq}}$
  2. the acidity constant is the relation which links the concentrations of the species present (and not the reverse): if we add the salt $ \ce {A^-} $, then according to the relation you see that the pH increases. You can also see it by reasoning thus: you add a base so the pH can only increase.
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