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At equilibrium in pure water, we have $$\ce{[H_3O+][OH-]} = 10^{-14}$$

Since $\ce{H3O+}$ and $\ce{OH-}$ ions are produced in pairs, we may conclude $$\ce{[H_3O+]}=\ce{[OH-]} = 10^{-7}$$

So far so good. But shouldn't things change when we introduce a new substance into water ? I mean why does the first equation above hold no matter what ?

Also when I introduce $\ce{H2SO4}$ into the water, it doesn't just give a $\ce{H+}$ ion, it also gives $\ce{HSO4-}$ ion. Shouldn't these new negative ions change the behavior of water? Why does my textbook never talk about these new negative ions? Help appreciated. Thanks!

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    $\begingroup$ The ionic product of water ($K_w$) has constant value of $10^ {-14} $at temperature of $25^o $ C. Unless you change the temperature, nothing is going to happen to $K_w$. $\endgroup$ – Soumik Das Sep 13 '18 at 6:17
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Your question title is a bit misleading, but i try to answer all the small questions in you question text.

The equation $[\ce{H3O^+}][\ce{OH^-}]$ holds true, if other parameters (like T) are constant. Keep in mind, the power of hydroxide decreases, whereas the power oxonium increases. Being equal in the equation and considering how logs are being computed,bthey will add up to 14 every time.

Regarding the introduced $\ce{HSO4^-}$, they don't contribute to pH by definition. On the other hand, they alter the behaviour of the water, by increasing its conductivity.

Your textbooks don't talk about the other negative ions in acidic or alkaline solutions, because they don't directly contribute to the values of pH or pOH by definition. In cases of polyacids like sulfuric acid, ions like $\ce{HSO4^-}$ are accounted for by using a different formula to calculate the actual pH value, but the "not-hydrogen" part is largely irrelevant in the behaviour of the solution itself.

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  • $\begingroup$ Ohk since the $pH$ and $pOH$ must add up to $14$, I see why $pOH$ decreases with addition of acid. Thank you :) But I still don't get why we care so much about $pH$, and never bother about other ions that may present in the water like $pHSO_4$. Why change in $H^+$ is a big deal ? $\endgroup$ – rsadhvika Sep 13 '18 at 11:23
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    $\begingroup$ It is the "biggest deal" in many cases, because the concentration of $\ce{H^+}$ has a high impact on the reactivity (behaviour) on everything else in the solution. It is a bit like asking, why T-shirt sizes are only expressed as being "l" or "m", even though they do not account for the type of material used for the shirt. $\endgroup$ – Chris Sep 13 '18 at 11:26
  • $\begingroup$ Oh ty for the quick response! I think I get $H^+$ is a big deal, but I don't see how the concentration of other solutes like $HSO_4^-$ don't influence the reactivity ? $\endgroup$ – rsadhvika Sep 13 '18 at 11:29
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    $\begingroup$ Hydrogensulfate can donate another proton, but then it is a rather unreactive anion. It is fully oxidized sulfur with a very stable (empty) electron shell. Becoming a stable and unreactive compound is actually an important part of the driving force behind the deprotonation reaction itself. $\endgroup$ – Chris Sep 13 '18 at 11:32
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    $\begingroup$ Not quite. $\ce{HSO4^-}$ is still active, because it can deprotonate. Also, sulfate is not unreactive in an absolute sense, but in contrast to the rest of the solution, if you only add sulfuric acid to water. There are many reactions that include sulfate. For instance, you can reduce sulfate to sulfite. But i think we are losing focus of what you were originally asking. $\endgroup$ – Chris Sep 13 '18 at 11:45
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As long as you stay in the dilute regime, that is if water molecules have most of the time to interact just with each other, $K_w$ won't change. Add 20% of anything soluble (salt, acid, alcohol, acetone, ...) and things look different.

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  • $\begingroup$ Oh, so if I add NaCl or other solutes to the water, the $H_3O^+$ and $OH^-$ ions will not notice the newly added solutes ? Is this because the solute concentration is low ? I think I get it... Thank you :) $\endgroup$ – rsadhvika Sep 13 '18 at 11:24
  • $\begingroup$ Yes. Added OH- or H3O+ ions take part in the autoprotolysis equillibrium, the general properties of the solvent are however mostly uninfluenced. $\endgroup$ – Karl Sep 13 '18 at 16:54

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