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My teacher tried to teach this concept to us in 5 minutes. I'm lost but I've made an effort to understand it all (but I need clarification.)

If we have a compound such as LiOH, what element would be needed to neutralize it? As in, I understand that this element has to be a base but, there are SO MANY bases. Can you maybe teach how to do these using the LiOH example and another example using HCl (aq). Any examples, really. I just need to understand that.

Secondly, we haven't learned solubility tables yet, I can identify an acid (H ...) and bases (... OH) but what if there's a compound like MgCl2, what do I do now? Do I use a table to look up the states or the pH? Is there another approach?

I learned that Alkali Metal + O2 = Oxides but I can't really apply that here, can I?

Literally any help will be appreciated, I'm not cramming rather just don't understand the concept.

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closed as too broad by M.A.R. ಠ_ಠ, Todd Minehardt, jerepierre, bon, Wildcat Nov 11 '15 at 9:22

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ Your question is really bad - what are we supposed to do do, teach you whole general and inorganic chemistry? $\endgroup$ – Mithoron Nov 10 '15 at 19:57
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As we are discussing acid-base neutralisation reactions they will be in solution (mixed with water).

First consider LiOH when mixed with water: $$\ce{LiOH -> Li+ + OH-}$$

As we can see $\ce{OH-}$ ions are formed, meaning this solution is a base. This means we need an acid to neutralise it. (Acids and bases react to neutralise each other).

Lets say we use $\ce{HCl}$ as our acid. Consider it mixed with water: $$\ce{HCl -> H+ + Cl-}$$

When we mix these two solutions together we get: $$\ce{Li+ + Cl- + H+ + OH- -> LiCl + H2O}$$

Notice how the $\ce{OH-}$ and $\ce{H+}$ react to form $\ce{H2O}$. The $\ce{LiOH}$ is now neutralised.

This can be written in one equation like this: $$\ce{LiOH + HCl -> LiCl + H2O}$$

Other acids could also be used to neutralise the $\ce{LiOH}$ such as $\ce{H2SO4}$: $$\ce{2LiOH + H2SO4 -> Li2SO4 + 2H2O}$$

I am uncertain of your second question. There is no way of neutralising $\ce{MgCl2}$ as it is a neutral salt (no $\ce{OH-}$ or $\ce{H+}$). Solubility tables (at this level) also do not affect the pH of your solution. If you haven't learned about it yet then don't attempt to now, you will likely just confuse yourself for your test tomorrow.

I hope this has been at an appropriate level for you.

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