0
$\begingroup$

Lets say I have a metal oxide and I put it in water, and it reacts. As far as I know, only metal oxides from group 1 and 2 react in water, making hydroxides.

Now here is my question: If I add HCl to the hydroxide, can I be sure that it will make a salt and water? Or are my previous assumptions wrong?

An example that works would be Sodium Oxide which reacted with water and then with HCl produces NaCl and water. Does it work the same way with all metals from this group though?

$\endgroup$
  • $\begingroup$ Some salts are dangerously toxic, not every salt is goody-goody $\ce{NaCl}$ you know. If you still have your wits with you, Do Not Drink. $\endgroup$ – Pritt says Reinstate Monica May 27 '17 at 14:04
  • $\begingroup$ I'm not actually going to drink anything. Are my other assumptions correct though? Are there any specific salts from group 1 and 2 I should know about (in the sense that they are very toxic)? Also, why are people downvoting the question? $\endgroup$ – user3450456 May 27 '17 at 14:28
  • 1
    $\begingroup$ I can't say why the DV's for sure, but in part it probably had to do with the "would I die if I drank it" phrase, which I removed; otherwise the question would have been closed as personal medical. Also, you could improve the quality of the question by giving a specific example. That's my 2 cents anyway. $\endgroup$ – airhuff May 27 '17 at 15:11
1
$\begingroup$

tl;dr: Yes, depending on extent of reaction

The group 1 and (group 2) hydroxides for example: $\ce{NaOH, KOH, RbOH, CsOH, Ba(OH)2, Ca(OH)2}$ are strong bases, being essentially fully ionized in aqueous solution; (although some are weaker $\ce{LiOH}$, $\ce{Mg(OH)2}$) . These should react with $\ce{HCl}$ - a strong acid forming water and a neutralisation product a typical corresponding salt $\ce{MCl}$ or $\ce{MCl2}$ .

Being soluble in aqueous envronment implies the salt is also obtained in this phase, so the question will be on extration method to use, in order to get this salt product in solid form.

Why I said depending on extent of reaction, is beacuse some metal oxides e.g $\ce{MgO}$ hardly form the hydroxide in the first place, and also insolubility with render the second reaction effectively out of question.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.