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One kg of each substance below is dissolved in a bathtub full of water. Will it be acidic, basic, or neutral?
$\ce{KNO2}$
$\ce{HF}$
$\ce{(CH3)3NHBr}$
$\ce{HO-CH2-CH2-O-CH2-Cl}$

$\ce{KNO2}$, $\ce{K+}$ $\ce{NO2-}$ $\ce{K+}$ is inert and $\ce{NO2-}$ is the base of strong acid, so it should be basic overall. $\ce{HF}$, $\ce{H+}$ $\ce{F-}$ $\ce{H+}$ is strong acid and $\ce{F-}$ is strong base so it should be neutral overall. $\ce{(CH3)3NHBr}$, Molecule acts like a $\ce{H3NHBr}$ or $\ce{NH4+}$ $\ce{Br-}$ so $\ce{NH4+}$ is acidic $\ce{Br-}$ is neutral, so overall solution is acidic.
$\ce{HO-CH2-CH2-O-CH2-Cl}$, No idea.

Can someone please explain them, if they're wrong? Thank you.

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  • $\begingroup$ $\ce{NO2^{-}}$ is the conjugate base of a weak acid, nitrous acid ($\ce{HNO2}$). Also, $\ce{F^{-}}$ is a weak base in water. $\endgroup$ – Nicolau Saker Neto Apr 27 '15 at 1:12
  • $\begingroup$ So KNO2 is still basic overall. HF is basic because of F-. I'm lost on the last two. $\endgroup$ – user15831 Apr 27 '15 at 1:19
  • $\begingroup$ I'm assuming the last one is a weak acid. $\endgroup$ – Asker123 Apr 27 '15 at 1:58
  • $\begingroup$ What is the last molecule comparable to? HOCl? $\endgroup$ – user15831 Apr 27 '15 at 2:00
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The $\ce{KNO2}$ is basic, the HF is acid, the $\ce{(CH3)3NHBr}$ being an ammonium salt is acid too, the last one is an alcohol, I would consider it neutral for your purpose. But you are missing one step to get your answer quite right. Last part, in the pH world, everything turn around two ions H+ (or H3O+) and OH- and the equilibrium between those two. A substance is considered acid when it either give a H+ as disolving or takes an OH-. On the other hand a substance is considered basic when it give an OH- or takes a H+. Like HF is an acid (H+, F-) and NaOH is basic (Na+, OH-). For KNO2, it can take an H+ to go back to HNO2, so it is basic. For organic compounds, alcool groups are usually considered inerte as they don't ionate in normal conditions. Now, maybe I'm wrong but your question sound like if you put all those in a bathtub what'S the final result? Is it right or not?

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  • $\begingroup$ I don't get what I'm supposed to do? $\endgroup$ – user15831 Apr 27 '15 at 2:28
  • $\begingroup$ (CH3)3NHBr isn't ammonium salt, but bromoamine and and alcohol hydrolyzes to HCl @Asker123 $\endgroup$ – Mithoron Apr 27 '15 at 21:55
  • $\begingroup$ So @mithoron could you tell me the difference between those two molecules BrNH2 vs NH4Br? $\endgroup$ – Bruno Apr 27 '15 at 22:13
  • $\begingroup$ en.wikipedia.org/wiki/Chloramine, These two last are much harder than first ones. Major difference is covalent vs ionic. $\endgroup$ – Mithoron Apr 27 '15 at 23:27
  • $\begingroup$ Oops, I thought it's tert-buthyl there, but it's trimethyl ammonium... But still this organic molecule is chloroether, which hydrolyses and therefore solution will be acidic. $\endgroup$ – Mithoron Apr 30 '15 at 17:57
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Assuming your question is how to determine the acidity or basicity of salts.

  1. Acidic Salt: A salt formed between a strong acid and a weak base is an acid salt.

$$\ce{HCl + NH4OH -> NH4Cl + HOH}$$ The $\ce{NH4Cl}$ is an acidic salt and if it dissociates it will be acidic.

  1. Basic Salt: A basic salt is formed between a weak acid and a strong base.

$$\ce{NaOH + HAc -> NaAc + HOH}$$ The Sodium Acetate the basic salt that dissociated into acetate ions and takes an H+ in water.

Applying this logic to your question gives us: $\ce{KNO2}$ is a basic salt, the $\ce{HF}$ is an acid, the $\ce{(CH3)3NHBr}$ being an ammonium salt is acidic too, the last one is an alcohol which has no effect.

Thanks.

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  • $\begingroup$ OK. That makes sense. But can you also figure out whether each anion/cation is a product of a strong/weak acid/base and determine the acidity/basicity of the salt using that? Ex: KNO2, The NO2- comes from HNO2, a weak acid, so it is basic overall (and K+ is an inert ion)? $\endgroup$ – user15831 Apr 27 '15 at 10:54
  • $\begingroup$ @user15831 That is exactly what you are supposed to do. But I doubt that will help with hydrocarbons. $\endgroup$ – Asker123 Apr 27 '15 at 21:05

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