What are the products for this reaction: NaOH + Na2S ?

Jun 25, 2017

No reaction should occur..

Explanation:

I'm not sure they can form anything....

Switching the cations (or anions) just yields the same species, so I don't think a reaction occurs here..

In theory:

$\text{NaOH"(aq) + "Na"_2"S"(aq) rarr "Na"_2"S"(aq) + "NaOH} \left(a q\right)$

On top of that, all the ions remain unchanged, so there is no net ionic equation, and every ion is a spectator ion, so no reaction would occur even then.

Jun 26, 2017

No Reaction => No Driving Force

Explanation:

For metathesis reactions to proceed to completion there must be a driving force. That is, the tendency for the system to form a compound that removes itself from the reaction and prevents it from reversing to reform reactants. Driving force products are precipitating salts, formation of weak electrolytes or gas decomposition products.

Precipitating Salts => need to know basic solubility rules of ionic compounds ...

Weak Electrolytes => formation of weak acid or weak base => requires an understanding of acid-base equilibria and

Gas Decomposition Products => requires an understanding of labile weak electrolytes that spontaneously decompose to form compounds like CO2, SO2, NO, NO2, NH3, etc.

Examples:

Salt Ppt => $A g C l \left(s\right)$
$N a C l \left(a q\right) + A g N {O}_{3} \left(a q\right) \implies N a N {O}_{3} \left(a q\right) + A g C l \left(s\right)$

Wk Electrolyge => $H - O H \left(l\right)$
$H C l \left(a q\right) + N a O H \left(a q\right) \implies N a C l \left(a q\right) + H - O H \left(l\right)$

Gas Decomp Product => $C {O}_{2} \left(g\right)$
$C a C {O}_{3} \left(a q\right) + 2 H C l \left(a q\right) \implies C a C {l}_{2} \left(a q\right) + {H}_{2} C {O}_{3} \left(l\right)$
${H}_{2} C {O}_{3} \left(l\right) \implies C {O}_{2} \left(g\right) + {H}_{2} O \left(l\right)$
$C a C {O}_{3} \left(a q\right) + 2 H C l \left(a q\right) \implies C a C {l}_{2} \left(a q\right) + C {O}_{2} \left(g\right) + {H}_{2} O \left(l\right)$