Chemistry, 26.01.2021 15:10 aorilneedshelp6636

The following chemical reaction takes place in aqueous solution: ZnBr2(aq)+2 NaOH(aq) --Zn(OH),(s)+2 NaBr(aq)
Write the net ionic equation for this reaction.

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Answer from: faithtunison

${\rm Zn^{2+}}\, (aq) + {\rm 2\; OH^{-}}\, (aq) \to {\rm Zn(OH)_2}\, (s)$ .

Explanation:

Ionic Equation for this reaction

Rewrite only the species that exist as ions. Those species typically include:

soluble salts,strong acids, and soluble bases.

In this reaction, both ${\rm ZnBr_2}\, (aq)$ and ${\rm NaBr}\, (aq)$ are salts. The state symbol " (aq) " suggests that both of these salts are soluble. Hence, both of these salts exist as ions and should be rewritten:

Each ${\rm ZnBr_2}\, (aq)$ formula unit would exist as one ${\rm {Zn}^{2+}}\, (aq)$ and $2\; {\rm Br^{-}}\, (aq)$ . Notice how there are twice as many ${\rm Br^{-}}$ ions as ${\rm {Zn}^{2+}}$ ions.Each ${\rm NaBr}\, (aq)$ formula unit would exist as one ${\rm Na^{+}}\, (aq)$ and one ${\rm Br^{-}}\, (aq)$ .

Similarly, the state symbol " (aq) " suggests that the base $\rm NaOH$ is also soluble:

Each ${\rm NaOH}\, (aq)$ formula unit would exist as one ${\rm Na^{+}}\, (aq)$ and one ${\rm OH^{-}}\, (aq)$ .

On the other hand, the state symbol " (s) " suggests that the base ${\rm Zn(OH)_2}\, (s)$ is a precipitate and is not soluble. Rather, the bonds within ${\rm Zn(OH)_2}$ stay mostly intact, and this species would not exist as ions. Hence, do not rewrite ${\rm Zn(OH)_2}\, (s)\!$ when deriving the ionic equation for this reaction.

Hence, the ionic equation for this reaction would be:

$\begin{aligned}&\underbrace{{\rm Zn^{2+}}\, (aq) + 2\, {\rm Br^{-}}\, (aq)}_{\text{from ${\rm ZnBr_2}\, (aq)$}} + \underbrace{2\, {\rm Na^{+}}\, (aq) + 2\, {\rm OH^{-}}\, (aq)}_{\text{from $2\, {\rm NaOH}\, (aq)$}} \\ & \to \underbrace{{\rm Zn(OH)_2}\, (s)}_{\text{precipitate}} + \underbrace{2\, {\rm Na^{+}}\, (aq) + 2\, {\rm Br^{-}}\, (aq)}_{\text{from $2\, {\rm NaBr}\, (aq)$}}\end{aligned}$ .

Net Ionic Equation for this reaction

Eliminate species that are present on both sides of the ionic equation to obtain the net ionic equation. A species should be eliminated if only if an equal number of this species are found on both sides of the ionic equation. Otherwise, subtract from the side with a larger number of that species.

For this reaction, the net ionic equation would be:

${\rm Zn^{2+}}\, (aq) + {\rm 2\; OH^{-}}\, (aq) \to {\rm Zn(OH)_2}\, (s)$ .

Answer from: Quest

explanation:

metals have excess of electrons whereas non-metals have deficiency of electrons. thus, in order to gain stability metals easily donate their electrons to the non-metals.

for example, electronic configuration of sodium is $1s^{2}2s^{2}2p^{6}3s^{1}$ and electronic configuration of phosphorous is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{3}$ .

when a phosphorous atom and sodium atom combine together then sodium will donate is extra one electron to the phosphorous atom. as a result there will be formation of ionic bond between sodium and phosphorous.

the reaction will be as follows.

$3na + p \rightarrow na_{3}p$

there, are three sodium sodium atoms and each of them will donate its one extra electron to phosphorous atom. thus, the sub-shell of phosphorous gets completely filled.

thus, we can conclude that electrons are transferred from atoms of sodium to atoms of phosphorous. this transfer makes the sodium atoms donate one electron and the phosphorous atoms to gain one electron. as a result, the sodium and phosphorus atoms strongly attract each other.