Chapter U4, Problem C17.1RE

Interpretation Introduction

Interpretation: The complete ionic equation and net ionic equation for the formation of solid nickel compound is to be written.

Concept introduction: In ionic equation, all the aqueous species are written in their ionic form. The net ionic equation is the equation in which the species which are actually taking part in the reaction are written.

Answer to Problem C17.1RE

The complete ionic equation for the formation of solid nickel compound is written below.

  ${\text{Ni}}^{2+}\left(aq\right)+2{\text{Cl}}^{-}\left(aq\right)+2{\text{Na}}^{+}+2{\text{OH}}^{-}\left(aq\right)\to \text{Ni}{\left(\text{OH}\right)}_{\text{2}}\left(s\right)+2{\text{Na}}^{+}\left(aq\right)+2{\text{Cl}}^{-}\left(aq\right)$ The net ionic equation for the formation of solid nickel compound is written below.

  ${\text{Ni}}^{2+}\left(aq\right)+2{\text{OH}}^{-}\left(aq\right)\to \text{Ni}{\left(\text{OH}\right)}_{\text{2}}\left(s\right)$

The complete ionic equation for the formation of solid nickel compound is written below.

  ${\text{Ni}}^{2+}\left(aq\right)+2{\text{Cl}}^{-}\left(aq\right)+2{\text{H}}^{+}+{\text{S}}^{2-}\left(aq\right)\to \text{NiS}\left(s\right)+2{\text{H}}^{+}\left(aq\right)+2{\text{Cl}}^{-}\left(aq\right)$ The net ionic equation for the formation of solid nickel compound is written below.

  ${\text{Ni}}^{2+}\left(aq\right)+{\text{S}}^{2-}\left(aq\right)\to \text{NiS}\left(s\right)$

The complete ionic equation for the formation of solid nickel compound is written below.

  ${\text{6Na}}^{+}\left(aq\right)+2{\text{PO}}_4^{3-}\left(aq\right)+3{\text{Ni}}^{2+}\left(aq\right)+6{\text{NO}}_3^{-}\left(aq\right)\to {\text{Ni}}_3{\left({\text{PO}}_4\right)}_2\left(s\right)+6{\text{Na}}^{+}\left(aq\right)+6{\text{NO}}_3^{-}\left(aq\right)$ The net ionic equation for the formation of solid nickel compound is written below.

  $2{\text{PO}}_4^{3-}\left(aq\right)+3{\text{Ni}}^{2+}\left(aq\right)\to {\text{Ni}}_3{\left({\text{PO}}_4\right)}_2\left(s\right)$

Explanation of Solution

Let’s take an example of solid nickel compound which is nickel hydroxide. Solid nickel is formed by the reaction of sodium hydroxide $\left(\text{NaOH}\right)$ with nickel chloride $\left({\text{NiCl}}_{\text{2}}\right)$ .

  ${\text{NiCl}}_{\text{2}}\left(aq\right)+2\text{NaOH}\left(aq\right)\to \text{Ni}{\left(\text{OH}\right)}_{\text{2}}\left(s\right)+2\text{NaCl}\left(aq\right)$

The compounds, present in the aqueous form, dissociate in their ions. The complete ionic equation for the formation of solid nickel compound is written below.

  ${\text{Ni}}^{2+}\left(aq\right)+2{\text{Cl}}^{-}\left(aq\right)+2{\text{Na}}^{+}+2{\text{OH}}^{-}\left(aq\right)\to \text{Ni}{\left(\text{OH}\right)}_{\text{2}}\left(s\right)+2{\text{Na}}^{+}\left(aq\right)+2{\text{Cl}}^{-}\left(aq\right)$ Here, ${\text{Na}}^{+}$ and ${\text{Cl}}^{-}$ is present at both side of the equation and termed as spectator ions. These ions can be canceled out. Therefore, the net ionic equation for the formation of solid nickel compound is written below.

  ${\text{Ni}}^{2+}\left(aq\right)+2{\text{OH}}^{-}\left(aq\right)\to \text{Ni}{\left(\text{OH}\right)}_{\text{2}}\left(s\right)$

Let’s take another example of solid nickel compound which is nickel sulfide. Solid nickel sulfide is formed by the reaction of hydrogen sulfide $\left({\text{H}}_2\text{S}\right)$ with nickel chloride $\left({\text{NiCl}}_{\text{2}}\right)$ .

  ${\text{NiCl}}_{\text{2}}\left(aq\right)+{\text{H}}_2\text{S}\left(aq\right)\to \text{NiS}\left(s\right)+2\text{HCl}\left(aq\right)$

The compounds, present in the aqueous form, dissociate in their ions. The complete ionic equation for the formation of solid nickel compound is written below.

  ${\text{Ni}}^{2+}\left(aq\right)+2{\text{Cl}}^{-}\left(aq\right)+2{\text{H}}^{+}+{\text{S}}^{2-}\left(aq\right)\to \text{NiS}\left(s\right)+2{\text{H}}^{+}\left(aq\right)+2{\text{Cl}}^{-}\left(aq\right)$

Here, ${\text{H}}^{+}$ and ${\text{Cl}}^{-}$ is present at both side of the equation and termed as spectator ions. These ions can be canceled out. Therefore, the net ionic equation for the formation of solid nickel compound is written below.

  ${\text{Ni}}^{2+}\left(aq\right)+{\text{S}}^{2-}\left(aq\right)\to \text{NiS}\left(s\right)$

Let’s take another example of solid nickel compound which is nickel phosphate. Solid nickel phosphate is formed by the reaction of sodium phosphate $\left({\text{Na}}_3{\text{PO}}_4\right)$ with nickel nitrate $\left(\text{Ni}{\left({\text{NO}}_3\right)}_{\text{2}}\right)$ .

  ${\text{2Na}}_3{\text{PO}}_{\text{4}}\left(aq\right)+3\text{Ni}{\left({\text{NO}}_3\right)}_2\left(aq\right)\to {\text{Ni}}_3{\left({\text{PO}}_4\right)}_2\left(s\right)+6{\text{NaNO}}_3\left(aq\right)$

The compounds, present in the aqueous form, dissociate in their ions. The complete ionic equation for the formation of solid nickel compound is written below.

  ${\text{6Na}}^{+}\left(aq\right)+2{\text{PO}}_4^{3-}\left(aq\right)+3{\text{Ni}}^{2+}\left(aq\right)+6{\text{NO}}_3^{-}\left(aq\right)\to {\text{Ni}}_3{\left({\text{PO}}_4\right)}_2\left(s\right)+6{\text{Na}}^{+}\left(aq\right)+6{\text{NO}}_3^{-}\left(aq\right)$ Here, ${\text{Na}}^{+}$ and ${\text{NO}}_3^{-}$ is present at both side of the equation and termed as spectator ions. These ions can be canceled out. Therefore, the net ionic equation for the formation of solid nickel compound is written below.

  $2{\text{PO}}_4^{3-}\left(aq\right)+3{\text{Ni}}^{2+}\left(aq\right)\to {\text{Ni}}_3{\left({\text{PO}}_4\right)}_2\left(s\right)$

Conclusion

The complete ionic equation for the formation of solid nickel compound is written below.

  ${\text{Ni}}^{2+}\left(aq\right)+2{\text{Cl}}^{-}\left(aq\right)+2{\text{Na}}^{+}+2{\text{OH}}^{-}\left(aq\right)\to \text{Ni}{\left(\text{OH}\right)}_{\text{2}}\left(s\right)+2{\text{Na}}^{+}\left(aq\right)+2{\text{Cl}}^{-}\left(aq\right)$ The net ionic equation for the formation of solid nickel compound is written below.

  ${\text{Ni}}^{2+}\left(aq\right)+2{\text{OH}}^{-}\left(aq\right)\to \text{Ni}{\left(\text{OH}\right)}_{\text{2}}\left(s\right)$

The complete ionic equation for the formation of solid nickel compound is written below.

  ${\text{Ni}}^{2+}\left(aq\right)+2{\text{Cl}}^{-}\left(aq\right)+2{\text{H}}^{+}+{\text{S}}^{2-}\left(aq\right)\to \text{NiS}\left(s\right)+2{\text{H}}^{+}\left(aq\right)+2{\text{Cl}}^{-}\left(aq\right)$ The net ionic equation for the formation of solid nickel compound is written below.

  ${\text{Ni}}^{2+}\left(aq\right)+{\text{S}}^{2-}\left(aq\right)\to \text{NiS}\left(s\right)$

The complete ionic equation for the formation of solid nickel compound is written below.

  ${\text{6Na}}^{+}\left(aq\right)+2{\text{PO}}_4^{3-}\left(aq\right)+3{\text{Ni}}^{2+}\left(aq\right)+6{\text{NO}}_3^{-}\left(aq\right)\to {\text{Ni}}_3{\left({\text{PO}}_4\right)}_2\left(s\right)+6{\text{Na}}^{+}\left(aq\right)+6{\text{NO}}_3^{-}\left(aq\right)$ The net ionic equation for the formation of solid nickel compound is written below.

  $2{\text{PO}}_4^{3-}\left(aq\right)+3{\text{Ni}}^{2+}\left(aq\right)\to {\text{Ni}}_3{\left({\text{PO}}_4\right)}_2\left(s\right)$