Question

Write a balanced equation for the dissociation of each of the following strong electrolytes in water:

a.$LiBr$

b.$NaN{O}_3$

c.$CuC{l}_2$

d.$K_{2}C{O}_3$

Short answer

(part a) The dissociation of the strong electrolyte $LiBr$is written as

$\mathrm{LiBr}\left(s\right)\stackrel{{\mathrm{H}}_2\mathrm{O}}{⟶}{\mathrm{Li}}^{+}\left(aq\right)+{\mathrm{Br}}^{-}\left(aq\right)$and it is a balanced equation.

(part b) The dissociation of the strong electrolyte $NaN{O}_3$is represented as

${\mathrm{NaNO}}_3\left(s\right)\stackrel{{\mathrm{H}}_2\mathrm{O}}{⟶}{\mathrm{Na}}^{+}\left(aq\right)+{\mathrm{NO}}_3^{-}\left(aq\right)$and is a Balanced equation.

(part c) The dissociation of the strong electrolyte $CuC{l}_2$is defined as follows

localid="1652520252941" ${\mathrm{CuCl}}_2\left(s\right)\stackrel{{\mathrm{H}}_2\mathrm{O}}{⟶}{\mathrm{Cu}}^{2+}\left(aq\right)+2{\mathrm{Cl}}^{-}\left(aq\right)$

(part d) The dissociation of the strong electrolyte $K_{2}C{O}_3$is as

${\mathrm{K}}_2{\mathrm{CO}}_3\left(s\right)\stackrel{{\mathrm{H}}_2\mathrm{O}}{⟶}2{\mathrm{K}}^{+}\left(aq\right)+{\mathrm{CO}}_3{}^{2-}(aq)$

Step-by-step solution

Introduction (part a)

A strong electrolyte is a solution which contains a huge proportion of a dissolved solute as ionic species. Ionic and polar substances are completely dissolved into ionic species and therefore engage in the activity very well, making them strong electrolytes.

Given Information (part a).

Find the balanced equation for the dissociation of the strong electrolyte:

$LiBr$

Explanation (part a). 

Electrolytes are substances that detach dissociate in water. $LiBr$is a powerful electrolyte. As a result, it totally detaches in water to create $L{i}^{+}$ions and $B{r}^{-}$ions.

This is written as,

$\mathrm{LiBr}\left(s\right)\stackrel{{\mathrm{H}}_2\mathrm{O}}{⟶}{\mathrm{Li}}^{+}\left(aq\right)+{\mathrm{Br}}^{-}\left(aq\right)$

Each element has an equal number of particles on both sides of the arrow in the equation. As just a result, it is an equitable equation.

Given Information (part b). 

Find the balanced equation for the dissociation of the strong electrolyte:

$NaN{O}_3$

Explanation (part b). 

$NaN{O}_3$ is a powerful electrolyte. As an outcome, it totally separates in water to create $N{a}^{+}$ and $NO{3}^{-}$ ions. The formula can be expressed as

${\mathrm{NaNO}}_3\left(s\right)\stackrel{{\mathrm{H}}_2\mathrm{O}}{⟶}{\mathrm{Na}}^{+}\left(aq\right)+{\mathrm{NO}}_3^{-}\left(aq\right)$

Each element has an equal number of particles on both sides of the arrow in the equation. As just a result, it is an equitable equation.

Given Information (part c). 

To find the balanced equation for the dissociation of the strong electrolyte:

$CuC{l}_2$

Explanation (part c). 

$CuC{l}_2$is a powerful electrolyte. As just an outcome, it totally decomposes throughout water to create $Cu{2}^{+}$and$C{l}^{-}$ ions. The formula can be written as

${\mathrm{CuCl}}_2\left(s\right)\stackrel{{\mathrm{H}}_2\mathrm{O}}{⟶}{\mathrm{Cu}}^{2+}\left(aq\right)+{\mathrm{Cl}}^{-}\left(aq\right)$

There's really one $Cu$atom on the both sides of the divide in this formula; however, there seem to be two chlorine $\left(Cl\right)$atoms just on left side of the equation but only one chlorine atom on the right side. As a matter of fact, we'll put a $2$next to $C{l}^{-}$.

As a matter of fact, the equilibrium constant for dissociation is

${\mathrm{CuCl}}_2\left(s\right)\stackrel{{\mathrm{H}}_2\mathrm{O}}{⟶}{\mathrm{Cu}}^{2+}\left(aq\right)+2{\mathrm{Cl}}^{-}\left(aq\right)$

Given Information (part d). 

Find out the balanced equation for the dissociation of the strong electrolyte:

$K_{2}C{O}_3$

Explanation (part d). 

${\mathrm{K}}_2{\mathrm{CO}}_3$is an extremely powerful electrolyte. As a consequence, in water, it completely detaches to construct ${{\mathrm{K}}_2}^{+}$ions and ${{\mathrm{CO}}_3}^{-}$ions. The equation is as follows:

${\mathrm{K}}_2{\mathrm{CO}}_3\left(s\right)\stackrel{{\mathrm{H}}_2\mathrm{O}}{⟶}{\mathrm{K}}^{+}\left(aq\right)+{\mathrm{CO}}_3{}^{2-}(aq)$

There's really one $C$particle as well as three oxygen atoms on the both sides of the issue in this equation; but even so, there really are two potassium$\left(K\right)$atoms just on left side of the equation only and one potassium atom on the right - hand side. As a necessary consequence, we'll put a $2$next to '$K^{+}$.

As an outcome, the equation for dissociation is

${\mathrm{K}}_2{\mathrm{CO}}_3\left(s\right)\stackrel{{\mathrm{H}}_2\mathrm{O}}{⟶}2{\mathrm{K}}^{+}\left(aq\right)+{\mathrm{CO}}_3{}^{2-}(aq)$