Q. 9.10 - An Introduction to General, Organic, and Biological Chemistry

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. $K_{2}C{O}_3$

Step-by-Step Solution

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

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

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

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

Step1: 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.

Step2: Given Information (part a).

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

$LiBr$

Step3: 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.

Step4: Given Information (part b).

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

$NaN{O}_3$

Step5: 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.

Step6: Given Information (part c).

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

$CuC{l}_2$

Step7: 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

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

Step8: Given Information (part d).

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

$K_{2}C{O}_3$

Step9: 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

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