Q4.1P - Chemistry: Molecular Nature Of Matter And Change

Question

Which of the following scenes best represents how the ions occur in an aqueous solution of:

                     

a. $\mathit{C}\mathit{a}\mathit{C}{\mathit{l}}_{\mathbf{2}}$ b. $\mathit{L}{\mathit{i}}_{\mathbf{2}}\mathit{S}{\mathit{O}}_{\mathbf{4}}$ c. $\mathit{N}{\mathit{H}}_{\mathbf{4}}\mathit{B}\mathit{r}\mathbf{?}$

Step-by-Step Solution

Answer

When ionic compounds dissolve in the aqueous medium it dissociates into their ions since they form an ionic bond and when it breaks, ions contain some positive or negative charge which depends upon the valency of that atom.

Step 1: a. Dissociation reaction of C a C l 2 in aqueous solution

In the case of $CaC{l}_2$ In an aqueous medium, $CaC{l}_2$ is dissociated as $C{a}^{2+}$ and $C{l}^{-}$ ions which means it produces one ion of $C{a}^{2+}$ and $C{l}^{-}$ two ions of in the solution, therefore, the number of anions is twice the number of cation

$CaC{l}_2\left(s\right)\to C{a}^{2+}\left(aq\right)+2C{l}^{-}\left(aq\right)$ 

In scene (B) four cations are having +2 charge and eight anions have a -1 charge Hence, scene (B) is the best representation of this case. 

Step 2: b. Dissociation reaction of L i 2 S O 4 in aqueous solution

In the case of $L{i}_{2}S{O}_4:$ In an aqueous medium, $L{i}_{2}S{O}_4$ is dissociated as $L{i}^{+}$ and $S{O_4}^{-2}$ ions which means it produces two ions of $L{i}^{+}$ and one ion of $S{O_4}^{-2}$ in the solution therefore the number of cation is twice the number of the anion.

$L{i}_{2}S{O}_4\left(s\right)\to 2L{i}^{+}\left(aq\right)+S{O_4}^{-2}\left(aq\right)$

In scene (C) four anions are having -2 charge and eight cations have a +1 charge.

Hence, scene (C) is the best representation of this case.

Step 3: Dissociation reaction of N H 4 B r in aqueous solution

In the case of $N{H}_{4}Br:$ In an aqueous medium, $N{H}_{4}Br$is dissociated as $N{H_4}^{+}$ and $B{r}^{-}$ ions which means it produces one ion of $N{H_4}^{+}$ and $B{r}^{-}$ one ion of in the solution therefore the number of cations is equal to number of the anions

$N{H}_{4}Br\left(s\right)\to N{H_4}^{+}\left(aq\right)+B{r}^{-}\left(aq\right)$

In scene (A) four anions are having -1 charge and four cations have a +1 charge.

Hence, scene (A) is the best representation of this case.