# Each of the following reactions shows a solute dissolved in water. Classify each solute as a strong electrolyte, a weak electrolyte, or a nonelectrolyte?

## $\text{M"(l)→"M} \left(a q\right)$ "AC"(aq)⇌"A"^(+)(aq) + "C"^(−)(aq) "BD"(s)→"B"^(+)(aq) + "D"^(−)(aq) "PR"(aq) → "P"^(+)(aq) + "R"^(−)(aq) $\text{N"(s)→"N} \left(a q\right)$

Jun 11, 2017

Here's what I got.

#### Explanation:

For starters, you should know that an electrolyte is simply a substance that dissolves in water to produce a solution that can conduct electricity.

Simply put, an electrolyte is a substance that produces ions when dissolved in water. Consequently, a non-electrolyte is a substance that does not produce ions when is dissolved in water. Now, the first thing to look out for here is the equilibrium arrow, $r i g h t \le f t h a r p \infty n s$.

We use this symbol to show that the reaction does not go to completion, which basically means that, at equilibrium, the reaction vessel contains both reactants and products.

${\text{AC"_ ((aq)) rightleftharpoons "A"_ ((aq))^(+) + "C}}_{\left(a q\right)}^{-}$

The fact that an equilibrium arrow is used tells you that a solution of $\text{AC}$ will contain both undissociated $\text{AC}$ and dissociated ${\text{A}}^{+}$ and ${\text{C}}^{-}$ ions.

In other words, $\text{AC}$ does not ionize completely, i.e. most of the compound exists as molecules of $\text{AC}$, not as ions, which implies that it is a weak electrolyte.

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\text{weak electrolyte = partially ionizes to produce ions}}}}$

By comparison, we use a regular reaction arrow, $\to$, to show that the reaction goes very, very close to completion, i.e. the reaction vessel contains almost exclusively products.

${\text{BD"_ ((s)) -> "B"_ ((aq))^(+) + "D}}_{\left(a q\right)}^{-}$

This means that when $\text{BD}$ is dissolved in water, it dissociates completely to produce ${\text{B}}^{+}$ and ${\text{D}}^{-}$ ions. In other words, the reaction vessel will contain almost exclusively ions, which implies that $\text{BD}$ is a strong electrolyte.

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\text{strong electrolyte = completely ionizes to produce ions}}}}$

The same can be said about $\text{PR}$, which ionizes completely to produce ${\text{P}}^{+}$ and ${\text{R}}^{-}$ ions.

${\text{PR"_ ((aq)) -> "P"_ ((aq))^(+) + "R}}_{\left(a q\right)}^{-}$

You can thus say that $\text{PR}$ is a strong electrolyte.

Finally, notice that $\text{M}$ and $\text{N}$ dissolve in water and do not produce ions.

${\text{M"_ ((l)) -> "N}}_{\left(a q\right)}$

${\text{N"_ ((s)) -> "N}}_{\left(a q\right)}$

This implies that they are non-electrolytes.

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\text{non-electrolyte = does not produce ions}}}}$