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

Answer:

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.

In your case, you have

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

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.

In your case you have

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

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.

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