# 7.8: Solution Equations: Examples

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$

( \newcommand{\kernel}{\mathrm{null}\,}\) $$\newcommand{\range}{\mathrm{range}\,}$$

$$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$

$$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$

$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$

$$\newcommand{\Span}{\mathrm{span}}$$

$$\newcommand{\id}{\mathrm{id}}$$

$$\newcommand{\Span}{\mathrm{span}}$$

$$\newcommand{\kernel}{\mathrm{null}\,}$$

$$\newcommand{\range}{\mathrm{range}\,}$$

$$\newcommand{\RealPart}{\mathrm{Re}}$$

$$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$

$$\newcommand{\Argument}{\mathrm{Arg}}$$

$$\newcommand{\norm}[1]{\| #1 \|}$$

$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$

$$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$

$$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$$

$$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$$

$$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vectorC}[1]{\textbf{#1}}$$

$$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$$

$$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$$

$$\newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}}$$

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

Learning Objectives
• Write a solution equation that represents the electrolyte behavior of a solute.

The previous three sections presented and discussed the solution equation patterns that represent the dissociative behaviors of non-, strong, and weak electrolytes, respectively.  The following paragraphs will describe how these generic patterns can be modified to reflect the chemical composition of a particular solution.

For example, write a balanced solution equation that represents the dissociation of potassium phosphate, which exhibits the characteristics of a strong electrolyte when dissolved in water.  (States of matter are not required.)

In order to apply the strong electrolyte solution equation pattern that was presented in Section 7.6, each substance that is referenced in the given statement must first be classified as a solute or a solvent.  Because the indicator word "in" is present in the given statement, the chemical that is mentioned after this word, water, H2O, is the solvent in this solution, and the remaining substance, potassium phosphate, K3PO4, is the solute, "by default."

In order to indicate the point at which the dissociation of a strong electrolyte occurs, a "forward," or left-to-right, arrow is utilized in the solution equation for this type of solute.  The chemical formula of the solvent, water, H2O, is written over this arrow, and the chemical formula of the solute, potassium phosphate, K3PO4, is written on the left side of the arrow.  Because a strong electrolyte completely dissociates into its constituent cations and anions as it dissolves, the ion symbol for each of these particles is written on the right side of the arrow.  The cationic component of potassium phosphate, K3PO4, is the potassium ion, K+1, and the anionic component of this solute is the phosphate ion, PO43, which is a polyatomic anion.   Finally, in order to indicate that these ions are unique particles, a plus sign is used to separate their symbols.  The information that is described in this paragraph is reflected in the unbalanced solution equation that is shown below.

___ $$\ce{K_3PO_4}$$ $$\overset{\ce{H_2O}}{\longrightarrow}$$ ___ $$\ce{K^{+1}}$$ + ___ $$\ce{PO_4^{–3}}$$

Since the left and right sides of this solution equation contain equal amounts of phosphate ions, PO43, that ion is balanced.  However, the potassium ion, K+1, is not balanced, and, therefore, a coefficient of 3 should be written in the "blank" that corresponds to this ion on the right side of the arrow, as shown below.  Because all of the components in the following solution equation are balanced, the equation that is shown below is the chemically-correct representation of the dissociation of potassium phosphate in water.

___ $$\ce{K_3PO_4}$$ $$\overset{\ce{H_2O}}{\longrightarrow}$$ 3 $$\ce{K^{+1}}$$ + ___ $$\ce{PO_4^{–3}}$$

Example $$\PageIndex{1}$$

Write a balanced solution equation that represents the solvation of sucrose, C12H22O11, which exhibits the characteristics of a non-electrolyte when dissolved in ethanol, C2H5OH.  (States of matter are not required.)

Solution

In order to apply the non-electrolyte solution equation pattern that was presented in Section 7.5, each substance that is referenced in the given statement must first be classified as a solute or a solvent.  Because the indicator word "in" is present in the given statement, the chemical that is mentioned after this word, ethanol, C2H5OH, is the solvent in this solution, and the remaining substance, sucrose, C12H22O11, is the solute, "by default."

In order to indicate the point at which the solvation of a non-electrolyte occurs, a "forward," or left-to-right, arrow is utilized in the solution equation for this type of solute.  The chemical formula of the solvent, ethanol, C2H5OH, is written over this arrow, and the chemical formula of the solute, sucrose, C12H22O11, is written on the left side of the arrow.  Furthermore, because a non-electrolyte does not dissociate as it dissolves, the chemical formula of the solute, sucrose, C12H22O11, is unchanged during the solvation process and, therefore, is also written on the right side of the arrow.  The information that is described in this paragraph is reflected in the solution equation that is shown below.

___ $$\ce{C_{12}H_{22}O_{11}}$$ $$\overset{\ce{C_2H_5OH}}{\longrightarrow}$$ ___ $$\ce{C_{12}H_{22}O_{11}}$$

Since the same chemical formula is written on both sides of the arrow, the equation that is written above is already balanced and, therefore, is the chemically-correct representation of the solvation of sucrose, C12H22O11, in ethanol, C2H5OH.

Exercise $$\PageIndex{1}$$

Write a balanced solution equation that represents the dissociation of nickel (III) sulfide, which exhibits the characteristics of a weak electrolyte when dissolved in water.  (States of matter are not required.)

In order to apply the weak electrolyte solution equation pattern that was presented in Section 7.7, each substance that is referenced in the given statement must first be classified as a solute or a solvent.  Because the indicator word "in" is present in the given statement, the chemical that is mentioned after this word, water, H2O, is the solvent in this solution, and the remaining substance, nickel (III) sulfide, Ni2S3, is the solute, "by default."

In order to indicate the point at which the "forward" dissociation and "reverse" recombination processes simultaneously occur during the solvation of a weak electrolyte, an equilibrium arrow is utilized in the solution equation for this type of solute.  The chemical formula of the solvent, water, H2O, is written over this arrow, and the chemical formula of the solute, nickel (III) sulfide, Ni2S3, is written on the left side of the arrow.  Because a weak electrolyte partially dissociates into its constituent cations and anions as it dissolves, the ion symbol for each of these particles is written on the right side of the arrow.  The cationic component of nickel (III) sulfide, Ni2S3, is the nickel (III) ion, Ni+3, which is a transition metal cation, and the anionic component of this solute is the sulfide ion, S2.   Finally, in order to indicate that these ions are unique particles, a plus sign is used to separate their symbols.  The information that is described in this paragraph is reflected in the unbalanced solution equation that is shown below.

___ $$\ce{Ni_2S_3}$$ $$\overset{\ce{H_2O}}{\longrightleftharpoons}$$ ___ $$\ce{Ni^{+3}}$$ + ___ $$\ce{S^{–2}}$$

Neither the nickel (III) ion, Ni+3, nor the sulfide ion, S2, is balanced, because each of these ions is present in different quantities on the left and right sides of the equilibrium arrow.  Therefore, coefficients of 2 and 3, respectively, should be written in the "blanks" that corresponds to each of these ions on the right side of the arrow, as shown below.  Because all of the components in the following solution equation are balanced, the equation that is shown below is the chemically-correct representation of the dissociation of nickel (III) sulfide in water.

___ $$\ce{Ni_2S_3}$$ $$\overset{\ce{H_2O}}{\longrightleftharpoons}$$ 2 $$\ce{Ni^{+3}}$$ + 3 $$\ce{S^{–2}}$$

7.8: Solution Equations: Examples is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.