21.21: Hydrolysis of Salts - Equations
- Page ID
- 53951
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Baking seems easy with all the pre-mixed items available. ("just add water and stir"). However, there is a good amount of chemistry involved in baking with ingredients that are measured out. One important ingredient is baking powder. The fluffiness in the final product of a non-yeast recipe is usually due to the carbon dioxide formed by baking powder. One popular brand uses a mix of sodium bicarbonate and sodium aluminum sulfate to produce the \(\ce{CO_2}\). The reaction is:
\[3 \ce{NaHCO_3} + \ce{NaAl(SO_4)_2} \rightarrow \ce{Al(OH)_3} + 2 \ce{Na_2SO_4} + 3 \ce{CO_2}\nonumber \]
If all goes well, the biscuits rise, the pancakes are fluffy, and everybody is happy.
Hydrolysis of Salts: Equations
A salt is an ionic compound that is formed when an acid and a base neutralize each other. While it may seem that salt solutions are always neutral, they can frequently be either acidic or basic.
Consider the salt formed when the weak acid hydrofluoric acid is neutralized by the strong base sodium hydroxide. The molecular and net ionic equations are shown below.
\[\begin{align*} &\ce{HF} \left( aq \right) + \ce{NaOH} \left( aq \right) \rightarrow \ce{NaF} \left( aq \right) + \ce{H_2O} \left( l \right) \\ &\ce{HF} \left( aq \right) + \ce{OH^-} \left( aq \right) \rightarrow \ce{F^-} \left( aq \right) + \ce{H_2O} \left( l \right) \end{align*}\nonumber \]
Since sodium fluoride is soluble, the sodium ion is a spectator ion in the neutralization reaction. The fluoride ion is capable of reacting, to a small extent, with water, accepting a proton.
\[\ce{F^-} \left( aq \right) + \ce{H_2O} \left( l \right) \rightleftharpoons \ce{HF} \left( aq \right) + \ce{OH^-} \left( aq \right)\nonumber \]
The fluoride ion is acting as a weak Brønsted-Lowry base. The hydroxide ion that is produced as a result of the above reaction makes the solution slightly basic. Salt hydrolysis is a reaction in which one of the ions from a salt reacts with water, forming either an acidic or basic solution.
Salts That Form Basic Solutions
When solid sodium fluoride is dissolved into water, it completely dissociates into sodium ions and fluoride ions. The sodium ions do not have any capability of hydrolyzing, but the fluoride ions hydrolyze to produce a small amount of hydrofluoric acid and hydroxide ion.
\[\ce{F^-} \left( aq \right) + \ce{H_2O} \left( l \right) \rightleftharpoons \ce{HF} \left( aq \right) + \ce{OH^-} \left( aq \right)\nonumber \]
Salts that are derived from the neutralization of a weak acid \(\left( \ce{HF} \right)\) by a strong base \(\left( \ce{NaOH} \right)\) will always produce salt solutions that are basic.
Salts That Form Acidic Solutions
Ammonium chloride \(\left( \ce{NH_4Cl} \right)\) is a salt that is formed when the strong acid \(\ce{HCl}\) is neutralized by the weak base \(\ce{NH_3}\). Ammonium chloride is soluble in water. The chloride ion produced is incapable of hydrolyzing because it is the conjugate base of the strong acid \(\ce{HCl}\). In other words, the \(\ce{Cl^-}\) ion cannot accept a proton from water to form \(\ce{HCl}\) and \(\ce{OH^-}\), as the fluoride ion did in the previous section. However, the ammonium ion is capable of reacting slightly with water, donating a proton and so acting as an acid.
\[\ce{NH_4^+} \left( aq \right) + \ce{H_2O} \left( l \right) \rightleftharpoons \ce{H_3O^+} \left( aq \right) + \ce{NH_3} \left( aq \right)\nonumber \]
Salts That Form Neutral Solutions
A salt that is derived from the reaction of a strong acid with a strong base forms a solution that has a pH of 7. An example is sodium chloride, formed from the neutralization of \(\ce{HCl}\) by \(\ce{NaOH}\). A solution of \(\ce{NaCl}\) in water has no acidic or basic properties, since neither ion is capable of hydrolyzing. Other salts that form neutral solutions include potassium nitrate \(\left( \ce{KNO_3} \right)\) and lithium bromide \(\left( \ce{LiBr} \right)\). The table below summarizes how to determine the acidity or basicity of a salt solution.
Salt formed from: | Salt Solution |
---|---|
Strong acid \(+\) Strong base | Neutral |
Strong acid \(+\) Weak base | Acidic |
Weak acid \(+\) Strong base | Basic |
Salts formed from the reaction of a weak acid and a weak base are more difficult to analyze due to competing hydrolysis reactions between the cation and the anion. These salts are not considered in this chapter's concept.
Summary
- Salt hydrolysis is a reaction in which one of the ions from a salt reacts with water, forming either an acidic or basic solution.
- Salts that are derived from the neutralization of a weak acid by a strong base will always produce salt solutions that are basic.
- Salts that are derived from the neutralization of a strong acid by a weak base will always produce salt solutions that are acidic.
- A salt that is derived from the reaction of a strong acid with a strong base forms a solution that has a pH of 7.