Why does a solution of sodium chloride remain neutral in water, while a solution of ammonium chloride turns acidic? What makes some salt solutions taste bitter, some sour, and others neutral? Why does the taste of some salts vary? Well, we will get these answers after reading this article on the hydrolysis of salt. Hydrolysis of salts is the process in which a salt reacts with water to produce an acidic or basic solution.
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The idea of salt hydrolysis was discovered by several scientists over time, but important contributions include Svante Arrhenius in 1887. Arrhenius made important contributions to the understanding of acids and bases, which contain the idea of the hydrolysis of salt in water. He says that salts break into ions in solution and that these ions can react with water. Van 't Hoff formulated the understanding of Ionic Equilibrium and its contribution to hydrolysis, which connects to the colligative properties. And last but not least is Gilbert N. Lewis in the 1920s.
When a salt is added to water, ions of the salt interact with water to cause acidity or basicity in an aqueous solution. This ionic interaction is called salt hydrolysis. The interaction of a cation is cationic hydrolysis, and the interaction ofan anion is anionic hydrolysis.
It is defined as the fraction of total salt that has undergone hydrolysis on attainment of equilibrium. It is denoted by h.
Let c be the concentration of salt and h be its degree of hydrolysis.
$\mathrm{A}^{-}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{OH}^{-}+\mathrm{HA}$
c
c - ch ch ch
$\mathrm{K}_{\mathrm{h}}=\frac{\left[\mathrm{OH}^{-}\right][\mathrm{HA}]}{\left[\mathrm{A}^{-}\right]}=\frac{(\mathrm{ch})(\mathrm{ch})}{\mathrm{c}-\mathrm{ch}}=\frac{\mathrm{ch}^2}{1-\mathrm{h}}$
Such salts give alkaline solutions in water. Some of such salts are CH3COONa, Na2CO3, K2CO3, KCN, etc. For our discussion, we consider CH3COONa (sodium acetate) in water. When CH3COONa is put in water, it completely ionizes to give CH3COO- (acetate) ions and Na+ ions. Now acetate ions (CH3COO-) absorb some H+ ions from weakly dissociated H2O molecules to form undissociated CH3COOH. Na+ remains in the ionic state in water. Now for Kw (ionic product) of water to remain constant, H2O further ionizes to produce more H+ and OH- ions. H+ ions are taken up by CH3COO- ions leaving OH- ions in excess and hence an alkaline solution is obtained.
Let BA represents such a salt. As it is put in water;
$\mathrm{BA}(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(\mathrm{aq}) \rightleftharpoons \mathrm{BOH}(\mathrm{aq})+\mathrm{HA}(\mathrm{aq})$
BA dissociates into ions and BOH being strong base also ionises.
$\mathrm{B}^{+}+\mathrm{A}^{-}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{B}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})+\mathrm{HA}(\mathrm{aq})$
Thus, the net reaction is:$\mathrm{A}^{-}(\mathrm{aq})+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{OH}^{-}(\mathrm{aq})+\mathrm{HA}(\mathrm{aq})$
Thus, the hydrolysis constant(Kh) is given as:$\mathrm{K}_{\mathrm{h}}=\frac{\left[\mathrm{OH}^{-}\right][\mathrm{HA}]}{\left[\mathrm{A}^{-}\right]}$
pH of a basic solution is given as:
$\mathrm{pH}=14+\log \left[\mathrm{OH}^{-}\right]$and $\left[\mathrm{OH}^{-}\right]=\mathrm{ch}=\sqrt{\mathrm{K}_{\mathrm{h}} \mathrm{c}}$
Thus, substituting for Kh, we get:
$\begin{aligned} & {\left[\mathrm{OH}^{-}\right]=\sqrt{\frac{\mathrm{K}_{\mathrm{w}} \mathrm{c}}{\mathrm{K}_{\mathrm{a}}}}} \\ & \mathrm{pH}=14+\log _{10} \sqrt{\frac{\mathrm{K}_{\mathrm{w}} \mathrm{c}}{\mathrm{K}_{\mathrm{a}}}} \\ & \text { Thus, } \mathrm{pH}=\frac{1}{2}\left(\mathrm{pK}_{\mathrm{w}}+\mathrm{pK}_{\mathrm{a}}+\log _{10} \mathrm{c}\right) \\ & \text { Hence, } \mathrm{pH}=7+\frac{1}{2}\left(\mathrm{pK}_{\mathrm{a}}+\log _{10} \mathrm{c}\right)\end{aligned}$
Such salts give acidic solutions in water. Some of such salts are NH4Cl, ZnCl2, FeCl3, etc. For the purpose of discussion, we will consider the hydrolysis of NH4Cl. When NH4Cl is put in water, it completely ionises in water to give NH4+ and Cl- ions. NH4+ ions combine with OH- ions furnished by weakly dissociated water to form NH4OH (a weak base). Now for keeping Kw constant, water further ionises to give H+ and OH- ions, where OH- ions are consumed by NH4+ ions leaving behind H+ ions in solution to give an acidic solution.
Let BA be one of such salts. When it is put into water, the reaction is as follows.$\mathrm{B}^{+}+\mathrm{A}^{-}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{BOH}+\mathrm{H}^{+}(\mathrm{aq})+\mathrm{A}^{-}(\mathrm{aq})$
Thus the net reaction of hydrolysis is as follows:
$\mathrm{B}^{+}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{BOH}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq})$
c - ch ch ch
$\begin{aligned} & \mathrm{K}_{\mathrm{h}}=\frac{[\mathrm{BOH}]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{B}^{+}\right]}=\frac{(\mathrm{ch})(\mathrm{ch})}{\mathrm{c}-\mathrm{ch}}=\frac{\mathrm{ch}^2}{1-\mathrm{h}^2}=\mathrm{ch}^2 \quad(\text { assuming } \mathrm{h}<<1) \\ & \text { Thus, } \mathrm{h}=\sqrt{\frac{\mathrm{K}_{\mathrm{h}}}{\mathrm{c}}}\end{aligned}$
Considering ionisation of weak base BOH and H2O.
$\begin{aligned} & \mathrm{BOH} \rightleftharpoons \mathrm{B}^{+}+\mathrm{OH}^{-} \quad \Rightarrow \quad \mathrm{K}_{\mathrm{b}}=\frac{\left[\mathrm{B}^{+}\right]\left[\mathrm{OH}^{-}\right]}{[\mathrm{BOH}]} \\ & \mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}^{+}+\mathrm{OH}^{-} \quad \Rightarrow \quad \mathrm{K}_{\mathrm{w}}=\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]\end{aligned}$
From expressions for Kh, Kb and Kw, we have :
$\mathrm{K}_{\mathrm{h}}=\frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{b}}} \Rightarrow \mathrm{h}=\sqrt{\frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{b}} \mathrm{c}}}$
Now, pH = - log [H+]and $\left[\mathrm{H}^{+}\right]=\mathrm{ch}=\mathrm{c} \sqrt{\frac{\mathrm{K}_{\mathrm{h}}}{\mathrm{c}}}=\sqrt{\mathrm{K}_{\mathrm{h}} \mathrm{c}} \Rightarrow\left[\mathrm{H}^{+}\right]=\sqrt{\frac{\mathrm{K}_{\mathrm{W}} \mathrm{c}}{\mathrm{K}_{\mathrm{b}}}}$
$\Rightarrow \quad \mathrm{pH}=-\log _{10} \sqrt{\frac{\mathrm{K}_{\mathrm{w}} \mathrm{c}}{\mathrm{K}_{\mathrm{b}}}}$
$\Rightarrow \quad \mathrm{pH}=\frac{1}{2}\left(\mathrm{pK}_{\mathrm{w}}-\mathrm{pK}_{\mathrm{b}}-\log _{10} c\right)$
Hence, $\mathrm{pH}\left(\right.$ at $\left.25^{\circ} \mathrm{C}\right)=7-\frac{1}{2}\left(\mathrm{pK}_{\mathrm{b}}+\log _{10} c\right)$
Example 1: The definition of acids and bases as acids are substances that dissociate in water to give $H^{+}$ ions, and bases are substances that produce $O H^{-}$ ions was given by
1) (correct)Arrhenius
2)Lewis
3)Bronsted Lowry
4)None of these
Solution
The definition of acid and bases as acids are substances that dissociate in water to give $H^{+}$ ions and bases are substances that produce OH- ions, was given by Arrhenius.
Hence, the answer is option(1).
Example 2: The pH of 0.1 M solution of the following salts increases in the order:
1)$\mathrm{NaCl}<\mathrm{NH}_4 \mathrm{Cl}<\mathrm{NaCN}<\mathrm{HCl}$
2) (correct)$\mathrm{HCl}<\mathrm{NH}_4 \mathrm{Cl}<\mathrm{NaCl}<\mathrm{NaCN}$
3)$\mathrm{HCl}<\mathrm{NaCl}<\mathrm{NaCN}<\mathrm{NH}_4 \mathrm{Cl}$
4)$\mathrm{NaCN}<\mathrm{NH}_4 \mathrm{Cl}<\mathrm{NaCl}<\mathrm{HCl}$
Solution
Solution of HCl and $\mathrm{NH}_4 \mathrm{Cl}$ will be acidic, NaCl solution of neutral whereas solution of NaCN will be basic.
Hence, the answer is option (2).
Example 3: An aqueous solution of aluminum sulfate would show:
1)Basic
2) (correct)Acidic
3)Neutral
4)Any of the above
Solution
$\mathrm{Al}_2\left(\mathrm{SO}_4\right)_3$ is a salt of $W_B / S_A \cdot\left[\mathrm{Al}(\mathrm{OH})_3+\mathrm{H}_2 \mathrm{SO}_4\right]$ it hydrolyses and gives the acidic solution.
Hence, the answer is option (2).
Example 4: A certain weak acid has Ka = 1.0 x 10-4. Calculate the equilibrium constant for its reaction with a strong base.
1)10-10
2) (correct)1010
3)104
4)108
Solution
Given,
$\mathrm{HA} \rightleftharpoons \mathrm{H}^{+}+\mathrm{A}^{-}, \mathrm{K}_{\mathrm{a}}=10^{-4}$
and we know that
$\mathrm{H}^{+}+\mathrm{OH}^{-} \rightleftharpoons \mathrm{H}_2 \mathrm{O}, \mathrm{K}=\frac{1}{\mathrm{~K}_{\mathrm{w}}}=10^{14}$
Adding the above two reactions we get
$\mathrm{HA}+\mathrm{OH}^{-} \rightleftharpoons \mathrm{A}^{-}+\mathrm{H}_2 \mathrm{O}, \mathrm{K}_{\mathrm{net}}=\frac{\mathrm{K}_{\mathrm{a}}}{\mathrm{K}_{\mathrm{w}}}$
$\therefore \mathrm{K}_{\text {net }}=\frac{\mathrm{K}_{\mathrm{a}}}{\mathrm{K}_{\mathrm{w}}}=\frac{10^{-4}}{10^{-14}}=10^{10}$
Hence, the answer is option (2).
Example 5: Which one of the following is not an acid salt?
1)$\mathrm{NaH}_2 \mathrm{PO}_4$
2)${ }^1 \mathrm{NaH}_2 \mathrm{PO}_3$
3)$\mathrm{Na}_2 \mathrm{SO}_3$
4) (correct)All of the above are acid salts
Solution
Salts of $S_B / W_A$ are called acid salts.
$\mathrm{NaH}_2 \mathrm{PO}_4$ ( Salt of $\left.\mathrm{NaOH}+\mathrm{H}_3 \mathrm{PO}_4\right)\left(S_B / W_A\right)$
$\mathrm{NaH}_2 \mathrm{PO}_2$ (Salt of $\left.\mathrm{NaOH}+\mathrm{H}_3 \mathrm{PO}_2\right)\left(S_B / W_A\right)$
$\mathrm{NaH}_2 \mathrm{PO}_3$ (Salt of $\left.\mathrm{NaOH}+\mathrm{H}_3 \mathrm{PO}_3\right)\left(S_B / W_A\right)$
All are acid salts.
Hence, the answer is option (4).
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The acidic or basic solution are formed by the hydrolyze of acid and bases, depending on the nature of their salt. Acidic Salts formed from strong acids and weak bases are Useful in Buffer Solutions such as Acidic salts can help create buffer solutions that resist changes in pH. pH Adjustment in which can be used to acidify solutions in various industrial processes. acid-base hydrolysis has its application in corrosion control such that the Acidic conditions can be used in controlling corrosion rates in metal processing.
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