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Salt Hydrolysis - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:35 AM | #JEE Main

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  • 32 Questions around this concept.

Solve by difficulty

QuestionMEDIUM

Directions: In the following questions, a statement of Assertion (A) is followed by a statement of reason (R).

Assertion: PH value of HCN solution increases when NaCN is added to it.

Reason: NaCN provides a common ion CN- which shifts the equilibrium of HCN (weak electrolyte) in backward direction.

Mark the correct choice as:

 

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Which of the following acids forms three series of salts?

 

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In base vs. acid titration, at the end point methyl orange is present as 

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20 mL of $0.1 \mathrm{M} \mathrm{NH}_4 \mathrm{OH}{\text { is mixed with }} 40 \mathrm{~mL}$ of 0.05 M HCl . The pH of the mixture is nearest to :
(Given : $\mathrm{K}_{\mathrm{b}}\left(\mathrm{NH}_4 \mathrm{OH}\right)=1 \times 10^{-5}, \log 2=0.30, \log 3=0.48, \log 5=0.69, \log 7$ $0.84, \log 11=1.04)$

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The plot of pH-metric titration of weak base$\mathrm{NH}_4 \mathrm{OH}_{\text {vs strong acid }} \mathrm{HCl}$ looks like :

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100ml of 1N NH4OH (Kb=5x10-5) is neutralized to equivalence point by 1N HCl. The PH of solution at equivalence point is

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Given below are two statement :

Statement (I) : Aqueous solution of ammonium carbonate is basic

Statement (II) : Acidic/basic nature of salt solution of a salt of weak acid and weak base depends on Ka and Kb value of acid and the base forming it.

In the light of the above statements, choose the most appropriate answer from the options given below:

View Solution
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Concepts Covered - 4

Salt Hydrolysis

Salt Hydrolysis

When a salt is added in water ions of the salt interact with water to cause acidity or basicity in aqueous solution. This ionic interaction is called salt hydrolysis. Interaction of cation is cationic hydrolysis and interaction of anion is anionic hydrolysis.

  • Hydrolysis is reverse of neutralization and an endothermic process.
  • If hydrolysis constant is Kh and neutralization constant is Kn. Then,
    Kn = 1/Kh
  • A solution of the salt of strong acid and weak base is acidic and for it pH < 7 or [H+] > 10-7. For example, FeCl3 (weak base + strong acid): Solution is acidic and involves cationic hydrolysis.
  • A solution of the salt of strong base and weak acid is basic and for it pH > 7 or [H+] < 10-7. For example, KCN (strong base + weak acid): Solution is basic and involves anionic hydrolysis.
  • A solution of the salt of weak acid and weak
    base, then:
    If Ka > Kb, it is acidic
    If Ka < Kb, it is basic
    If Ka = Kb, it is neutral
  • CH3COONH4 (weak acid + weak base): Solution is neutral and involves both cationic and anionic hydrolysis.
  • A solution of the salt of strong acid and strong base is neutral or pH = 7 or [H+] = 10-7 
  • A salt of strong acid and strong base is never hydrolyzed however, ions are hydrated. For example, K2SO4 (however, strong base + strong acid).

Degree of 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_{h}\: =\: \frac{[OH^{-}][HA]}{[A^{-}]}\: =\: \frac{(ch)(ch)}{c-ch}\: =\: \frac{ch^{2}}{1-h}}

\\\mathrm{Thus,\: K_{h}\: =\: ch^{2}\: \: \: \: \: \: (assuming\: h<<1)}\\\\\mathrm{h\: =\: \sqrt{\frac{K_{h}}{c}}}

Salt hydrolysis: Weak Acid and Strong Base

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 ionises 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 ionises 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_{h}\: =\: \frac{[OH^{-}][HA]}{[A^{-}]}}

pH of Solution

pH of a basic solution is given as:

\mathrm{pH}=14\: +\: \log \left[\mathrm{OH}^{-}\right]    \mathrm{and}\: \: \: \: [\mathrm{OH}^{-}]=\mathrm{ch}=\sqrt{\mathrm{K}_{\mathrm{h}} \mathrm{c}}

Thus, substituting for Kh, we get:

\mathrm{[OH^{-}]\: =\: \sqrt{\frac{K_{w}c}{K_{a}}}}

\mathrm{pH\: =\: 14\: +\: log_{10}\sqrt{\frac{K_{w}c}{K_{a}}}}

\mathrm{Thus,\: pH\: =\: \frac{1}{2}(pK_{w}\: +\: pK_{a}\: +\: log_{10}\, c)}

\mathrm{Hence,\: pH\: =\: 7\: +\: \frac{1}{2}(pK_{a}\: +\: log_{10}\, c)}

Salt hydrolysis: Weak Base and Strong Acid

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

\\\mathrm{K_{h}\: =\: \frac{[BOH][H^{+}]}{[B^{+}]}\: =\: \frac{(ch)(ch)}{c-ch}\: =\: \frac{ch^{2}}{1-h^{2}}\: =\: ch^{2}\: \: \: \: \: \: (assuming\: h<<1)}\\\\\mathrm{\mathbf{Thus},\: h\: =\: \sqrt{\frac{K_{h}}{c}}}

Considering ionisation of weak base BOH and H2O.

\\\mathrm{BOH} \rightleftharpoons \mathrm{B}^{+}+\mathrm{OH}^{-}\: \: \: \: \: \Rightarrow\: \: \: \: \mathrm{K_{b}\: =\: \frac{[B^{+}][OH^{-}]}{[BOH]}}\\\\\mathrm{H_{2}O} \rightleftharpoons \mathrm{H}^{+}+\mathrm{OH}^{-}\: \: \: \: \: \Rightarrow\: \: \: \: \mathrm{K_{w}\: =\:[H^{+}][OH^{-}]}

From expressions for Kh, Kb and Kw, we have :

\mathrm{K}_{\mathrm{h}}=\frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{b}}}\mathrm{\: \: \: \: \Rightarrow\: \: \: \mathrm{h}=\sqrt{\frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{b}} \mathrm{c}}}}

pH of Solution
Now, pH = - log [H+]
\text { 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 \quad\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)

\mathrm{\mathbf{Hence},\: pH}\left(\text { at } 25^{\circ} \mathrm{C}\right)=7-\frac{1}{2}\left(\mathrm{pK}_{\mathrm{b}}+\log _{10} c\right)

Salt hydrolysis: Weak Acid and Weak Base

Let us consider ammonium acetate (CH3COONH4) for our discussion. Both NH4+ ions and CH3COO- ions react respectively with OH- and H+ ions furnished by water to form NH4OH (a weak base) and CH3COOH (acetic acid).

Let BA represents such a salt.

                \mathrm{B}^{+}+\: \mathrm{A}^{-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{BOH}\, +\quad \mathrm{HA}
                                                     weak base    weak acid
Initially:      c          c                             0                 0
At equil:   c - ch   c - ch                       ch               ch

\Rightarrow \quad \mathrm{K}_{\mathrm{h}}=\frac{[\mathrm{BOH}][\mathrm{HA}]}{\left[\mathrm{B}^{+}\right]\left[\mathrm{A}^{-}\right]}=\frac{(\mathrm{ch})(\mathrm{ch})}{(\mathrm{c}-\mathrm{ch})^{2}}

\Rightarrow \quad \mathrm{K}_{\mathrm{h}}=\frac{\mathrm{h}^{2}}{(1-\mathrm{h})^{2}} \, ; \quad \text { Taking square root on both sides to get: }

\Rightarrow\: \: \:\: \: \mathrm{h}=\frac{\sqrt{\mathrm{K}_{\mathrm{h}}}}{1+\sqrt{\mathrm{K}_{\mathrm{h}}}}

Hence, the degree of hydrolysis of such salts is independent of concentration of salt solution.

Now considering the dissociation of both weak base and acid.

\mathrm{HA} \rightleftharpoons \mathrm{H}^{+}+\mathrm{A}^{-} \quad ; \quad \mathrm{K}_{\mathrm{a}}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}

\mathrm{BOH} \rightleftharpoons \mathrm{B}^{+}+\mathrm{OH}^{-} \quad ;\mathrm{\quad K_{b}=\frac{\left[B^{+}\right]\left[O H^{-}\right]}{[B O H]}}

\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}^{+}+\mathrm{OH}^{-} \quad ; \quad \mathrm{K}_{\mathrm{w}}=\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]

Combining Kh, Kb, Ka and Kw, we get:

\mathrm{K}_{\mathrm{h}}=\frac{\mathrm{K}_{\mathrm{W}}}{\mathrm{K}_{\mathrm{a}} \mathrm{K}_{\mathrm{b}}} \quad \text { and } \quad \mathrm{h}={\frac{\mathrm{\sqrt K}_{\mathrm{h}}}{1+\sqrt{\mathrm{K}_{\mathrm{h}}}}}

pH of Solution

Considering Ka for acid, we have:

\mathrm{K}_{\mathrm{a}}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]} \quad \Rightarrow \quad\left[\mathrm{H}^{+}\right]=\mathrm{K}_{\mathrm{a}} \frac{[\mathrm{HA}]}{\left[\mathrm{A}^{-}\right]}

Since, base and acids are weaker, hence,

[\mathrm{BOH}]=[\mathrm{HA}] \quad \Rightarrow \quad\left[\mathrm{B}^{+}\right]=\left[\mathrm{A}^{-}\right]

\Rightarrow \quad \mathrm{K}_{\mathrm{h}}=\frac{[\mathrm{BOH}][\mathrm{HA}]}{\left[\mathrm{B}^{+}\right]\left[\mathrm{A}^{-}\right]}=\frac{[\mathrm{HA}]^{2}}{\left[\mathrm{A}^{-}\right]^{2}} \quad \Rightarrow \quad\left[\mathrm{H}^{+}\right]=\mathrm{K}_{\mathrm{a}} \sqrt{\mathrm{K}_{\mathrm{h}}}=\sqrt{\frac{\mathrm{K}_{\mathrm{w}} \mathrm{K}_{\mathrm{a}}}{\mathrm{K}_{\mathrm{b}}}}

          \mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]=-\log \sqrt{\frac{\mathrm{K}_{\mathrm{W}} \mathrm{K}_{\mathrm{a}}}{\mathrm{K}_{\mathrm{b}}}}

\Rightarrow \quad \mathrm{pH}=\frac{1}{2}\left(\mathrm{pK}_{\mathrm{w}}+\mathrm{pK}_{\mathrm{a}}-\mathrm{pK}_{\mathrm{b}}\right)

\mathrm{\mathbf{Hence},\: pH}(\text {at } 25^{\circ} \mathrm{C}), \quad \mathrm{pH}=7+\frac{1}{2}\left(\mathrm{pK}_{\mathrm{a}}-\mathrm{pK}_{\mathrm{b}}\right)

Study it with Videos

Salt Hydrolysis
Salt hydrolysis: Weak Acid and Strong Base
Salt hydrolysis: Weak Base and Strong Acid

Study other Related Concepts

Salt Hydrolysis Current Topic

Equilibrium
Law Of Mass Action
Equilibrium Constant
Relation between Kp and Kc
Degree of Dissociation
Le Chatelier's Principles on Equilibrium
Ionic Equilibrium
Bronsted Lowry and Lewis Acid-Base theory
Ionization Of Acids And Bases
Ka and Kb Relationship

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